Chapter 9 Straight Lines

We need to find the slope of the given lines based on the provided information.

The formula for the slope ($m$) of a line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$ is:

$m = \frac{y_2 - y_1}{x_2 - x_1}$, provided $x_1 \neq x_2$.

The formula for the slope ($m$) of a line making an inclination $\theta$ with the positive direction of the x-axis is:

$m = \tan(\theta)$

(a) Passing through the points (3, – 2) and (–1, 4)

Let $(x_1, y_1) = (3, -2)$ and $(x_2, y_2) = (-1, 4)$.

Using the two-point slope formula:

$m = \frac{y_2 - y_1}{x_2 - x_1}$

$m = \frac{4 - (-2)}{-1 - 3}$

$m = \frac{4 + 2}{-4}$

$m = \frac{6}{-4}$

Simplify the fraction:

$m = -\frac{\cancel{6}^{3}}{\cancel{4}_{2}}$

$m = -\frac{3}{2}$

The slope of the line is $-\frac{3}{2}$.

(b) Passing through the points (3, – 2) and (7, – 2)

Let $(x_1, y_1) = (3, -2)$ and $(x_2, y_2) = (7, -2)$.

Using the two-point slope formula:

$m = \frac{y_2 - y_1}{x_2 - x_1}$

$m = \frac{-2 - (-2)}{7 - 3}$

$m = \frac{-2 + 2}{4}$

$m = \frac{0}{4}$

$m = 0$

The slope of the line is 0. This indicates a horizontal line.

(c) Passing through the points (3, – 2) and (3, 4)

Let $(x_1, y_1) = (3, -2)$ and $(x_2, y_2) = (3, 4)$.

Using the two-point slope formula:

$m = \frac{y_2 - y_1}{x_2 - x_1}$

$m = \frac{4 - (-2)}{3 - 3}$

$m = \frac{4 + 2}{0}$

$m = \frac{6}{0}$

Division by zero is undefined. The slope of the line is undefined. This indicates a vertical line.

(d) Making inclination of 60° with the positive direction of x-axis

The inclination of the line is $\theta = 60^\circ$.

Using the formula $m = \tan(\theta)$:

$m = \tan(60^\circ)$

The value of $\tan(60^\circ)$ is $\sqrt{3}$.

$m = \sqrt{3}$

The slope of the line is $\sqrt{3}$.

Given:

To Find:

The slope of the other line (let's denote it as $m_2$).

Solution:

The formula for the tangent of the angle $\theta$ between two lines with slopes $m_1$ and $m_2$ is given by:

$\tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right|$

Substitute the given values $\theta = \frac{\pi}{4}$ and $m_1 = \frac{1}{2}$ into the formula.

We know that $\tan(\frac{\pi}{4}) = 1$.

$1 = \left| \frac{m_2 - \frac{1}{2}}{1 + \frac{1}{2} m_2} \right|$

To solve for $m_2$, we consider two cases because of the absolute value:

Case 1: $\frac{m_2 - \frac{1}{2}}{1 + \frac{1}{2} m_2} = 1$

Multiply both sides by $(1 + \frac{1}{2} m_2)$, assuming $1 + \frac{1}{2} m_2 \neq 0$:

$m_2 - \frac{1}{2} = 1 \left(1 + \frac{1}{2} m_2\right)$

$m_2 - \frac{1}{2} = 1 + \frac{1}{2} m_2$

Collect terms involving $m_2$ on the left side and constant terms on the right side:

$m_2 - \frac{1}{2} m_2 = 1 + \frac{1}{2}$

Combine the terms:

$\left(1 - \frac{1}{2}\right) m_2 = \frac{2 + 1}{2}$

$\frac{1}{2} m_2 = \frac{3}{2}$

Multiply both sides by 2 to solve for $m_2$:

$m_2 = \frac{3}{2} \times 2$

$m_2 = \frac{3}{\cancel{2}} \times \cancel{2} = 3$

Let's check if the denominator was zero for $m_2=3$: $1 + \frac{1}{2}(3) = 1 + \frac{3}{2} = \frac{5}{2} \neq 0$. So, $m_2 = 3$ is a valid solution.

Case 2: $\frac{m_2 - \frac{1}{2}}{1 + \frac{1}{2} m_2} = -1$

Multiply both sides by $(1 + \frac{1}{2} m_2)$, assuming $1 + \frac{1}{2} m_2 \neq 0$:

$m_2 - \frac{1}{2} = -1 \left(1 + \frac{1}{2} m_2\right)$

$m_2 - \frac{1}{2} = -1 - \frac{1}{2} m_2$

Collect terms involving $m_2$ on the left side and constant terms on the right side:

$m_2 + \frac{1}{2} m_2 = -1 + \frac{1}{2}$

Combine the terms:

$\left(1 + \frac{1}{2}\right) m_2 = \frac{-2 + 1}{2}$

$\frac{3}{2} m_2 = -\frac{1}{2}$

Multiply both sides by $\frac{2}{3}$ to solve for $m_2$:

$m_2 = -\frac{1}{2} \times \frac{2}{3}$

$m_2 = -\frac{1}{\cancel{2}} \times \frac{\cancel{2}}{3} = -\frac{1}{3}$

Let's check if the denominator was zero for $m_2 = -\frac{1}{3}$: $1 + \frac{1}{2}\left(-\frac{1}{3}\right) = 1 - \frac{1}{6} = \frac{5}{6} \neq 0$. So, $m_2 = -\frac{1}{3}$ is a valid solution.

Since there are two possible cases for the absolute value, there are two possible values for the slope of the other line.

Answer:

The slope of the other line is $\mathbf{3}$ or $\mathbf{-\frac{1}{3}}$.

Given:

Line 1 passes through points $A(-2, 6)$ and $B(4, 8)$.

Line 2 passes through points $C(8, 12)$ and $D(x, 24)$.

Line 1 is perpendicular to Line 2.

To Find:

The value of $x$.

Solution:

The slope of a line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the formula $m = \frac{y_2 - y_1}{x_2 - x_1}$.

Let $m_1$ be the slope of Line 1 passing through $A(-2, 6)$ and $B(4, 8)$.

Here, $(x_1, y_1) = (-2, 6)$ and $(x_2, y_2) = (4, 8)$.

$m_1 = \frac{8 - 6}{4 - (-2)}$

$m_1 = \frac{2}{4 + 2}$

$m_1 = \frac{2}{6}$

$m_1 = \frac{1}{3}$

Let $m_2$ be the slope of Line 2 passing through $C(8, 12)$ and $D(x, 24)$.

Here, $(x_1, y_1) = (8, 12)$ and $(x_2, y_2) = (x, 24)$.

$m_2 = \frac{24 - 12}{x - 8}$

$m_2 = \frac{12}{x - 8}$

Since Line 1 is perpendicular to Line 2, the product of their slopes is $-1$, provided neither line is vertical or horizontal.

The slope of Line 1 is $\frac{1}{3}$, which is not zero or undefined, so Line 1 is neither horizontal nor vertical. This implies Line 2 must also be neither horizontal nor vertical, so its slope $m_2$ is defined and non-zero. Thus, $x - 8 \neq 0$ (i.e., $x \neq 8$) and $\frac{12}{x-8} \neq 0$ (which is true since $12 \neq 0$).

So, the condition for perpendicular lines is $m_1 \times m_2 = -1$.

Substitute the slopes we found:

Simplify the left side of the equation:

$\frac{12}{3(x - 8)} = -1$

$\frac{4}{x - 8} = -1$

Multiply both sides by $(x - 8)$:

$4 = -1(x - 8)$

$4 = -x + 8$

Solve for $x$. Add $x$ to both sides:

$4 + x = 8$

Subtract 4 from both sides:

$x = 8 - 4$

$x = 4$

This value $x=4$ satisfies the condition $x \neq 8$.

Answer:

The value of $x$ is $\mathbf{4}$.

Given:

The vertices of the quadrilateral are $A(-4, 5)$, $B(0, 7)$, $C(5, -5)$, and $D(-4, -2)$.

To Draw:

A quadrilateral in the Cartesian plane with the given vertices.

To Find:

The area of the quadrilateral ABCD.

Drawing the Quadrilateral:

Plot the points $A(-4, 5)$, $B(0, 7)$, $C(5, -5)$, and $D(-4, -2)$ on a Cartesian plane. Connect the points in order A to B, B to C, C to D, and D to A to form the quadrilateral ABCD.

Finding the Area of the Quadrilateral:

We can find the area of the quadrilateral by dividing it into two triangles using a diagonal. Let's use the diagonal AC to divide the quadrilateral ABCD into triangle ABC and triangle ADC.

The formula for the area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is:

Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$

1. Area of Triangle ABC:

The vertices are $A(-4, 5)$, $B(0, 7)$, $C(5, -5)$.

Let $(x_1, y_1) = (-4, 5)$, $(x_2, y_2) = (0, 7)$, $(x_3, y_3) = (5, -5)$.

Area(ABC) $= \frac{1}{2} |(-4)(7 - (-5)) + (0)((-5) - 5) + (5)(5 - 7)|$

Area(ABC) $= \frac{1}{2} |(-4)(7 + 5) + (0)(-10) + (5)(-2)|$

Area(ABC) $= \frac{1}{2} |(-4)(12) + 0 - 10|$

Area(ABC) $= \frac{1}{2} |-48 - 10|$

Area(ABC) $= \frac{1}{2} |-58|$

Area(ABC) $= \frac{1}{2} \times 58$

2. Area of Triangle ADC:

The vertices are $A(-4, 5)$, $D(-4, -2)$, $C(5, -5)$. For the area calculation using the formula, it's often helpful to list the vertices in a cyclic order, say $A(-4, 5)$, $C(5, -5)$, $D(-4, -2)$.

Let $(x_1, y_1) = (-4, 5)$, $(x_2, y_2) = (5, -5)$, $(x_3, y_3) = (-4, -2)$.

Area(ADC) $= \frac{1}{2} |(-4)((-5) - (-2)) + (5)((-2) - 5) + (-4)(5 - (-5))|$

Area(ADC) $= \frac{1}{2} |(-4)(-5 + 2) + (5)(-7) + (-4)(5 + 5)|$

Area(ADC) $= \frac{1}{2} |(-4)(-3) - 35 - 4(10)|$

Area(ADC) $= \frac{1}{2} |12 - 35 - 40|$

Area(ADC) $= \frac{1}{2} |12 - 75|$

Area(ADC) $= \frac{1}{2} |-63|$

Area(ADC) $= \frac{1}{2} \times 63$

3. Total Area of Quadrilateral ABCD:

The area of the quadrilateral is the sum of the areas of the two triangles it is divided into by the diagonal AC.

Area(ABCD) = Area(ABC) + Area(ADC)

Area(ABCD) $= 29 + 31.5$

Area(ABCD) $= 60.5$ square units.

Answer:

The area of the quadrilateral is $60.5$ square units.

Given:

An equilateral triangle with side length $2a$.

Its base lies along the y-axis.

The mid-point of the base is at the origin (0, 0).

To Find:

The vertices of the triangle.

Solution:

Let the base of the equilateral triangle be the line segment connecting two vertices on the y-axis. Let these vertices be $B$ and $C$.

Since the base lies along the y-axis, the x-coordinates of B and C are both 0. Let the coordinates of B and C be $(0, y_1)$ and $(0, y_2)$.

The length of the base is the distance between these two points, which is $|y_2 - y_1|$. We are given that the side length of the equilateral triangle is $2a$, so the length of the base is $2a$.

$|y_2 - y_1| = 2a$

The mid-point of the base is at the origin (0, 0). The coordinates of the midpoint of a segment with endpoints $(x_1, y_1)$ and $(x_2, y_2)$ are $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$.

Given the midpoint is $(0, 0)$ and the endpoints are $(0, y_1)$ and $(0, y_2)$:

$\frac{0 + 0}{2} = 0$ (This is consistent for the x-coordinate)

$\frac{y_1 + y_2}{2} = 0 \implies y_1 + y_2 = 0 \implies y_2 = -y_1$

Now, using the length of the base: $|y_2 - y_1| = |(-y_1) - y_1| = |-2y_1| = 2a$.

$|2y_1| = 2a$

$2|y_1| = 2a$

$|y_1| = a$

This means $y_1 = a$ or $y_1 = -a$.

If $y_1 = a$, then $y_2 = -a$. The vertices on the y-axis are $(0, a)$ and $(0, -a)$.

If $y_1 = -a$, then $y_2 = a$. The vertices on the y-axis are $(0, -a)$ and $(0, a)$.

So, the two vertices forming the base on the y-axis are $(0, a)$ and $(0, -a)$. Let these be $B=(0, a)$ and $C=(0, -a)$.

Now, let the third vertex be $A=(x, y)$. Since the base lies on the y-axis and its midpoint is the origin, the third vertex must lie on the x-axis (or at the same x-coordinate distance from the y-axis). For an equilateral triangle with its base center at the origin on the y-axis, the third vertex lies on the x-axis. So, $y=0$. Let the third vertex be $A=(x, 0)$.

The distance from $A$ to $B$ and from $A$ to $C$ must both be equal to the side length, $2a$.

Using the distance formula for the distance between $A(x, 0)$ and $B(0, a)$:

$AB^2 = (x - 0)^2 + (0 - a)^2 = x^2 + (-a)^2 = x^2 + a^2$

We know $AB = 2a$, so $AB^2 = (2a)^2 = 4a^2$.

$x^2 + a^2 = 4a^2$

$x^2 = 4a^2 - a^2$

$x^2 = 3a^2$

$x = \pm \sqrt{3a^2} = \pm a\sqrt{3}$

So, the x-coordinate of the third vertex can be $a\sqrt{3}$ or $-a\sqrt{3}$.

The third vertex can be $(a\sqrt{3}, 0)$ or $(-a\sqrt{3}, 0)$.

Thus, the vertices of the equilateral triangle are $(0, a)$, $(0, -a)$, and either $(a\sqrt{3}, 0)$ or $(-a\sqrt{3}, 0)$.

Answer:

The vertices of the triangle are $\mathbf{(0, a), (0, -a),}$ and $\mathbf{(a\sqrt{3}, 0)}$ (or $\mathbf{(-a\sqrt{3}, 0)}$).

Given:

Two points $P(x_1, y_1)$ and $Q(x_2, y_2)$.

To Find:

The distance between P and Q under specific conditions.

Solution:

The distance between two points $P(x_1, y_1)$ and $Q(x_2, y_2)$ in the Cartesian plane is given by the distance formula:

$PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

(i) PQ is parallel to the y-axis

If the line segment PQ is parallel to the y-axis, it means that the x-coordinates of the two points are the same.

So, $x_1 = x_2$.

Substitute $x_1 = x_2$ into the distance formula:

$PQ = \sqrt{(x_2 - x_2)^2 + (y_2 - y_1)^2}$

$PQ = \sqrt{(0)^2 + (y_2 - y_1)^2}$

$PQ = \sqrt{(y_2 - y_1)^2}$

The square root of a squared value is the absolute value.

$PQ = |y_2 - y_1|$

This represents the absolute difference in the y-coordinates.

(ii) PQ is parallel to the x-axis

If the line segment PQ is parallel to the x-axis, it means that the y-coordinates of the two points are the same.

So, $y_1 = y_2$.

Substitute $y_1 = y_2$ into the distance formula:

$PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_2)^2}$

$PQ = \sqrt{(x_2 - x_1)^2 + (0)^2}$

$PQ = \sqrt{(x_2 - x_1)^2}$

The square root of a squared value is the absolute value.

$PQ = |x_2 - x_1|$

This represents the absolute difference in the x-coordinates.

Answer:

(i) When PQ is parallel to the y-axis, the distance is $\mathbf{|y_2 - y_1|}$.

(ii) When PQ is parallel to the x-axis, the distance is $\mathbf{|x_2 - x_1|}$.

Given:

Two points $A(7, 6)$ and $B(3, 4)$.

A point on the x-axis that is equidistant from points A and B.

To Find:

The coordinates of the point on the x-axis.

Solution:

Let the required point on the x-axis be $P$. Since the point lies on the x-axis, its y-coordinate is 0. Let the coordinates of P be $(x, 0)$.

The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the distance formula:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

The point P(x, 0) is equidistant from A(7, 6) and B(3, 4).

This means the distance PA is equal to the distance PB:

Using the distance formula:

Distance PA $= \sqrt{(x - 7)^2 + (0 - 6)^2} = \sqrt{(x - 7)^2 + (-6)^2} = \sqrt{(x - 7)^2 + 36}$

Distance PB $= \sqrt{(x - 3)^2 + (0 - 4)^2} = \sqrt{(x - 3)^2 + (-4)^2} = \sqrt{(x - 3)^2 + 16}$

Equating PA and PB:

$\sqrt{(x - 7)^2 + 36} = \sqrt{(x - 3)^2 + 16}$

Square both sides to eliminate the square roots:

$(x - 7)^2 + 36 = (x - 3)^2 + 16$

Expand the squared terms using the formula $(a-b)^2 = a^2 - 2ab + b^2$:

$(x^2 - 14x + 49) + 36 = (x^2 - 6x + 9) + 16$

Combine the constants on each side:

$x^2 - 14x + 85 = x^2 - 6x + 25$

Subtract $x^2$ from both sides:

$-14x + 85 = -6x + 25$

Move the terms involving $x$ to one side and the constants to the other side. Add $14x$ to both sides and subtract 25 from both sides:

$85 - 25 = -6x + 14x$

$60 = 8x$

Divide by 8 to solve for $x$:

$x = \frac{60}{8}$

Simplify the fraction by dividing the numerator and denominator by 4:

$x = \frac{\cancel{60}^{15}}{\cancel{8}_{2}}$

$x = \frac{15}{2}$

The x-coordinate of the point is $\frac{15}{2}$. Since the point is on the x-axis, its y-coordinate is 0.

The coordinates of the point are $\left(\frac{15}{2}, 0\right)$.

Answer:

The point on the x-axis equidistant from the points (7, 6) and (3, 4) is $\mathbf{\left(\frac{15}{2}, 0\right)}$.

Given:

Point 1: The origin $(0, 0)$.

Point 2: The mid-point of the line segment joining the points $P(0, -4)$ and $B(8, 0)$.

To Find:

The slope of the line passing through these two points.

Solution:

First, we need to find the coordinates of the mid-point of the line segment joining the points $P(0, -4)$ and $B(8, 0)$.

The formula for the coordinates of the mid-point $(x_m, y_m)$ of a line segment with endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is:

$x_m = \frac{x_1 + x_2}{2}$

$y_m = \frac{y_1 + y_2}{2}$

Here, let $(x_1, y_1) = (0, -4)$ and $(x_2, y_2) = (8, 0)$.

The x-coordinate of the mid-point is $x_m = \frac{0 + 8}{2} = \frac{8}{2} = 4$.

The y-coordinate of the mid-point is $y_m = \frac{-4 + 0}{2} = \frac{-4}{2} = -2$.

So, the mid-point of the line segment joining P and B is $(4, -2)$. Let's call this point M.

The line we need to find the slope of passes through two points: the origin $O(0, 0)$ and the mid-point $M(4, -2)$.

The formula for the slope ($m$) of a line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$ is:

$m = \frac{y_2 - y_1}{x_2 - x_1}$

Let $(x_1, y_1) = (0, 0)$ (the origin) and $(x_2, y_2) = (4, -2)$ (the mid-point).

Using the slope formula:

$m = \frac{-2 - 0}{4 - 0}$

$m = \frac{-2}{4}$

Simplify the fraction:

$m = -\frac{1}{2}$

Answer:

The slope of the line is $\mathbf{-\frac{1}{2}}$.

Given:

Three points $A(4, 4)$, $B(3, 5)$, and $C(-1, -1)$.

To Show:

That these points are the vertices of a right-angled triangle, without using the Pythagoras theorem.

Solution:

We can determine if a triangle is right-angled by examining the slopes of its sides. Two lines are perpendicular if the product of their slopes is $-1$ (provided neither is vertical or horizontal).

We will find the slopes of the line segments AB, BC, and AC.

The formula for the slope ($m$) of a line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$ is $m = \frac{y_2 - y_1}{x_2 - x_1}$.

1. Slope of AB:

Points are $A(4, 4)$ and $B(3, 5)$.

Let $(x_1, y_1) = (4, 4)$ and $(x_2, y_2) = (3, 5)$.

$m_{AB} = \frac{5 - 4}{3 - 4} = \frac{1}{-1} = -1$

2. Slope of BC:

Points are $B(3, 5)$ and $C(-1, -1)$.

Let $(x_1, y_1) = (3, 5)$ and $(x_2, y_2) = (-1, -1)$.

$m_{BC} = \frac{-1 - 5}{-1 - 3} = \frac{-6}{-4} = \frac{6}{4} = \frac{3}{2}$

3. Slope of AC:

Points are $A(4, 4)$ and $C(-1, -1)$.

Let $(x_1, y_1) = (4, 4)$ and $(x_2, y_2) = (-1, -1)$.

$m_{AC} = \frac{-1 - 4}{-1 - 4} = \frac{-5}{-5} = 1$

Now, let's check the product of the slopes of any two sides.

Product of slopes of AB and BC: $m_{AB} \times m_{BC} = (-1) \times \frac{3}{2} = -\frac{3}{2}$ (Not -1)

Product of slopes of BC and AC: $m_{BC} \times m_{AC} = \frac{3}{2} \times 1 = \frac{3}{2}$ (Not -1)

Product of slopes of AB and AC: $m_{AB} \times m_{AC} = (-1) \times 1 = -1$

Since the product of the slopes of AB and AC is $-1$, the line segment AB is perpendicular to the line segment AC.

This means that the angle between sides AB and AC is $90^\circ$. This angle is at vertex A.

Therefore, the triangle ABC is a right-angled triangle with the right angle at vertex A.

Conclusion:

By calculating the slopes of the sides and showing that the product of the slopes of two sides (AB and AC) is $-1$, we have shown that the points (4, 4), (3, 5), and (–1, –1) are the vertices of a right-angled triangle, without using the Pythagoras theorem.

Given:

The line makes an angle of $30^\circ$ with the positive direction of the y-axis, measured anticlockwise.

To Find:

The slope of the line.

Solution:

The slope of a line is defined as the tangent of the angle it makes with the positive direction of the x-axis, measured anticlockwise.

Let $\theta$ be the angle the line makes with the positive direction of the x-axis, measured anticlockwise. The slope of the line is $m = \tan(\theta)$.

The positive y-axis is at an angle of $90^\circ$ from the positive x-axis, measured anticlockwise.

The problem states that the line makes an angle of $30^\circ$ with the positive y-axis, measured anticlockwise from the positive y-axis.

Therefore, the angle $\theta$ from the positive x-axis to the line, measured anticlockwise, is the sum of the angle from the positive x-axis to the positive y-axis ($90^\circ$) and the angle from the positive y-axis to the line ($30^\circ$).

$\theta = 90^\circ + 30^\circ = 120^\circ$

Now, we can find the slope of the line using the formula $m = \tan(\theta)$.

$m = \tan(120^\circ)$

We can calculate $\tan(120^\circ)$ using the property $\tan(180^\circ - A) = -\tan(A)$ or $\tan(90^\circ + A) = -\cot(A)$.

Using $\tan(120^\circ) = \tan(180^\circ - 60^\circ)$:

$m = \tan(180^\circ - 60^\circ) = -\tan(60^\circ)$

We know that $\tan(60^\circ) = \sqrt{3}$.

$m = -\sqrt{3}$

Alternatively, using $\tan(120^\circ) = \tan(90^\circ + 30^\circ)$:

$m = \tan(90^\circ + 30^\circ) = -\cot(30^\circ)$

We know that $\cot(30^\circ) = \sqrt{3}$.

$m = -\sqrt{3}$

Answer:

The slope of the line is $\mathbf{-\sqrt{3}}$.

Given:

The four points are $A(-2, -1)$, $B(4, 0)$, $C(3, 3)$, and $D(-3, 2)$.

To Show:

That these points are the vertices of a parallelogram, without using the distance formula.

Solution:

A quadrilateral is a parallelogram if and only if its opposite sides are parallel.

Two lines are parallel if they have the same slope.

The formula for the slope ($m$) of a line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $m = \frac{y_2 - y_1}{x_2 - x_1}$ (where $x_1 \neq x_2$).

We will calculate the slopes of the four sides of the quadrilateral ABCD.

1. Slope of side AB:

Using points $A(-2, -1)$ and $B(4, 0)$.

$m_{AB} = \frac{0 - (-1)}{4 - (-2)} = \frac{0 + 1}{4 + 2} = \frac{1}{6}$

2. Slope of side BC:

Using points $B(4, 0)$ and $C(3, 3)$.

$m_{BC} = \frac{3 - 0}{3 - 4} = \frac{3}{-1} = -3$

3. Slope of side CD:

Using points $C(3, 3)$ and $D(-3, 2)$.

$m_{CD} = \frac{2 - 3}{-3 - 3} = \frac{-1}{-6} = \frac{1}{6}$

4. Slope of side DA:

Using points $D(-3, 2)$ and $A(-2, -1)$.

$m_{DA} = \frac{-1 - 2}{-2 - (-3)} = \frac{-3}{-2 + 3} = \frac{-3}{1} = -3$

Now, let's compare the slopes of the opposite sides:

Slope of AB ($m_{AB}$) = $\frac{1}{6}$

Slope of CD ($m_{CD}$) = $\frac{1}{6}$

Since $m_{AB} = m_{CD}$, the side AB is parallel to the side CD.

Slope of BC ($m_{BC}$) = $-3$

Slope of DA ($m_{DA}$) = $-3$

Since $m_{BC} = m_{DA}$, the side BC is parallel to the side DA.

Since both pairs of opposite sides (AB parallel to CD and BC parallel to DA) are parallel, the quadrilateral ABCD is a parallelogram.

Conclusion:

By showing that the opposite sides have equal slopes, we have proven that the points (– 2, – 1), (4, 0), (3, 3) and (–3, 2) are the vertices of a parallelogram, without using the distance formula.

Alternate Solution (Using Midpoints):

A quadrilateral is a parallelogram if and only if its diagonals bisect each other. This means the midpoints of the two diagonals coincide.

The formula for the midpoint $(x_m, y_m)$ of a line segment with endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $x_m = \frac{x_1 + x_2}{2}$ and $y_m = \frac{y_1 + y_2}{2}$.

We will find the midpoint of the diagonal AC and the midpoint of the diagonal BD.

1. Midpoint of diagonal AC:

Using points $A(-2, -1)$ and $C(3, 3)$.

Midpoint of AC = $\left(\frac{-2 + 3}{2}, \frac{-1 + 3}{2}\right) = \left(\frac{1}{2}, \frac{2}{2}\right) = \left(\frac{1}{2}, 1\right)$

2. Midpoint of diagonal BD:

Using points $B(4, 0)$ and $D(-3, 2)$.

Midpoint of BD = $\left(\frac{4 + (-3)}{2}, \frac{0 + 2}{2}\right) = \left(\frac{1}{2}, \frac{2}{2}\right) = \left(\frac{1}{2}, 1\right)$

Since the midpoint of diagonal AC is $\left(\frac{1}{2}, 1\right)$ and the midpoint of diagonal BD is also $\left(\frac{1}{2}, 1\right)$, the midpoints coincide.

This means the diagonals AC and BD bisect each other. Therefore, the quadrilateral ABCD is a parallelogram.

Conclusion (Alternate Method):

By showing that the midpoints of the diagonals coincide, we have proven that the points (– 2, – 1), (4, 0), (3, 3) and (–3, 2) are the vertices of a parallelogram, without using the distance formula.

Given:

The line joining the points $A(3, -1)$ and $B(4, -2)$.

To Find:

The angle between the x-axis and the line joining the given points.

Solution:

The angle between a line and the positive direction of the x-axis (measured anticlockwise) is called the inclination of the line. The slope of a line is equal to the tangent of its inclination.

First, we find the slope of the line joining the points $A(3, -1)$ and $B(4, -2)$.

The formula for the slope ($m$) of a line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$ is:

$m = \frac{y_2 - y_1}{x_2 - x_1}$

Let $(x_1, y_1) = (3, -1)$ and $(x_2, y_2) = (4, -2)$.

Slope of the line $m = \frac{-2 - (-1)}{4 - 3}$

$m = \frac{-2 + 1}{1}$

$m = \frac{-1}{1}$

$m = -1$

Let $\theta$ be the angle between the line and the positive direction of the x-axis. The slope of the line is $m = \tan \theta$.

We have $m = -1$, so $\tan \theta = -1$.

We need to find the angle $\theta$ such that $\tan \theta = -1$. The tangent function is negative in the second and fourth quadrants.

The angle in the first quadrant whose tangent is 1 is $45^\circ$ or $\frac{\pi}{4}$ radians.

Since the tangent is negative, the angle $\theta$ lies in the second or fourth quadrant. Conventionally, the angle of a line with the positive x-axis is taken in the range $[0^\circ, 180^\circ)$.

In the second quadrant, the angle $\theta$ can be found using the relation $\tan(180^\circ - \alpha) = -\tan(\alpha)$.

Here, $\alpha = 45^\circ$.

So, $\theta = 180^\circ - 45^\circ = 135^\circ$.

In radians, $\theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.

Answer:

The angle between the x-axis and the line joining the points (3, –1) and (4, –2) is $\mathbf{135^\circ}$ or $\mathbf{\frac{3\pi}{4}}$ radians.

Given:

The slope of one line is double the slope of another line.

The tangent of the angle between the two lines is $\frac{1}{3}$.

To Find:

The slopes of the lines.

Solution:

Let the slopes of the two lines be $m_1$ and $m_2$.

According to the problem statement, the slope of one line is double the slope of the other. We can express this relationship as:

Let $\theta$ be the angle between the two lines. The tangent of the angle between two lines with slopes $m_1$ and $m_2$ is given by the formula:

$\tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right|$

We are given that $\tan \theta = \frac{1}{3}$. Substitute this value and the relationship from (1) into the formula:

$\frac{1}{3} = \left| \frac{(2m_1) - m_1}{1 + m_1 (2m_1)} \right|$

$\frac{1}{3} = \left| \frac{m_1}{1 + 2m_1^2} \right|$

To solve for $m_1$, we consider two cases based on the absolute value:

Case 1: $\frac{m_1}{1 + 2m_1^2} = \frac{1}{3}$

Cross-multiply:

$3m_1 = 1 + 2m_1^2$

Rearrange the equation into a quadratic form:

$2m_1^2 - 3m_1 + 1 = 0$

We can factor this quadratic equation:

$2m_1^2 - 2m_1 - m_1 + 1 = 0$

$2m_1(m_1 - 1) - 1(m_1 - 1) = 0$

$(2m_1 - 1)(m_1 - 1) = 0$

This gives two possible values for $m_1$ in this case:

$2m_1 - 1 = 0 \implies m_1 = \frac{1}{2}$

$m_1 - 1 = 0 \implies m_1 = 1$

If $m_1 = \frac{1}{2}$, then $m_2 = 2m_1 = 2 \times \frac{1}{2} = 1$. The slopes are $\frac{1}{2}$ and 1.

If $m_1 = 1$, then $m_2 = 2m_1 = 2 \times 1 = 2$. The slopes are 1 and 2.

Case 2: $\frac{m_1}{1 + 2m_1^2} = -\frac{1}{3}$

Cross-multiply:

$3m_1 = -1(1 + 2m_1^2)$

$3m_1 = -1 - 2m_1^2$

Rearrange the equation into a quadratic form:

$2m_1^2 + 3m_1 + 1 = 0$

We can factor this quadratic equation:

$2m_1^2 + 2m_1 + m_1 + 1 = 0$

$2m_1(m_1 + 1) + 1(m_1 + 1) = 0$

$(2m_1 + 1)(m_1 + 1) = 0$

This gives two possible values for $m_1$ in this case:

$2m_1 + 1 = 0 \implies m_1 = -\frac{1}{2}$

$m_1 + 1 = 0 \implies m_1 = -1$

If $m_1 = -\frac{1}{2}$, then $m_2 = 2m_1 = 2 \times (-\frac{1}{2}) = -1$. The slopes are $-\frac{1}{2}$ and -1.

If $m_1 = -1$, then $m_2 = 2m_1 = 2 \times (-1) = -2$. The slopes are -1 and -2.

We have found four pairs of slopes $(m_1, m_2)$ that satisfy the condition that $m_2 = 2m_1$. These pairs are $(\frac{1}{2}, 1)$, $(1, 2)$, $(-\frac{1}{2}, -1)$, and $(-1, -2)$. Each of these pairs represents the slopes of the two lines.

Answer:

The slopes of the lines are $\mathbf{\frac{1}{2}}$ and $\mathbf{1}$, or $\mathbf{1}$ and $\mathbf{2}$, or $\mathbf{-\frac{1}{2}}$ and $\mathbf{-1}$, or $\mathbf{-1}$ and $\mathbf{-2}$.

Given:

To Show:

Show that $k - y_1 = m (h - x_1)$.

Solution:

The slope of a line passing through two points $(x_a, y_a)$ and $(x_b, y_b)$ is defined as the change in the y-coordinates divided by the change in the x-coordinates.

The formula for the slope ($m$) is:

$m = \frac{\text{Change in y}}{\text{Change in x}} = \frac{y_b - y_a}{x_b - x_a}$

In this problem, the two given points are $(x_1, y_1)$ and $(h, k)$. Let's take $(x_a, y_a) = (x_1, y_1)$ and $(x_b, y_b) = (h, k)$.

Using the slope formula with these points, we get:

This formula for slope is defined when the denominator $(h - x_1)$ is not equal to zero, i.e., when $h \neq x_1$. If $h = x_1$, the line is vertical, and its slope is undefined.

Assuming the slope $m$ is a finite value (which means the line is not vertical, so $h \neq x_1$), we can multiply both sides of the equation by the denominator $(h - x_1)$:

$m \times (h - x_1) = \frac{k - y_1}{h - x_1} \times (h - x_1)$

On the right side, the term $(h - x_1)$ in the numerator and denominator cancels out:

$m (h - x_1) = k - y_1$

Rearranging the equation to match the required form:

This equation is a rearrangement of the slope formula and is known as the point-slope form of the equation of a line, where $(x_1, y_1)$ is a point on the line and $m$ is the slope.

Hence, it is shown that $k - y_1 = m (h - x_1)$ using the definition of the slope of a line passing through two given points.

Given:

A line passes through the point $(-2, 3)$.

The line is parallel to one of the coordinate axes.

To Find:

The equations of the lines parallel to the x-axis and the y-axis that pass through the point $(-2, 3)$.

Solution:

Let the given point be $P(-2, 3)$.

A line parallel to the x-axis has a constant y-coordinate for all points on the line. The equation of a line parallel to the x-axis is of the form $y = c$, where $c$ is a constant.

Since the line passes through the point $P(-2, 3)$, the y-coordinate of this point must satisfy the equation of the line.

So, $y = 3$.

The equation of the line parallel to the x-axis and passing through $(-2, 3)$ is $y = 3$.

A line parallel to the y-axis has a constant x-coordinate for all points on the line. The equation of a line parallel to the y-axis is of the form $x = c$, where $c$ is a constant.

Since the line passes through the point $P(-2, 3)$, the x-coordinate of this point must satisfy the equation of the line.

So, $x = -2$.

The equation of the line parallel to the y-axis and passing through $(-2, 3)$ is $x = -2$.

Answer:

The equation of the line parallel to the x-axis and passing through (– 2, 3) is $\mathbf{y = 3}$.

The equation of the line parallel to the y-axis and passing through (– 2, 3) is $\mathbf{x = -2}$.

Given:

The line passes through the point $(x_1, y_1) = (-2, 3)$.

The slope of the line is $m = -4$.

To Find:

The equation of the line.

Solution:

We can use the point-slope form of the equation of a line, which is given by:

$y - y_1 = m(x - x_1)$

Substitute the coordinates of the given point $(x_1, y_1) = (-2, 3)$ and the given slope $m = -4$ into this formula:

$y - 3 = -4(x - (-2))$

Simplify the expression inside the parenthesis:

$y - 3 = -4(x + 2)$

This is the equation of the line in point-slope form.

We can further simplify this equation to other standard forms, such as the general form $Ax + By + C = 0$.

Distribute the $-4$ on the right side:

$y - 3 = -4x - 8$

Move all terms to one side of the equation to get the general form:

$y + 4x - 3 + 8 = 0$

$4x + y + 5 = 0$

This is the equation of the line in the general form.

Answer:

The equation of the line through (– 2, 3) with slope – 4 is $\mathbf{y - 3 = -4(x + 2)}$ (point-slope form) or $\mathbf{4x + y + 5 = 0}$ (general form).

Given:

The line passes through the points $(x_1, y_1) = (1, -1)$ and $(x_2, y_2) = (3, 5)$.

To Find:

The equation of the line.

Solution:

We can find the equation of the line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$ using the two-point form formula:

$\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}$

Substitute the coordinates of the given points $(1, -1)$ and $(3, 5)$ into this formula.

Let $(x_1, y_1) = (1, -1)$ and $(x_2, y_2) = (3, 5)$.

$\frac{y - (-1)}{5 - (-1)} = \frac{x - 1}{3 - 1}$

Simplify the denominators:

$\frac{y + 1}{5 + 1} = \frac{x - 1}{2}$

$\frac{y + 1}{6} = \frac{x - 1}{2}$

Now, we can cross-multiply to eliminate the denominators:

$2(y + 1) = 6(x - 1)$

Divide both sides of the equation by 2:

$y + 1 = 3(x - 1)$

This is the equation of the line in point-slope form with point $(1, -1)$ and a calculated slope of 3.

To express the equation in the general form ($Ax + By + C = 0$), distribute the 3 on the right side:

$y + 1 = 3x - 3$

Move all terms to one side:

$0 = 3x - y - 3 - 1$

$0 = 3x - y - 4$

So, the equation of the line is $3x - y - 4 = 0$.

Answer:

The equation of the line through the points (1, –1) and (3, 5) is $\mathbf{3x - y - 4 = 0}$.

Given:

The tangent of the inclination of the line is $\tan \theta = \frac{1}{2}$.

The slope of the line is $m = \tan \theta = \frac{1}{2}$.

We need to find the equation of the line under two different conditions regarding intercepts.

(i) y-intercept is $-\frac{3}{2}$

The equation of a line with slope $m$ and y-intercept $c$ is given by the slope-intercept form:

$y = mx + c$

We are given the slope $m = \frac{1}{2}$ and the y-intercept $c = -\frac{3}{2}$.

Substitute these values into the slope-intercept form:

$y = \frac{1}{2}x + \left(-\frac{3}{2}\right)$

$y = \frac{1}{2}x - \frac{3}{2}$

To remove the fraction, multiply the entire equation by 2:

$2y = x - 3$

Rearrange the terms to get the general form $Ax + By + C = 0$:

$0 = x - 2y - 3$

So, the equation of the line is $x - 2y - 3 = 0$.

(ii) x-intercept is 4

The x-intercept is the x-coordinate of the point where the line crosses the x-axis. At this point, the y-coordinate is 0. So, the line passes through the point $(4, 0)$.

We have the slope $m = \frac{1}{2}$ and a point $(x_1, y_1) = (4, 0)$ on the line.

We can use the point-slope form of the equation of a line:

$y - y_1 = m(x - x_1)$

Substitute the values $m = \frac{1}{2}$, $x_1 = 4$, and $y_1 = 0$ into the formula:

$y - 0 = \frac{1}{2}(x - 4)$

$y = \frac{1}{2}(x - 4)$

To remove the fraction, multiply both sides by 2:

$2y = x - 4$

Rearrange the terms to get the general form $Ax + By + C = 0$:

$0 = x - 2y - 4$

So, the equation of the line is $x - 2y - 4 = 0$.

Answer:

(i) The equation of the line with slope $\frac{1}{2}$ and y-intercept $-\frac{3}{2}$ is $\mathbf{x - 2y - 3 = 0}$.

(ii) The equation of the line with slope $\frac{1}{2}$ and x-intercept 4 is $\mathbf{x - 2y - 4 = 0}$.

Given:

The x-intercept of the line is $-3$.

The y-intercept of the line is $2$.

To Find:

The equation of the line.

Solution:

The equation of a line which makes intercepts 'a' and 'b' on the x-axis and y-axis respectively is given by the intercept form:

$\frac{x}{a} + \frac{y}{b} = 1$

In this problem, the x-intercept is $a = -3$ and the y-intercept is $b = 2$.

Substitute these values into the intercept form:

$\frac{x}{-3} + \frac{y}{2} = 1$

To eliminate the denominators, multiply the entire equation by the least common multiple (LCM) of 3 and 2, which is 6.

$6 \times \left(\frac{x}{-3}\right) + 6 \times \left(\frac{y}{2}\right) = 6 \times 1$

$\cancel{6}^{-2} \times \frac{x}{\cancel{-3}_{1}} + \cancel{6}^{3} \times \frac{y}{\cancel{2}_{1}} = 6$

$-2x + 3y = 6$

To express the equation in the general form $Ax + By + C = 0$, move all terms to one side:

$-2x + 3y - 6 = 0$

Multiplying by $-1$ to make the coefficient of x positive is optional but common practice:

$2x - 3y + 6 = 0$

Answer:

The equation of the line is $\mathbf{2x - 3y + 6 = 0}$.

We need to find the equations of the x-axis and the y-axis.

Equation of the x-axis:

The x-axis is a horizontal line in the Cartesian coordinate system. Every point on the x-axis has a y-coordinate of 0, regardless of its x-coordinate.

For any point $(x, y)$ on the x-axis, $y = 0$.

Therefore, the equation of the x-axis is $\mathbf{y = 0}$.

Equation of the y-axis:

The y-axis is a vertical line in the Cartesian coordinate system. Every point on the y-axis has an x-coordinate of 0, regardless of its y-coordinate.

For any point $(x, y)$ on the y-axis, $x = 0$.

Therefore, the equation of the y-axis is $\mathbf{x = 0}$.

Given:

The line passes through the point $(x_1, y_1) = (-4, 3)$.

The slope of the line is $m = \frac{1}{2}$.

To Find:

The equation of the line.

Solution:

We can use the point-slope form of the equation of a line, which is given by:

$y - y_1 = m(x - x_1)$

Substitute the coordinates of the given point $(x_1, y_1) = (-4, 3)$ and the given slope $m = \frac{1}{2}$ into this formula:

$y - 3 = \frac{1}{2}(x - (-4))$

Simplify the expression inside the parenthesis:

$y - 3 = \frac{1}{2}(x + 4)$

This is the equation of the line in point-slope form.

We can further simplify this equation to the general form $Ax + By + C = 0$ by multiplying both sides by 2 to remove the fraction:

$2(y - 3) = 1(x + 4)$

$2y - 6 = x + 4$

Move all terms to one side of the equation:

$0 = x - 2y + 4 + 6$

$x - 2y + 10 = 0$

This is the equation of the line in the general form.

Answer:

The equation of the line passing through the point (– 4, 3) with slope $\frac{1}{2}$ is $\mathbf{x - 2y + 10 = 0}$.

Given:

The line passes through the origin $(0, 0)$.

The slope of the line is $m$.

To Find:

The equation of the line.

Solution:

We can use the point-slope form of the equation of a line, which is given by:

$y - y_1 = m(x - x_1)$

Substitute the coordinates of the given point $(x_1, y_1) = (0, 0)$ and the given slope $m$ into this formula:

$y - 0 = m(x - 0)$

Simplify the equation:

$y = mx$

This is the equation of the line in the slope-intercept form, where the y-intercept is $c=0$ (since the line passes through the origin).

To express the equation in the general form $Ax + By + C = 0$, move all terms to one side:

$mx - y = 0$

or

$mx - y + 0 = 0$

Answer:

The equation of the line passing through (0, 0) with slope $m$ is $\mathbf{y = mx}$ or $\mathbf{mx - y = 0}$.

Given:

The line passes through the point $(x_1, y_1) = (2, 2\sqrt{3})$.

The line is inclined with the x-axis at an angle of $\theta = 75^\circ$.

To Find:

The equation of the line.

Solution:

The slope ($m$) of a line that makes an inclination $\theta$ with the positive direction of the x-axis is given by $m = \tan(\theta)$.

In this case, $\theta = 75^\circ$. So, the slope is $m = \tan(75^\circ)$.

We can find the value of $\tan(75^\circ)$ using the tangent addition formula: $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$.

Let $A = 45^\circ$ and $B = 30^\circ$. Then $A+B = 45^\circ + 30^\circ = 75^\circ$.

We know the values of $\tan(45^\circ)$ and $\tan(30^\circ)$:

$\tan(45^\circ) = 1$

$\tan(30^\circ) = \frac{1}{\sqrt{3}}$

Substitute these values into the tangent addition formula:

$m = \tan(75^\circ) = \frac{\tan 45^\circ + \tan 30^\circ}{1 - \tan 45^\circ \tan 30^\circ} = \frac{1 + \frac{1}{\sqrt{3}}}{1 - 1 \times \frac{1}{\sqrt{3}}}$

$m = \frac{1 + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}}$

To simplify, find a common denominator in the numerator and the denominator:

$m = \frac{\frac{\sqrt{3}+1}{\sqrt{3}}}{\frac{\sqrt{3}-1}{\sqrt{3}}}$

$m = \frac{\sqrt{3}+1}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}-1}$

$m = \frac{\sqrt{3}+1}{\sqrt{3}-1}$

Rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator, which is $\sqrt{3}+1$:

$m = \frac{\sqrt{3}+1}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}$

$m = \frac{(\sqrt{3}+1)^2}{(\sqrt{3})^2 - 1^2}$

Expand the numerator and simplify the denominator:

$m = \frac{(\sqrt{3})^2 + 2(\sqrt{3})(1) + 1^2}{3 - 1}$

$m = \frac{3 + 2\sqrt{3} + 1}{2}$

$m = \frac{4 + 2\sqrt{3}}{2}$

Factor out 2 from the numerator:

$m = \frac{2(2 + \sqrt{3})}{2}$

$m = 2 + \sqrt{3}$

Now that we have the slope $m = 2 + \sqrt{3}$ and a point $(x_1, y_1) = (2, 2\sqrt{3})$ that the line passes through, we can use the point-slope form of the equation of a line:

$y - y_1 = m(x - x_1)$

Substitute the values:

$y - 2\sqrt{3} = (2 + \sqrt{3})(x - 2)$

This is the equation of the line. We can expand and rearrange it into the general form $Ax + By + C = 0$ if desired.

Expand the right side:

$y - 2\sqrt{3} = (2 + \sqrt{3})x - 2(2 + \sqrt{3})$

$y - 2\sqrt{3} = (2 + \sqrt{3})x - 4 - 2\sqrt{3}$

Move all terms to one side:

$(2 + \sqrt{3})x - y - 4 - 2\sqrt{3} + 2\sqrt{3} = 0$

$(2 + \sqrt{3})x - y - 4 = 0$

Answer:

The equation of the line is $\mathbf{y - 2\sqrt{3} = (2 + \sqrt{3})(x - 2)}$ or $\mathbf{(2 + \sqrt{3})x - y - 4 = 0}$.

Given:

The line intersects the x-axis at a distance of 3 units to the left of the origin.

The slope of the line is $m = -2$.

To Find:

The equation of the line.

Solution:

The x-axis is the line where the y-coordinate is 0. A point on the x-axis at a distance of 3 units to the left of the origin (0, 0) has coordinates $(-3, 0)$.

So, the line passes through the point $(x_1, y_1) = (-3, 0)$.

We are given the slope of the line, $m = -2$.

We can use the point-slope form of the equation of a line, which is given by:

$y - y_1 = m(x - x_1)$

Substitute the coordinates of the given point $(x_1, y_1) = (-3, 0)$ and the given slope $m = -2$ into this formula:

$y - 0 = -2(x - (-3))$

Simplify the expression inside the parenthesis:

$y = -2(x + 3)$

This is the equation of the line in point-slope form with y1 = 0.

We can further simplify this equation to the general form $Ax + By + C = 0$ by distributing the $-2$ on the right side:

$y = -2x - 6$

Move all terms to one side of the equation:

$2x + y + 6 = 0$

This is the equation of the line in the general form.

Answer:

The equation of the line is $\mathbf{2x + y + 6 = 0}$.

Given:

The line intersects the y-axis at a distance of 2 units above the origin.

The line makes an angle of $30^\circ$ with the positive direction of the x-axis.

To Find:

The equation of the line.

Solution:

The point where the line intersects the y-axis is called the y-intercept. Since the intersection is at a distance of 2 units above the origin, the coordinates of this point are $(0, 2)$.

The y-intercept is $c = 2$.

The angle the line makes with the positive direction of the x-axis is the inclination, denoted by $\theta$. We are given $\theta = 30^\circ$.

The slope ($m$) of a line is the tangent of its inclination:

$m = \tan \theta$

Substitute the given value of $\theta$:

$m = \tan(30^\circ)$

The value of $\tan(30^\circ)$ is $\frac{1}{\sqrt{3}}$.

So, the slope of the line is $m = \frac{1}{\sqrt{3}}$.

Now that we have the slope ($m$) and the y-intercept ($c$), we can use the slope-intercept form of the equation of a line, which is:

$y = mx + c$

Substitute the values $m = \frac{1}{\sqrt{3}}$ and $c = 2$ into the formula:

$y = \frac{1}{\sqrt{3}}x + 2$

This is one form of the equation of the line.

To remove the fraction and express the equation in the general form $Ax + By + C = 0$, we can multiply the entire equation by $\sqrt{3}$:

$\sqrt{3}y = \sqrt{3} \left(\frac{1}{\sqrt{3}}x\right) + \sqrt{3}(2)$

$\sqrt{3}y = x + 2\sqrt{3}$

Move all terms to one side:

$0 = x - \sqrt{3}y + 2\sqrt{3}$

So, the equation of the line is $x - \sqrt{3}y + 2\sqrt{3} = 0$.

Answer:

The equation of the line is $\mathbf{y = \frac{1}{\sqrt{3}}x + 2}$ or $\mathbf{x - \sqrt{3}y + 2\sqrt{3} = 0}$.

Given:

The line passes through the points $(x_1, y_1) = (-1, 1)$ and $(x_2, y_2) = (2, -4)$.

To Find:

The equation of the line.

Solution:

We can find the equation of the line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$ using the two-point form formula:

$\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}$

Substitute the coordinates of the given points $(-1, 1)$ and $(2, -4)$ into this formula.

Let $(x_1, y_1) = (-1, 1)$ and $(x_2, y_2) = (2, -4)$.

$\frac{y - 1}{-4 - 1} = \frac{x - (-1)}{2 - (-1)}$

Simplify the denominators:

$\frac{y - 1}{-5} = \frac{x + 1}{2 + 1}$

$\frac{y - 1}{-5} = \frac{x + 1}{3}$

Now, we can cross-multiply to eliminate the denominators:

$3(y - 1) = -5(x + 1)$

Distribute the numbers on both sides:

$3y - 3 = -5x - 5$

Move all terms to one side to get the general form $Ax + By + C = 0$:

$5x + 3y - 3 + 5 = 0$

$5x + 3y + 2 = 0$

Answer:

The equation of the line passing through the points (–1, 1) and (2, – 4) is $\mathbf{5x + 3y + 2 = 0}$.

Given:

The vertices of $\Delta PQR$ are $P(2, 1)$, $Q(-2, 3)$, and $R(4, 5)$.

To Find:

The equation of the median through the vertex R.

Solution:

A median of a triangle from a vertex connects the vertex to the midpoint of the opposite side.

The median through vertex R connects R to the midpoint of the side PQ.

Let M be the midpoint of the line segment PQ. The coordinates of the midpoint of a segment with endpoints $(x_1, y_1)$ and $(x_2, y_2)$ are given by:

Midpoint $(x_m, y_m) = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$

Using the coordinates of P(2, 1) and Q(-2, 3), we find the coordinates of M:

$x_m = \frac{2 + (-2)}{2} = \frac{0}{2} = 0$

$y_m = \frac{1 + 3}{2} = \frac{4}{2} = 2$

So, the midpoint M is $(0, 2)$.

The median through vertex R is the line segment RM, which passes through the points $R(4, 5)$ and $M(0, 2)$.

We can find the equation of the line passing through these two points using the two-point form of the line equation:

$\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}$

Let $(x_1, y_1) = (4, 5)$ and $(x_2, y_2) = (0, 2)$.

Substitute these coordinates into the formula:

$\frac{y - 5}{2 - 5} = \frac{x - 4}{0 - 4}$

Simplify the denominators:

$\frac{y - 5}{-3} = \frac{x - 4}{-4}$

Multiply both sides by $-1$ to make the denominators positive:

$\frac{y - 5}{3} = \frac{x - 4}{4}$

Now, cross-multiply to eliminate the denominators:

$4(y - 5) = 3(x - 4)$

Distribute the numbers on both sides:

$4y - 20 = 3x - 12$

Rearrange the terms to get the general form $Ax + By + C = 0$:

$0 = 3x - 4y - 12 + 20$

$3x - 4y + 8 = 0$

Answer:

The equation of the median through the vertex R is $\mathbf{3x - 4y + 8 = 0}$.

Given:

The required line passes through the point $(-3, 5)$.

The required line is perpendicular to the line passing through the points $(2, 5)$ and $(-3, 6)$.

To Find:

The equation of the required line.

Solution:

First, we find the slope of the line passing through the points $(2, 5)$ and $(-3, 6)$. Let's call this line L1, and its slope $m_1$.

Using the slope formula $m = \frac{y_2 - y_1}{x_2 - x_1}$ for points $(x_1, y_1) = (2, 5)$ and $(x_2, y_2) = (-3, 6)$:

$m_1 = \frac{6 - 5}{-3 - 2}$

$m_1 = \frac{1}{-5}$

$m_1 = -\frac{1}{5}$

The required line (let's call it L2) is perpendicular to L1. If two non-vertical lines are perpendicular, the product of their slopes is $-1$. Let the slope of L2 be $m_2$.

$m_1 \times m_2 = -1$

Substitute the value of $m_1$:

$-\frac{1}{5} \times m_2 = -1$

Multiply both sides by $-5$ to solve for $m_2$:

$m_2 = (-1) \times (-5)$

$m_2 = 5$

The slope of the required line is 5.

Now we have the slope of the required line ($m_2 = 5$) and a point that it passes through $(x_1, y_1) = (-3, 5)$. We can use the point-slope form of the equation of a line:

$y - y_1 = m(x - x_1)$

Substitute the values:

$y - 5 = 5(x - (-3))$

$y - 5 = 5(x + 3)$

This is the equation in point-slope form. We can expand and rearrange it into the general form $Ax + By + C = 0$:

$y - 5 = 5x + 15$

Move all terms to one side:

$0 = 5x - y + 15 + 5$

$5x - y + 20 = 0$

Answer:

The equation of the line passing through (–3, 5) and perpendicular to the line through the points (2, 5) and (–3, 6) is $\mathbf{5x - y + 20 = 0}$.

Given:

Two points $A(1, 0)$ and $B(2, 3)$.

A line is perpendicular to the line segment AB.

This line divides the line segment AB in the ratio 1:n.

To Find:

The equation of the line.

Solution:

Let the line segment joining the points $A(1, 0)$ and $B(2, 3)$ be denoted by L1. First, we find the slope of L1.

The slope ($m_1$) of the line passing through $A(x_1, y_1) = (1, 0)$ and $B(x_2, y_2) = (2, 3)$ is given by:

$m_1 = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - 0}{2 - 1} = \frac{3}{1} = 3$

Let the required line be denoted by L2. Since L2 is perpendicular to L1, the product of their slopes is $-1$. Let the slope of L2 be $m_2$.

$m_1 \times m_2 = -1$

$3 \times m_2 = -1$

$m_2 = -\frac{1}{3}$

The required line L2 divides the line segment AB in the ratio 1:n. This means the point of intersection of L2 and the line segment AB is a point, let's call it P, that divides AB internally in the ratio 1:n.

Using the section formula, the coordinates of point P $(x_p, y_p)$ dividing the segment joining $(x_1, y_1)$ and $(x_2, y_2)$ in the ratio $m:n$ are:

$x_p = \frac{m x_2 + n x_1}{m + n}$

$y_p = \frac{m y_2 + n y_1}{m + n}$

Here, $(x_1, y_1) = (1, 0)$, $(x_2, y_2) = (2, 3)$, and the ratio is $1:n$ (so $m=1$ and the ratio value is $n$).

$x_p = \frac{1 \times 2 + n \times 1}{1 + n} = \frac{2 + n}{1 + n}$

$y_p = \frac{1 \times 3 + n \times 0}{1 + n} = \frac{3}{1 + n}$

So, the point P through which the line L2 passes is $\left(\frac{2+n}{1+n}, \frac{3}{1+n}\right)$.

Now we have the slope of the required line $m_2 = -\frac{1}{3}$ and a point $P\left(\frac{2+n}{1+n}, \frac{3}{1+n}\right)$ on the line. We can use the point-slope form of the equation of a line:

$y - y_p = m_2(x - x_p)$

Substitute the values:

$y - \frac{3}{1+n} = -\frac{1}{3} \left(x - \frac{2+n}{1+n}\right)$

To clear the denominators, multiply both sides of the equation by $3(1+n)$:

$3(1+n) \left(y - \frac{3}{1+n}\right) = 3(1+n) \left(-\frac{1}{3}\right) \left(x - \frac{2+n}{1+n}\right)$

$3(1+n)y - 3(1+n) \times \frac{3}{1+n} = -(1+n) \left(x - \frac{2+n}{1+n}\right)$

$3(1+n)y - 9 = -(1+n)x + (1+n)\frac{2+n}{1+n}$

$3(1+n)y - 9 = -(1+n)x + (2+n)$

Move all terms to one side to get the general form $Ax + By + C = 0$:

$(1+n)x + 3(1+n)y - 9 - (2+n) = 0$

$(1+n)x + 3(1+n)y - 9 - 2 - n = 0$

$(1+n)x + 3(1+n)y - (11 + n) = 0$

Answer:

The equation of the line is $\mathbf{(1+n)x + 3(1+n)y - (11 + n) = 0}$.

Given:

The line cuts off equal intercepts on the coordinate axes.

The line passes through the point $(2, 3)$.

To Find:

The equation of the line.

Solution:

Let the x-intercept and the y-intercept of the line be $a$ and $b$ respectively.

The problem states that the line cuts off equal intercepts on the coordinate axes, which means $a = b$.

The equation of a line in the intercept form is given by:

$\frac{x}{a} + \frac{y}{b} = 1$

Since $a = b$, we can substitute $b$ with $a$ in the equation:

$\frac{x}{a} + \frac{y}{a} = 1$

Combine the fractions on the left side (assuming $a \neq 0$):

$\frac{x + y}{a} = 1$

$x + y = a$

We are given that the line passes through the point $(2, 3)$. This means that the coordinates of this point must satisfy the equation of the line.

Substitute $x = 2$ and $y = 3$ into equation (1):

$2 + 3 = a$

$5 = a$

So, the value of the equal intercepts is 5. Since $a \neq 0$, our assumption was valid.

Substitute the value of $a$ back into equation (1):

$x + y = 5$

To express the equation in the general form $Ax + By + C = 0$, move the constant term to the left side:

$x + y - 5 = 0$

Check for the case where intercepts are zero ($a=b=0$):

If the intercepts are zero, the line passes through the origin $(0,0)$. The given line passes through $(2,3)$, which is not the origin. Therefore, the intercepts cannot be zero, and $a \neq 0$.

Answer:

The equation of the line is $\mathbf{x + y - 5 = 0}$.

Given:

The line passes through the point $(2, 2)$.

The sum of the intercepts on the coordinate axes is 9.

To Find:

The equation of the line.

Solution:

Let the x-intercept of the line be $a$ and the y-intercept be $b$.

The problem states that the sum of the intercepts is 9:

From this equation, we can express $b$ in terms of $a$:

$b = 9 - a$

The equation of a line in the intercept form is given by:

$\frac{x}{a} + \frac{y}{b} = 1$

Substitute $b = 9 - a$ into the intercept form (assuming $a \neq 0$ and $b \neq 0$, which means $a \neq 0$ and $a \neq 9$):

We are given that the line passes through the point $(2, 2)$. This means that the coordinates of this point must satisfy the equation of the line.

Substitute $x = 2$ and $y = 2$ into equation (2):

$\frac{2}{a} + \frac{2}{9 - a} = 1$

To solve for $a$, find a common denominator, which is $a(9 - a)$:

$\frac{2(9 - a) + 2a}{a(9 - a)} = 1$

Multiply both sides by $a(9 - a)$:

$2(9 - a) + 2a = a(9 - a)$

Distribute the terms:

$18 - 2a + 2a = 9a - a^2$

$18 = 9a - a^2$

Rearrange the equation into a quadratic form:

$a^2 - 9a + 18 = 0$

Factor the quadratic equation. We need two numbers that multiply to 18 and add up to -9. These numbers are -3 and -6.

$(a - 3)(a - 6) = 0$

This gives two possible values for $a$:

$a - 3 = 0 \implies a = 3$

$a - 6 = 0 \implies a = 6$

Both these values are not 0 and not 9, so our initial assumption was valid.

Now, find the corresponding values of $b$ using $b = 9 - a$ and substitute the pairs $(a, b)$ into the intercept form $\frac{x}{a} + \frac{y}{b} = 1$ to find the equations of the lines.

Case 1: $a = 3$

$b = 9 - 3 = 6$

The intercepts are $a = 3$ and $b = 6$.

The equation of the line is $\frac{x}{3} + \frac{y}{6} = 1$.

Multiply by the LCM of 3 and 6 (which is 6) to clear the denominators:

$6 \times \frac{x}{3} + 6 \times \frac{y}{6} = 6 \times 1$

$2x + y = 6$

In general form: $2x + y - 6 = 0$.

Case 2: $a = 6$

$b = 9 - 6 = 3$

The intercepts are $a = 6$ and $b = 3$.

The equation of the line is $\frac{x}{6} + \frac{y}{3} = 1$.

Multiply by the LCM of 6 and 3 (which is 6) to clear the denominators:

$6 \times \frac{x}{6} + 6 \times \frac{y}{3} = 6 \times 1$

$x + 2y = 6$

In general form: $x + 2y - 6 = 0$.

There are two possible lines that satisfy the given conditions.

Answer:

The equations of the lines are $\mathbf{2x + y - 6 = 0}$ and $\mathbf{x + 2y - 6 = 0}$.

We have two parts to this question.

Part 1: Find the equation of the line through (0, 2) making an angle $\frac{2\pi}{3}$ with the positive x-axis.

Given:

The line passes through the point $(x_1, y_1) = (0, 2)$.

The inclination of the line with the positive x-axis is $\theta = \frac{2\pi}{3}$.

To Find:

The equation of this line.

Solution:

The slope ($m$) of a line is the tangent of its inclination:

$m = \tan \theta$

Substitute the given value of $\theta$:

$m = \tan\left(\frac{2\pi}{3}\right)$

The angle $\frac{2\pi}{3}$ radians is equivalent to $\frac{2 \times 180^\circ}{3} = 2 \times 60^\circ = 120^\circ$.

$m = \tan(120^\circ)$

We know that $\tan(120^\circ) = -\sqrt{3}$.

So, the slope of the line is $m = -\sqrt{3}$.

We have the slope $m = -\sqrt{3}$ and a point $(x_1, y_1) = (0, 2)$ on the line. Since the point is $(0, 2)$, it is the y-intercept, so $c = 2$.

We can use the slope-intercept form of the equation of a line, $y = mx + c$:

$y = -\sqrt{3}x + 2$

To express the equation in the general form $Ax + By + C = 0$, move all terms to one side:

$\sqrt{3}x + y - 2 = 0$

Equation of the first line: $\mathbf{\sqrt{3}x + y - 2 = 0}$.

Part 2: Find the equation of a line parallel to the first line and crossing the y-axis at a distance of 2 units below the origin.

Given:

The second line is parallel to the first line (whose equation is $\sqrt{3}x + y - 2 = 0$).

The second line crosses the y-axis at a distance of 2 units below the origin.

To Find:

The equation of this second line.

Solution:

Parallel lines have the same slope. The slope of the first line is $m_1 = -\sqrt{3}$.

The slope of the second line, $m_2$, is equal to $m_1$.

$m_2 = -\sqrt{3}$

The second line crosses the y-axis at a distance of 2 units below the origin. This is the y-intercept of the second line. A distance of 2 units below the origin on the y-axis corresponds to the point $(0, -2)$.

So, the y-intercept of the second line is $c_2 = -2$.

We have the slope $m_2 = -\sqrt{3}$ and the y-intercept $c_2 = -2$. We can use the slope-intercept form of the equation of a line, $y = mx + c$:

$y = -\sqrt{3}x + (-2)$

$y = -\sqrt{3}x - 2$

To express the equation in the general form $Ax + By + C = 0$, move all terms to one side:

$\sqrt{3}x + y + 2 = 0$

Equation of the second line: $\mathbf{\sqrt{3}x + y + 2 = 0}$.

Given:

A line passes through the point $P(-2, 9)$.

The perpendicular from the origin $O(0, 0)$ to this line meets the line at point P.

To Find:

The equation of the line.

Solution:

The line segment connecting the origin $O(0, 0)$ and the point $P(-2, 9)$ is perpendicular to the required line at the point $P$.

First, let's find the slope of the line segment OP.

The formula for the slope ($m$) of a line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$ is $m = \frac{y_2 - y_1}{x_2 - x_1}$.

Using points $O(0, 0)$ and $P(-2, 9)$:

Slope of OP ($m_{OP}$) $= \frac{9 - 0}{-2 - 0} = \frac{9}{-2} = -\frac{9}{2}$

The required line passes through point P(-2, 9) and is perpendicular to the line segment OP.

If two non-vertical lines are perpendicular, the product of their slopes is $-1$. Let the slope of the required line be $m_{line}$.

$m_{OP} \times m_{line} = -1$

$-\frac{9}{2} \times m_{line} = -1$

Multiply both sides by $-\frac{2}{9}$ to solve for $m_{line}$:

$m_{line} = (-1) \times \left(-\frac{2}{9}\right)$

$m_{line} = \frac{2}{9}$

The slope of the required line is $\frac{2}{9}$.

Now we have the slope of the required line ($m_{line} = \frac{2}{9}$) and a point that it passes through $(x_1, y_1) = (-2, 9)$. We can use the point-slope form of the equation of a line:

$y - y_1 = m_{line}(x - x_1)$

Substitute the values:

$y - 9 = \frac{2}{9}(x - (-2))$

$y - 9 = \frac{2}{9}(x + 2)$

This is the equation in point-slope form. We can expand and rearrange it into the general form $Ax + By + C = 0$:

Multiply both sides by 9:

$9(y - 9) = 2(x + 2)$

$9y - 81 = 2x + 4$

Move all terms to one side:

$0 = 2x - 9y + 4 + 81$

$2x - 9y + 85 = 0$

Answer:

The equation of the line is $\mathbf{2x - 9y + 85 = 0}$.

Given:

To Express:

L in terms of C.

Solution:

Since L is a linear function of C, the relationship between L and C can be represented by the equation of a straight line. Let the equation be of the form $L = mC + k$, where $m$ is the slope of the line and $k$ is the L-intercept.

We are given two points $(C, L)$ that lie on this line: $(C_1, L_1) = (20, 124.942)$ and $(C_2, L_2) = (110, 125.134)$.

The slope ($m$) of the line passing through two points $(C_1, L_1)$ and $(C_2, L_2)$ is given by the formula:

$m = \frac{L_2 - L_1}{C_2 - C_1}$

Substitute the given values into the slope formula:

$m = \frac{125.134 - 124.942}{110 - 20}$

Calculate the difference in L values and the difference in C values:

$L_2 - L_1 = 125.134 - 124.942 = 0.192$

$C_2 - C_1 = 110 - 20 = 90$

So, the slope is:

$m = \frac{0.192}{90}$

Now we have the slope $m = \frac{0.192}{90}$ and a point that the line passes through, for example, $(C_1, L_1) = (20, 124.942)$. We can use the point-slope form of the linear equation, which is $L - L_1 = m(C - C_1)$.

Substitute the values of $L_1$, $C_1$, and $m$ into the point-slope form:

$L - 124.942 = \frac{0.192}{90}(C - 20)$

To express L in terms of C, we can add $124.942$ to both sides of the equation:

This equation expresses L as a function of C.

Note that the slope $\frac{0.192}{90}$ can also be written as a simplified fraction. Multiplying the numerator and denominator by 1000 gives $\frac{192}{90000}$, which simplifies to $\frac{4}{1875}$ as shown in the previous attempt.

So, the equation can also be written as $L = \frac{4}{1875}(C - 20) + 124.942$. Both forms are correct ways to express L in terms of C.

Answer:

The length L in terms of C can be expressed by the equation $L = \frac{0.192}{90}(C - 20) + 124.942$ or equivalently $L = \frac{4}{1875}(C - 20) + 124.942$.

Given:

The relationship between the selling price and demand for milk is linear.

When the price is $\textsf{₹}14$/litre, the demand is 980 litres/week.

When the price is $\textsf{₹}16$/litre, the demand is 1220 litres/week.

To Find:

The demand (litres sold weekly) when the selling price is $\textsf{₹}17$/litre.

Solution:

Let $P$ be the selling price per litre (in $\textsf{₹}$) and $D$ be the demand in litres per week. Since the relationship between selling price and demand is linear, we can express this relationship as the equation of a line. We can consider $(P, D)$ as points on this line.

We are given two points $(P_1, D_1) = (14, 980)$ and $(P_2, D_2) = (16, 1220)$.

We can use the two-point form of the equation of a line:

$\frac{D - D_1}{D_2 - D_1} = \frac{P - P_1}{P_2 - P_1}$

Substitute the given values:

$\frac{D - 980}{1220 - 980} = \frac{P - 14}{16 - 14}$

Simplify the denominators:

$\frac{D - 980}{240} = \frac{P - 14}{2}$

Multiply both sides by 240 to solve for $D - 980$:

$D - 980 = \frac{240}{2}(P - 14)$

$D - 980 = 120(P - 14)$

Distribute 120 on the right side:

$D - 980 = 120P - 120 \times 14$

$120 \times 14 = 1680$

$D - 980 = 120P - 1680$

Add 980 to both sides to express $D$ in terms of $P$:

$D = 120P - 1680 + 980$

This is the linear equation relating demand $D$ and price $P$.

We need to find the demand when the selling price is $\textsf{₹}17$/litre. Substitute $P = 17$ into equation (1):

$D = 120(17) - 700$

Calculate $120 \times 17$:

$120 \times 17 = 2040$

$D = 2040 - 700$

$D = 1340$

So, at a price of $\textsf{₹}17$/litre, the owner could sell 1340 litres of milk weekly.

Answer:

The owner could sell $\mathbf{1340}$ litres of milk weekly at $\textsf{₹}17$/litre.

Given:

A line segment has its endpoints on the coordinate axes (x-axis and y-axis).

The midpoint of this line segment is $P(a, b)$.

To Show:

The equation of the line containing this segment is $\frac{x}{a} + \frac{y}{b} = 2$.

Solution:

Let the line segment intersect the x-axis at point A and the y-axis at point B.

Since point A is on the x-axis, its y-coordinate is 0. Let its coordinates be $(x_1, 0)$. This point is the x-intercept of the line, so $x_1 = a'$ (using $a'$ to distinguish from the 'a' in the midpoint). So, $A = (a', 0)$.

Since point B is on the y-axis, its x-coordinate is 0. Let its coordinates be $(0, y_1)$. This point is the y-intercept of the line, so $y_1 = b'$ (using $b'$ to distinguish from the 'b' in the midpoint). So, $B = (0, b')$.

The midpoint of the line segment AB is given as $P(a, b)$.

Using the midpoint formula for the segment with endpoints $A(a', 0)$ and $B(0, b')$:

The x-coordinate of the midpoint is $\frac{a' + 0}{2} = \frac{a'}{2}$.

The y-coordinate of the midpoint is $\frac{0 + b'}{2} = \frac{b'}{2}$.

We are given that the midpoint is $(a, b)$. Therefore, we can equate the coordinates:

So, the x-intercept of the line is $2a$ and the y-intercept is $2b$. (This assumes $a \neq 0$ and $b \neq 0$. If $a=0$, the midpoint is on the y-axis, and the x-intercept cannot be defined as $2a$ unless the line is the y-axis itself. Similarly for $b=0$. If the line is the x or y-axis, the midpoint is the origin (0,0), so $a=0, b=0$. The equation $\frac{x}{0} + \frac{y}{0} = 2$ is undefined. The problem implies a line segment *between* the axes, suggesting it's not the axes themselves, so $a,b \neq 0$).

The equation of a line with x-intercept $a'$ and y-intercept $b'$ is given by the intercept form:

$\frac{x}{a'} + \frac{y}{b'} = 1$}

Substitute the values $a' = 2a$ and $b' = 2b$ into the intercept form:

$\frac{x}{2a} + \frac{y}{2b} = 1$

To show the desired equation, we can manipulate this form. Find a common denominator on the left side, which is $2ab$:

$\frac{bx + ay}{2ab} = 1$

Multiply both sides by $2ab$:

$bx + ay = 2ab$

Alternatively, from $\frac{x}{2a} + \frac{y}{2b} = 1$, we can multiply the entire equation by 2:

$2 \times \left(\frac{x}{2a} + \frac{y}{2b}\right) = 2 \times 1$

$\frac{\cancel{2}x}{\cancel{2}a} + \frac{\cancel{2}y}{\cancel{2}b} = 2$

$\frac{x}{a} + \frac{y}{b} = 2$

This is the required equation.

Conclusion:

By determining the x-intercept and y-intercept based on the midpoint being $(a, b)$ and then using the intercept form of the line equation, we have shown that the equation of the line is $\frac{x}{a} + \frac{y}{b} = 2$.

Given:

A line segment is between the coordinate axes.

Point $R(h, k)$ divides this line segment in the ratio 1:2.

To Find:

The equation of the line containing the segment.

Solution:

Let the line segment intersect the x-axis at point $A$ and the y-axis at point $B$.

Since point A is on the x-axis, its coordinates are $(x_1, 0)$. Let the x-intercept be $a'$, so $A = (a', 0)$.

Since point B is on the y-axis, its coordinates are $(0, y_2)$. Let the y-intercept be $b'$, so $B = (0, b')$.

The point $R(h, k)$ divides the line segment AB in the ratio 1:2. This means R is 1 part away from A and 2 parts away from B. The ratio is $m:n = 1:2$, where $m=1$ is associated with the coordinates of B and $n=2$ is associated with the coordinates of A.

Using the section formula, the coordinates of $R(h, k)$ are:

$h = \frac{m x_2 + n x_1}{m + n} = \frac{1 \cdot 0 + 2 \cdot a'}{1 + 2} = \frac{2a'}{3}$

$k = \frac{m y_2 + n y_1}{m + n} = \frac{1 \cdot b' + 2 \cdot 0}{1 + 2} = \frac{b'}{3}$

From these equations, we can express the intercepts $a'$ and $b'$ in terms of $h$ and $k$:

$h = \frac{2a'}{3} \implies 3h = 2a' \implies a' = \frac{3h}{2}$

$k = \frac{b'}{3} \implies b' = 3k$

So, the x-intercept of the line is $\frac{3h}{2}$ and the y-intercept is $3k$.

(This assumes the line segment is not on an axis itself, implying $h \neq 0$ and $k \neq 0$.)

The equation of a line with x-intercept $a'$ and y-intercept $b'$ is given by the intercept form:

Substitute the values $a' = \frac{3h}{2}$ and $b' = 3k$ into equation (1):

$\frac{x}{\frac{3h}{2}} + \frac{y}{3k} = 1$

Simplify the first term:

$\frac{x}{\frac{3h}{2}} = x \times \frac{2}{3h} = \frac{2x}{3h}$

The equation of the line is:

$\frac{2x}{3h} + \frac{y}{3k} = 1$

We can express this equation in the general form by clearing the denominators. Multiply the entire equation by the least common multiple of the denominators, $3hk$ (assuming $h \neq 0, k \neq 0$):

$3hk \left(\frac{2x}{3h}\right) + 3hk \left(\frac{y}{3k}\right) = 3hk \times 1$

$\cancel{3h}k \frac{2x}{\cancel{3h}} + 3h\cancel{k} \frac{y}{\cancel{3k}} = 3hk$

$2kx + 3hy = 3hk$

Rearrange to the general form $Ax + By + C = 0$:

$2kx + 3hy - 3hk = 0$

Answer:

The equation of the line is $\mathbf{\frac{2x}{3h} + \frac{y}{3k} = 1}$ or $\mathbf{2kx + 3hy - 3hk = 0}$.

Given:

The three points are $A(3, 0)$, $B(-2, -2)$, and $C(8, 2)$.

To Prove:

The three points (3, 0), (– 2, – 2), and (8, 2) are collinear.

Solution:

To prove that three points are collinear using the concept of the equation of a line, we can find the equation of the line passing through two of the points and then check if the third point lies on that line.

Let's find the equation of the line passing through points $A(3, 0)$ and $B(-2, -2)$.

Using the two-point form of the equation of a line, which passes through $(x_1, y_1)$ and $(x_2, y_2)$:

$\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}$

Let $(x_1, y_1) = (3, 0)$ and $(x_2, y_2) = (-2, -2)$.

Substitute these coordinates into the two-point formula:

$\frac{y - 0}{-2 - 0} = \frac{x - 3}{-2 - 3}$

Simplify the denominators:

$\frac{y}{-2} = \frac{x - 3}{-5}$

Cross-multiply to eliminate the denominators:

$-5 \times y = -2 \times (x - 3)$

$-5y = -2x + 6$

Rearrange the equation into the general form $Ax + By + C = 0$ by moving all terms to one side:

This is the equation of the line passing through points A and B.

Now, we need to check if the third point $C(8, 2)$ lies on the line represented by equation (1). A point lies on a line if its coordinates satisfy the equation of the line.

Substitute the coordinates of point $C(8, 2)$ (i.e., $x=8$ and $y=2$) into equation (1):

Left Hand Side (LHS) $= 2x - 5y - 6$

LHS $= 2(8) - 5(2) - 6$

LHS $= 16 - 10 - 6$

LHS $= 6 - 6$

LHS $= 0$

The Right Hand Side (RHS) of equation (1) is 0.

Since LHS = RHS ($0 = 0$), the coordinates of point $C(8, 2)$ satisfy the equation of the line passing through A and B.

This means that point C lies on the same line as points A and B.

Conclusion:

Since the third point (8, 2) lies on the line passing through the other two points (3, 0) and (– 2, – 2), the three points are collinear.

Given:

Point $(x_0, y_0) = (3, -5)$.

Line equation $3x - 4y - 26 = 0$.

To Find:

The distance of the point from the line.

Solution:

The distance $d$ of a point $(x_0, y_0)$ from a line $Ax + By + C = 0$ is given by the formula:

$d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$

In the given problem:

The point is $(x_0, y_0) = (3, -5)$.

The equation of the line is $3x - 4y - 26 = 0$.

Comparing the line equation with the general form $Ax + By + C = 0$, we have $A = 3$, $B = -4$, and $C = -26$.

Substitute these values into the distance formula:

$d = \frac{|(3)(3) + (-4)(-5) + (-26)|}{\sqrt{3^2 + (-4)^2}}$

Calculate the expression in the numerator:

$Ax_0 + By_0 + C = (3)(3) + (-4)(-5) + (-26)$

$= 9 + 20 - 26$

$= 29 - 26$

$= 3$

So the numerator is $|3| = 3$.

Calculate the expression in the denominator:

$\sqrt{A^2 + B^2} = \sqrt{3^2 + (-4)^2}$

$= \sqrt{9 + 16}$

$= \sqrt{25}$

$= 5$

Substitute the calculated values back into the distance formula:

$d = \frac{3}{5}$

The distance of the point (3, – 5) from the line 3x – 4y –26 = 0 is $\frac{3}{5}$ units.

Answer:

The distance is $\mathbf{\frac{3}{5}}$.

Given:

Line 1: $3x - 4y + 7 = 0$

Line 2: $3x - 4y + 5 = 0$

To Find:

The distance between the two parallel lines.

Solution:

The general form of the equations of two parallel lines is $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$.

The distance $d$ between these two parallel lines is given by the formula:

$d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$

From the given equations, $3x - 4y + 7 = 0$ and $3x - 4y + 5 = 0$, we can identify the values of A, B, C$_1$, and C$_2$:

$A = 3$

$B = -4$

$C_1 = 7$

$C_2 = 5$

Substitute these values into the distance formula:

$d = \frac{|7 - 5|}{\sqrt{3^2 + (-4)^2}}$

Calculate the value in the numerator:

$|7 - 5| = |2| = 2$

Calculate the value in the denominator:

$\sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$

Substitute the calculated values back into the distance formula:

$d = \frac{2}{5}$

The distance between the parallel lines is $\frac{2}{5}$ units.

Answer:

The distance between the parallel lines is $\mathbf{\frac{2}{5}}$.

We are asked to reduce the given equations into slope-intercept form ($y = mx + c$) and find their slopes ($m$) and y-intercepts ($c$).

(i) x + 7y = 0

To reduce this equation into the slope-intercept form, we need to isolate $y$ on one side of the equation.

Subtract $x$ from both sides:

$7y = -x$

Divide both sides by 7:

$y = \frac{-x}{7}$

This can be written as:

$y = -\frac{1}{7}x + 0$

Comparing this with the form $y = mx + c$, we have:

Slope ($m$) = $-\frac{1}{7}$

y-intercept ($c$) = $0$

(ii) 6x + 3y – 5 = 0

To reduce this equation into the slope-intercept form, we need to isolate $y$ on one side of the equation.

Subtract $6x$ and add 5 to both sides:

$3y = -6x + 5$

Divide both sides by 3:

$y = \frac{-6x + 5}{3}$

This can be written as:

$y = \frac{-6x}{3} + \frac{5}{3}$

$y = -2x + \frac{5}{3}$

Comparing this with the form $y = mx + c$, we have:

Slope ($m$) = $-2$

y-intercept ($c$) = $\frac{5}{3}$

(iii) y = 0

This equation is already in a form similar to the slope-intercept form. We can write it as:

$y = 0x + 0$

Comparing this with the form $y = mx + c$, we have:

Slope ($m$) = $0$

y-intercept ($c$) = $0$

We are asked to reduce the given equations into intercept form ($\frac{x}{a} + \frac{y}{b} = 1$) and find their intercepts on the axes (x-intercept $a$ and y-intercept $b$).

(i) 3x + 2y – 12 = 0

To reduce this equation into the intercept form, we need to rearrange it so that the constant term is on the right side and equal to 1.

Move the constant term to the right side:

$3x + 2y = 12$

Divide the entire equation by the constant term (12) to make the right side equal to 1:

$\frac{3x}{12} + \frac{2y}{12} = \frac{12}{12}$

Simplify the terms:

$\frac{\cancel{3}x}{\cancel{12}_{4}} + \frac{\cancel{2}y}{\cancel{12}_{6}} = 1$

$\frac{x}{4} + \frac{y}{6} = 1$

Comparing this with the intercept form $\frac{x}{a} + \frac{y}{b} = 1$, we have:

x-intercept ($a$) = $4$

y-intercept ($b$) = $6$

(ii) 4x – 3y = 6

To reduce this equation into the intercept form, we need to rearrange it so that the constant term is on the right side and equal to 1.

The constant term is already on the right side (6). Divide the entire equation by the constant term (6):

$\frac{4x}{6} - \frac{3y}{6} = \frac{6}{6}$

Simplify the terms:

$\frac{\cancel{4}^{2}x}{\cancel{6}_{3}} - \frac{\cancel{3}y}{\cancel{6}_{2}} = 1$

$\frac{2x}{3} - \frac{y}{2} = 1$

To match the standard intercept form $\frac{x}{a} + \frac{y}{b} = 1$, rewrite the terms with a positive sign in the middle:

$\frac{x}{3/2} + \frac{y}{-2} = 1$

Comparing this with the intercept form $\frac{x}{a} + \frac{y}{b} = 1$, we have:

x-intercept ($a$) = $\frac{3}{2}$

y-intercept ($b$) = $-2$

(iii) 3y + 2 = 0

To reduce this equation into the intercept form, we need to rearrange it so that the constant term is on the right side and equal to 1.

Move the constant term to the right side:

$3y = -2$

Divide the entire equation by the constant term (-2) to make the right side equal to 1:

$\frac{3y}{-2} = \frac{-2}{-2}$

$\frac{3y}{-2} = 1$

$\frac{y}{-2/3} = 1$

This equation can be written to explicitly show the x-term with a denominator:

$\frac{x}{\infty} + \frac{y}{-2/3} = 1$ (A line parallel to the x-axis does not intersect the x-axis unless it is the x-axis itself, in which case the x-intercept is undefined or considered $\pm \infty$.)

Alternatively, interpret the form $\frac{y}{b} = 1$ as the equation where the x-term has a coefficient of 0. $0x + \frac{y}{-2/3} = 1$.

In this case, the line is horizontal. It is parallel to the x-axis and intersects the y-axis at $y = -\frac{2}{3}$.

The y-intercept is $b = -\frac{2}{3}$.

The line is parallel to the x-axis, so it never intersects the x-axis unless it's the x-axis itself. Since $y = -\frac{2}{3}$ is not the x-axis ($y=0$), there is no x-intercept. The x-intercept is sometimes considered to be infinite for lines parallel to the x-axis.

Comparing $\frac{y}{-2/3} = 1$ with $\frac{x}{a} + \frac{y}{b} = 1$, we have:

x-intercept ($a$) = Undefined (or $\pm \infty$)

y-intercept ($b$) = $-\frac{2}{3}$

Given:

Point $(x_0, y_0) = (-1, 1)$.

Equation of the line: $12(x + 6) = 5(y – 2)$.

To Find:

The distance of the point from the line.

Solution:

First, we need to rewrite the equation of the line in the general form $Ax + By + C = 0$.

Start with the given equation:

$12(x + 6) = 5(y – 2)$

Distribute the numbers on both sides:

$12x + 72 = 5y - 10$}

Move all terms to one side of the equation:

$12x - 5y + 72 + 10 = 0$

$12x - 5y + 82 = 0$}

This equation is now in the general form $Ax + By + C = 0$, where $A = 12$, $B = -5$, and $C = 82$.

The distance $d$ of a point $(x_0, y_0)$ from a line $Ax + By + C = 0$ is given by the formula:

$d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$

Substitute the coordinates of the given point $(x_0, y_0) = (-1, 1)$ and the coefficients from the line equation ($A=12$, $B=-5$, $C=82$) into the distance formula:

$d = \frac{|(12)(-1) + (-5)(1) + (82)|}{\sqrt{(12)^2 + (-5)^2}}$

Calculate the value inside the absolute value in the numerator:

$Ax_0 + By_0 + C = 12(-1) + (-5)(1) + 82 = -12 - 5 + 82 = -17 + 82 = 65$

So the numerator is $|65| = 65$.

Calculate the value in the denominator:

$\sqrt{A^2 + B^2} = \sqrt{12^2 + (-5)^2} = \sqrt{144 + 25} = \sqrt{169} = 13$

Substitute the calculated values back into the distance formula:

$d = \frac{65}{13}$

Simplify the fraction:

$d = 5$

The distance of the point (–1, 1) from the line 12(x + 6) = 5(y – 2) is 5 units.

Answer:

The distance is $\mathbf{5}$.

Given:

The line equation is $\frac{x}{3} + \frac{y}{4} = 1$.

The distance from a point on the x-axis to this line is 4 units.

To Find:

The coordinates of the points on the x-axis.

Solution:

First, let's convert the equation of the line from the intercept form to the general form $Ax + By + C = 0$.

$\frac{x}{3} + \frac{y}{4} = 1$

Find a common denominator (LCM of 3 and 4 is 12):

$\frac{4x + 3y}{12} = 1$

Multiply both sides by 12:

$4x + 3y = 12$

Move the constant term to the left side:

$4x + 3y - 12 = 0$

This is the general form of the line equation, where $A = 4$, $B = 3$, and $C = -12$.

Let the required point on the x-axis be $(x_0, 0)$. Since the point is on the x-axis, its y-coordinate is 0.

The distance $d$ of a point $(x_0, y_0)$ from a line $Ax + By + C = 0$ is given by the formula:

$d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$

We are given that the distance $d = 4$, and the point is $(x_0, y_0) = (x_0, 0)$. Substitute these values into the distance formula:

$4 = \frac{|(4)(x_0) + (3)(0) + (-12)|}{\sqrt{4^2 + 3^2}}$

$4 = \frac{|4x_0 + 0 - 12|}{\sqrt{16 + 9}}$

$4 = \frac{|4x_0 - 12|}{\sqrt{25}}$

$4 = \frac{|4x_0 - 12|}{5}$

Multiply both sides by 5:

$20 = |4x_0 - 12|$

To solve for $x_0$, we consider the two possibilities of the absolute value:

Case 1: $4x_0 - 12 = 20$

Add 12 to both sides:

$4x_0 = 20 + 12$

$4x_0 = 32$

Divide by 4:

$x_0 = \frac{32}{4} = 8$

The point on the x-axis is $(8, 0)$.

Case 2: $4x_0 - 12 = -20$}

Add 12 to both sides:

$4x_0 = -20 + 12$

$4x_0 = -8$

Divide by 4:

$x_0 = \frac{-8}{4} = -2$

The point on the x-axis is $(-2, 0)$.

Thus, there are two points on the x-axis whose distances from the given line are 4 units.

Answer:

The points on the x-axis are $\mathbf{(8, 0)}$ and $\mathbf{(-2, 0)}$.

We need to find the distance between the given pairs of parallel lines.

The distance $d$ between two parallel lines in the form $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by the formula:

$d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$

(i) 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0

These equations are already in the general form $Ax + By + C = 0$.

Comparing with the standard form, we have:

For the first line ($15x + 8y – 34 = 0$): $A = 15$, $B = 8$, $C_1 = -34$.

For the second line ($15x + 8y + 31 = 0$): $A = 15$, $B = 8$, $C_2 = 31$.

The coefficients of x and y are the same (A and B), confirming that the lines are parallel.

Substitute these values into the distance formula:

$d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$

$d = \frac{|(-34) - (31)|}{\sqrt{(15)^2 + (8)^2}}$

$d = \frac{|-34 - 31|}{\sqrt{225 + 64}}$

$d = \frac{|-65|}{\sqrt{289}}$

$d = \frac{65}{17}$

The distance between the parallel lines is $\frac{65}{17}$ units.

(ii) l (x + y) + p = 0 and l (x + y) – r = 0

First, rewrite the equations in the general form $Ax + By + C = 0$ by distributing $l$:

For the first line ($l(x+y) + p = 0$): $lx + ly + p = 0$.

For the second line ($l(x+y) - r = 0$): $lx + ly - r = 0$.

Comparing with the standard form, we have:

For the first line: $A = l$, $B = l$, $C_1 = p$.

For the second line: $A = l$, $B = l$, $C_2 = -r$.

The coefficients of x and y are the same (A and B), confirming that the lines are parallel (assuming $l \neq 0$; if $l=0$, the equations become $p=0$ and $-r=0$, which represent no lines or coinciding lines depending on $p$ and $r$). We assume $l \neq 0$ for parallel distinct lines.

Substitute these values into the distance formula:

$d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$

$d = \frac{|p - (-r)|}{\sqrt{l^2 + l^2}}$

$d = \frac{|p + r|}{\sqrt{2l^2}}$

$d = \frac{|p + r|}{\sqrt{2} \sqrt{l^2}}$

$d = \frac{|p + r|}{\sqrt{2} |l|}$

To rationalize the denominator, multiply by $\frac{\sqrt{2}}{\sqrt{2}}$:

$d = \frac{|p + r|}{\sqrt{2} |l|} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2} |p + r|}{2 |l|}$

The distance between the parallel lines is $\frac{|p + r|}{\sqrt{2} |l|}$ or $\frac{\sqrt{2} |p + r|}{2 |l|}$ units (where $l \neq 0$).

Given:

The required line is parallel to the line $3x - 4y + 2 = 0$.

The required line passes through the point $(x_1, y_1) = (-2, 3)$.

To Find:

The equation of the required line.

Solution:

If two lines are parallel, they have the same slope.

First, find the slope of the given line $3x - 4y + 2 = 0$. We can rewrite this equation in the slope-intercept form $y = mx + c$ to find its slope.

$3x + 2 = 4y$

Divide by 4:

$y = \frac{3}{4}x + \frac{2}{4}$

$y = \frac{3}{4}x + \frac{1}{2}$

The slope of the given line is $m_1 = \frac{3}{4}$.

Since the required line is parallel to this line, its slope $m_2$ is equal to $m_1$.

$m_2 = \frac{3}{4}$

Now we have the slope of the required line ($m = \frac{3}{4}$) and a point it passes through $(x_1, y_1) = (-2, 3)$. We can use the point-slope form of the equation of a line, which is given by:

$y - y_1 = m(x - x_1)$

Substitute the values:

$y - 3 = \frac{3}{4}(x - (-2))$

$y - 3 = \frac{3}{4}(x + 2)$

This is the equation of the line in point-slope form. We can rearrange it into the general form $Ax + By + C = 0$ if desired.

Multiply both sides by 4 to eliminate the fraction:

$4(y - 3) = 3(x + 2)$

$4y - 12 = 3x + 6$

Move all terms to one side:

$0 = 3x - 4y + 6 + 12$

$3x - 4y + 18 = 0$

Answer:

The equation of the line parallel to the line 3x - 4y + 2 = 0 and passing through the point (–2, 3) is $\mathbf{3x - 4y + 18 = 0}$.

Given:

The required line is perpendicular to the line $x - 7y + 5 = 0$.

The required line has an x-intercept of 3.

To Find:

The equation of the required line.

Solution:

First, find the slope of the given line $x - 7y + 5 = 0$. We can rewrite this equation in the slope-intercept form $y = mx + c$ to find its slope.

Rearrange the terms to isolate $y$:

$x + 5 = 7y$

Divide by 7:

$y = \frac{1}{7}x + \frac{5}{7}$

The slope of the given line is $m_1 = \frac{1}{7}$.

The required line is perpendicular to this line. If two non-vertical lines are perpendicular, the product of their slopes is $-1$. Let the slope of the required line be $m_2$.

$m_1 \times m_2 = -1$

Substitute the value of $m_1$:

$\frac{1}{7} \times m_2 = -1$

Multiply both sides by 7 to solve for $m_2$:

$m_2 = -1 \times 7$

$m_2 = -7$

The slope of the required line is $-7$.

The required line has an x-intercept of 3. This means the line crosses the x-axis at the point where $x=3$ and $y=0$. So, the line passes through the point $(x_1, y_1) = (3, 0)$.

Now we have the slope of the required line ($m = -7$) and a point it passes through $(x_1, y_1) = (3, 0)$. We can use the point-slope form of the equation of a line, which is given by:

$y - y_1 = m(x - x_1)$

Substitute the values:

$y - 0 = -7(x - 3)$

$y = -7(x - 3)$

This is the equation of the line. We can expand it and rearrange it into the general form $Ax + By + C = 0$:

$y = -7x + 21$

Move all terms to one side:

$7x + y - 21 = 0$

Answer:

The equation of the line is $\mathbf{7x + y - 21 = 0}$.

Given:

Line 1: $\sqrt{3}x + y = 0$

Line 2: $x + \sqrt{3}y = 1$

To Find:

The angles between the two lines.

Solution:

First, we find the slopes of the two lines. We rewrite each equation in the slope-intercept form $y = mx + c$, where $m$ is the slope.

For Line 1: $\sqrt{3}x + y = 0$

Subtract $\sqrt{3}x$ from both sides:

$y = -\sqrt{3}x$

The slope of Line 1 is $m_1 = -\sqrt{3}$.

For Line 2: $x + \sqrt{3}y = 1$

Subtract $x$ from both sides:

$\sqrt{3}y = -x + 1$

Divide by $\sqrt{3}$:

$y = -\frac{1}{\sqrt{3}}x + \frac{1}{\sqrt{3}}$

The slope of Line 2 is $m_2 = -\frac{1}{\sqrt{3}}$.

Let $\theta$ be the angle between the two lines. The tangent of the angle $\theta$ between two lines with slopes $m_1$ and $m_2$ is given by the formula:

$\tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right|$

Substitute the values of $m_1$ and $m_2$ into the formula:

$m_2 - m_1 = -\frac{1}{\sqrt{3}} - (-\sqrt{3}) = -\frac{1}{\sqrt{3}} + \sqrt{3} = \frac{-1 + (\sqrt{3})(\sqrt{3})}{\sqrt{3}} = \frac{-1 + 3}{\sqrt{3}} = \frac{2}{\sqrt{3}}$

$1 + m_1 m_2 = 1 + (-\sqrt{3}) \left(-\frac{1}{\sqrt{3}}\right) = 1 + 1 = 2$

Now substitute these values into the tangent formula:

$\tan \theta = \left| \frac{\frac{2}{\sqrt{3}}}{2} \right|$

$\tan \theta = \left| \frac{2}{\sqrt{3}} \times \frac{1}{2} \right|$

$\tan \theta = \left| \frac{1}{\sqrt{3}} \right|$

$\tan \theta = \frac{1}{\sqrt{3}}$

Since $\tan \theta = \frac{1}{\sqrt{3}}$, one possible value for the acute angle $\theta$ is $30^\circ$.

In radians, $30^\circ = \frac{\pi}{6}$.

The angles between two lines are $\theta$ and $180^\circ - \theta$ (or $\pi - \theta$).

The acute angle is $\theta = 30^\circ$ (or $\frac{\pi}{6}$ radians).

The obtuse angle is $180^\circ - 30^\circ = 150^\circ$ (or $\pi - \frac{\pi}{6} = \frac{5\pi}{6}$ radians).

Answer:

The angles between the lines are $\mathbf{30^\circ}$ and $\mathbf{150^\circ}$ (or $\mathbf{\frac{\pi}{6}}$ and $\mathbf{\frac{5\pi}{6}}$ radians).

Given:

Line 1 passes through points $A(h, 3)$ and $B(4, 1)$.

Line 2 has the equation $7x - 9y - 19 = 0$.

Line 1 intersects Line 2 at a right angle (i.e., they are perpendicular).

To Find:

The value of $h$.

Solution:

First, find the slope of Line 1 passing through points $A(h, 3)$ and $B(4, 1)$. Let the slope be $m_1$.

Using the slope formula $m = \frac{y_2 - y_1}{x_2 - x_1}$ for points $(x_1, y_1) = (h, 3)$ and $(x_2, y_2) = (4, 1)$:

$m_1 = \frac{1 - 3}{4 - h} = \frac{-2}{4 - h}$

Note that $h \neq 4$, otherwise the denominator is 0, and Line 1 would be a vertical line with undefined slope.

Next, find the slope of Line 2 with the equation $7x - 9y - 19 = 0$. Let the slope be $m_2$. We can rewrite the equation in the slope-intercept form $y = mx + c$ to find the slope.

Rearrange the terms to isolate $y$:

$7x - 19 = 9y$

Divide by 9:

$y = \frac{7}{9}x - \frac{19}{9}$

The slope of Line 2 is $m_2 = \frac{7}{9}$.

Since Line 1 is perpendicular to Line 2, the product of their slopes is $-1$, provided neither line is vertical or horizontal.

The slope of Line 2 is $\frac{7}{9}$, which is not 0 or undefined, so Line 2 is neither horizontal nor vertical. This implies Line 1 must also be neither horizontal nor vertical, so its slope $m_1$ is defined and non-zero. Thus, $4 - h \neq 0$ (i.e., $h \neq 4$) and $\frac{-2}{4-h} \neq 0$ (which is true since $-2 \neq 0$).

The condition for perpendicular lines is $m_1 \times m_2 = -1$.

Substitute the slopes we found:

$\left(\frac{-2}{4 - h}\right) \times \left(\frac{7}{9}\right) = -1$

Multiply the terms on the left side:

$\frac{-14}{9(4 - h)} = -1$

Multiply both sides by $9(4 - h)$:

$-14 = -1 \times 9(4 - h)$

$-14 = -9(4 - h)$

$-14 = -36 + 9h$

Move the constant term to the left side by adding 36 to both sides:

$-14 + 36 = 9h$

$22 = 9h$

Divide by 9 to solve for $h$:

$h = \frac{22}{9}$

This value $h = \frac{22}{9}$ is not equal to 4, so our initial assumption ($h \neq 4$) was valid.

Answer:

The value of $h$ is $\mathbf{\frac{22}{9}}$.

Given:

A point $(x_1, y_1)$.

A line with equation $Ax + By + C = 0$.

A required line that passes through $(x_1, y_1)$ and is parallel to $Ax + By + C = 0$.

To Prove:

The equation of the required line is $A(x - x_1) + B(y - y_1) = 0$.

Solution:

A line is uniquely determined by a point it passes through and its slope (unless it is a vertical line). Two distinct lines are parallel if and only if they have the same slope. Vertical lines are parallel to each other and have undefined slopes.

Consider the given equation $A(x - x_1) + B(y - y_1) = 0$.

We can rewrite this equation as $Ax - Ax_1 + By - By_1 = 0$, which is $Ax + By - (Ax_1 + By_1) = 0$.

This is a linear equation in the form $Ax' + By' + C' = 0$ (where $x'=x$, $y'=y$, and $C' = -(Ax_1 + By_1)$), which represents a straight line, provided that A and B are not both zero (which must be true for $Ax + By + C = 0$ to represent a line).

Now, let's check if the point $(x_1, y_1)$ satisfies this equation. Substitute $x=x_1$ and $y=y_1$ into the equation $A(x - x_1) + B(y - y_1) = 0$:

Left Hand Side (LHS) = $A(x_1 - x_1) + B(y_1 - y_1)$

LHS = $A(0) + B(0)$

LHS = $0 + 0 = 0$

The Right Hand Side (RHS) is 0. Since LHS = RHS, the point $(x_1, y_1)$ satisfies the equation $A(x - x_1) + B(y - y_1) = 0$. Thus, the required line passes through the point $(x_1, y_1)$.

Next, let's determine the slope of the line $A(x - x_1) + B(y - y_1) = 0$. We can rewrite it as $Ax + By - (Ax_1 + By_1) = 0$.

The given line is $Ax + By + C = 0$.

If $B \neq 0$, the slope of the line $Ax + By + C = 0$ is $m_1 = -\frac{A}{B}$.

If $B \neq 0$, the slope of the line $A(x - x_1) + B(y - y_1) = 0$ (or $Ax + By - (Ax_1 + By_1) = 0$) is $m_2 = -\frac{A}{B}$.

Since $m_1 = m_2$, the two lines are parallel when $B \neq 0$.

If $B = 0$, then since A and B are not both zero for $Ax + By + C = 0$ to be a line, A must be non-zero ($A \neq 0$).

In this case, the given line $Ax + By + C = 0$ becomes $Ax + C = 0$, which is $x = -\frac{C}{A}$. This is a vertical line.

The required line is parallel to this vertical line and passes through $(x_1, y_1)$. A line parallel to a vertical line is also a vertical line. The equation of a vertical line passing through $(x_1, y_1)$ is $x = x_1$.

Let's check if the equation $A(x - x_1) + B(y - y_1) = 0$ simplifies to $x = x_1$ when $B=0$ and $A \neq 0$.

Substitute $B=0$ into $A(x - x_1) + B(y - y_1) = 0$:

$A(x - x_1) + 0(y - y_1) = 0$

$A(x - x_1) = 0$

Since $A \neq 0$, we can divide by A:

$x - x_1 = 0$

$x = x_1$

This matches the equation of the vertical line passing through $(x_1, y_1)$, which is parallel to $Ax + C = 0$.

In both cases ($B \neq 0$ and $B = 0, A \neq 0$), the equation $A(x - x_1) + B(y - y_1) = 0$ represents a line that passes through $(x_1, y_1)$ and is parallel to $Ax + By + C = 0$. Since a line is uniquely defined by these properties, this is the equation of the required line.

Conclusion:

The equation of the line through the point (x₁ , y₁ ) and parallel to the line Ax + By + C = 0 is indeed $\mathbf{A (x – x_1) + B (y – y_1) = 0}$.

Given:

Both lines pass through the point $(2, 3)$.

The angle between the two lines is $60^\circ$.

The slope of one line is $m_1 = 2$.

To Find:

The equation of the other line.

Solution:

Let the slope of the other line be $m_2$. The angle $\theta$ between two lines with slopes $m_1$ and $m_2$ is given by:

$\tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right|$

We are given $\theta = 60^\circ$ and $m_1 = 2$. We know that $\tan(60^\circ) = \sqrt{3}$.

Substitute these values into the formula:

$\sqrt{3} = \left| \frac{m_2 - 2}{1 + 2 m_2} \right|$

To remove the absolute value, we consider two cases:

Case 1: $\frac{m_2 - 2}{1 + 2 m_2} = \sqrt{3}$

Multiply both sides by $1 + 2 m_2$ (assuming $1 + 2 m_2 \neq 0$):

$m_2 - 2 = \sqrt{3}(1 + 2 m_2)$

$m_2 - 2 = \sqrt{3} + 2\sqrt{3} m_2$

Gather terms involving $m_2$ on one side and constants on the other:

$m_2 - 2\sqrt{3} m_2 = \sqrt{3} + 2$

$m_2(1 - 2\sqrt{3}) = \sqrt{3} + 2$

$m_2 = \frac{\sqrt{3} + 2}{1 - 2\sqrt{3}}$

To rationalize the denominator, multiply the numerator and denominator by the conjugate of the denominator, $1 + 2\sqrt{3}$:

$m_2 = \frac{\sqrt{3} + 2}{1 - 2\sqrt{3}} \times \frac{1 + 2\sqrt{3}}{1 + 2\sqrt{3}} = \frac{(\sqrt{3})(1) + (\sqrt{3})(2\sqrt{3}) + (2)(1) + (2)(2\sqrt{3})}{1^2 - (2\sqrt{3})^2}$

$m_2 = \frac{\sqrt{3} + 2(3) + 2 + 4\sqrt{3}}{1 - 4(3)} = \frac{\sqrt{3} + 6 + 2 + 4\sqrt{3}}{1 - 12} = \frac{8 + 5\sqrt{3}}{-11}$

$m_2 = -\frac{8 + 5\sqrt{3}}{11}$

This value satisfies $1 + 2 m_2 \neq 0$, so it's a valid slope.

Case 2: $\frac{m_2 - 2}{1 + 2 m_2} = -\sqrt{3}$

Multiply both sides by $1 + 2 m_2$ (assuming $1 + 2 m_2 \neq 0$):

$m_2 - 2 = -\sqrt{3}(1 + 2 m_2)$

$m_2 - 2 = -\sqrt{3} - 2\sqrt{3} m_2$

Gather terms involving $m_2$ on one side and constants on the other:

$m_2 + 2\sqrt{3} m_2 = 2 - \sqrt{3}$}

$m_2(1 + 2\sqrt{3}) = 2 - \sqrt{3}$}

$m_2 = \frac{2 - \sqrt{3}}{1 + 2\sqrt{3}}$

To rationalize the denominator, multiply the numerator and denominator by the conjugate of the denominator, $1 - 2\sqrt{3}$:

$m_2 = \frac{2 - \sqrt{3}}{1 + 2\sqrt{3}} \times \frac{1 - 2\sqrt{3}}{1 - 2\sqrt{3}} = \frac{(2)(1) + (2)(-2\sqrt{3}) + (-\sqrt{3})(1) + (-\sqrt{3})(-2\sqrt{3})}{1^2 - (2\sqrt{3})^2}$

$m_2 = \frac{2 - 4\sqrt{3} - \sqrt{3} + 2(3)}{1 - 4(3)} = \frac{2 - 5\sqrt{3} + 6}{1 - 12} = \frac{8 - 5\sqrt{3}}{-11}$}

$m_2 = -\frac{8 - 5\sqrt{3}}{11} = \frac{5\sqrt{3} - 8}{11}$

This value satisfies $1 + 2 m_2 \neq 0$, so it's a valid slope.

We have two possible slopes for the other line: $m_2 = -\frac{8 + 5\sqrt{3}}{11}$ or $m_2 = \frac{5\sqrt{3} - 8}{11}$. Both lines pass through the point $(2, 3)$. We can use the point-slope form $y - y_1 = m(x - x_1)$ to find the equations.

Equation for the first possible slope: $m_2 = -\frac{8 + 5\sqrt{3}}{11}$, point $(2, 3)$.

$y - 3 = -\frac{8 + 5\sqrt{3}}{11}(x - 2)$

Multiply by 11:

$11(y - 3) = -(8 + 5\sqrt{3})(x - 2)$

$11y - 33 = -(8x - 16 + 5\sqrt{3}x - 10\sqrt{3})$

$11y - 33 = -8x + 16 - 5\sqrt{3}x + 10\sqrt{3}$

Move all terms to one side:

$(8 + 5\sqrt{3})x + 11y - 33 - 16 - 10\sqrt{3} = 0$

$(8 + 5\sqrt{3})x + 11y - (49 + 10\sqrt{3}) = 0$

Equation for the second possible slope: $m_2 = \frac{5\sqrt{3} - 8}{11}$, point $(2, 3)$.

$y - 3 = \frac{5\sqrt{3} - 8}{11}(x - 2)$

Multiply by 11:

$11(y - 3) = (5\sqrt{3} - 8)(x - 2)$

$11y - 33 = (5\sqrt{3})x - 10\sqrt{3} - 8x + 16$

$11y - 33 = (5\sqrt{3} - 8)x - 10\sqrt{3} + 16$

Move all terms to one side:

$(8 - 5\sqrt{3})x + 11y - 33 - 16 + 10\sqrt{3} = 0$

$(8 - 5\sqrt{3})x + 11y - (49 - 10\sqrt{3}) = 0$

Answer:

The equation of the other line is $\mathbf{(8 + 5\sqrt{3})x + 11y - (49 + 10\sqrt{3}) = 0}$ or $\mathbf{(8 - 5\sqrt{3})x + 11y - (49 - 10\sqrt{3}) = 0}$.

Given:

The line segment joins the points $A(3, 4)$ and $B(-1, 2)$.

To Find:

The equation of the right bisector of the line segment AB.

Solution:

A right bisector of a line segment is a line that is perpendicular to the segment and passes through its midpoint.

Step 1: Find the midpoint of the line segment AB.

Let the midpoint of AB be $M(x_m, y_m)$. The formula for the midpoint of a segment with endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $x_m = \frac{x_1 + x_2}{2}$ and $y_m = \frac{y_1 + y_2}{2}$.

Using points $A(3, 4)$ and $B(-1, 2)$:

$x_m = \frac{3 + (-1)}{2} = \frac{2}{2} = 1$

$y_m = \frac{4 + 2}{2} = \frac{6}{2} = 3$

The midpoint of the line segment AB is $M(1, 3)$. The right bisector passes through this point.

Step 2: Find the slope of the line segment AB.

The formula for the slope ($m$) of a line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$ is $m = \frac{y_2 - y_1}{x_2 - x_1}$.

Using points $A(3, 4)$ and $B(-1, 2)$:

Slope of AB ($m_{AB}$) $= \frac{2 - 4}{-1 - 3} = \frac{-2}{-4} = \frac{1}{2}$

Step 3: Find the slope of the right bisector.

The right bisector is perpendicular to the line segment AB. If two non-vertical lines are perpendicular, the product of their slopes is $-1$. Let the slope of the right bisector be $m_{\text{bisector}}$.

$m_{AB} \times m_{\text{bisector}} = -1$

$\frac{1}{2} \times m_{\text{bisector}} = -1$

Multiply both sides by 2:

$m_{\text{bisector}} = -2$

The slope of the right bisector is $-2$.

Step 4: Find the equation of the right bisector.

We have the slope of the right bisector ($m = -2$) and a point it passes through ($M(1, 3)$). We can use the point-slope form of the equation of a line:

$y - y_m = m(x - x_m)$

Substitute the coordinates of the midpoint $M(1, 3)$ and the slope $m = -2$:

$y - 3 = -2(x - 1)$

This is the equation of the line in point-slope form. We can expand it and rearrange it into the general form $Ax + By + C = 0$:

$y - 3 = -2x + 2$}

Move all terms to one side:

$2x + y - 3 - 2 = 0$

$2x + y - 5 = 0$

Answer:

The equation of the right bisector of the line segment joining the points (3, 4) and (–1, 2) is $\mathbf{2x + y - 5 = 0}$.

Given:

Point $P(x_0, y_0) = (-1, 3)$.

Line $L_1$: $3x - 4y - 16 = 0$.

To Find:

The coordinates of the foot of the perpendicular from the point P to the line $L_1$. Let the foot of the perpendicular be point $Q(x, y)$.

Solution:

The foot of the perpendicular from point P to line $L_1$ is the point of intersection of line $L_1$ and the line ($L_2$) that passes through P and is perpendicular to $L_1$.

Step 1: Find the equation of the line ($L_2$) perpendicular to $L_1$ and passing through P.

First, we find the slope of the given line $L_1: 3x - 4y - 16 = 0$. We can rewrite this equation in the slope-intercept form $y = mx + c$ to find its slope.

$3x - 16 = 4y$

Divide by 4:

The slope of line $L_1$ is $m_1 = \frac{3}{4}$.

The required line $L_2$ is perpendicular to $L_1$. If two non-vertical lines are perpendicular, the product of their slopes is $-1$. Let the slope of $L_2$ be $m_2$.

$m_1 \times m_2 = -1$

$\frac{3}{4} \times m_2 = -1$

Multiply both sides by $\frac{4}{3}$:

Now we have the slope of line $L_2$ ($m_2 = -\frac{4}{3}$) and a point it passes through ($P(-1, 3)$). We can use the point-slope form of the equation of a line, $y - y_0 = m(x - x_0)$.

Substitute the values:

$y - 3 = -\frac{4}{3}(x - (-1))$

$y - 3 = -\frac{4}{3}(x + 1)$

To eliminate the fraction, multiply both sides by 3:

$3(y - 3) = -4(x + 1)$

$3y - 9 = -4x - 4$

Rearrange the equation into the general form $Ax + By + C = 0$:

Step 2: Find the point of intersection of $L_1$ and $L_2$.

The foot of the perpendicular $Q(x, y)$ is the solution to the system of equations formed by $L_1$ and $L_2$:

We can use the elimination method to solve this system. Multiply the first equation by 3 and the second equation by 4 to make the coefficients of $y$ equal and opposite:

Add equations (a) and (b):

$(9x - 12y) + (16x + 12y) = 48 + 20$

$25x = 68$

$x = \frac{68}{25}$

Now substitute the value of $x$ into either of the original equations to find $y$. Using $4x + 3y = 5$:

$4\left(\frac{68}{25}\right) + 3y = 5$

$\frac{272}{25} + 3y = 5$

$3y = 5 - \frac{272}{25}$

$3y = \frac{5 \times 25 - 272}{25} = \frac{125 - 272}{25} = \frac{-147}{25}$

$y = \frac{-147}{25 \times 3} = \frac{-\cancel{147}^{49}}{25 \times \cancel{3}_1} = -\frac{49}{25}$

The coordinates of the foot of the perpendicular are $\left(\frac{68}{25}, -\frac{49}{25}\right)$.

Answer:

The coordinates of the foot of perpendicular from the point (–1, 3) to the line 3x – 4y – 16 = 0 are $\mathbf{\left(\frac{68}{25}, -\frac{49}{25}\right)}$.

Given:

A line with the equation $y = mx + c$.

The perpendicular from the origin $O(0, 0)$ to this line meets it at the point $P(-1, 2)$.

To Find:

The values of $m$ and $c$.

Solution:

The point $P(-1, 2)$ is the foot of the perpendicular from the origin to the line $y = mx + c$. This means the line segment OP (joining the origin O(0, 0) and the point P(-1, 2)) is perpendicular to the line $y = mx + c$ at point P.

Step 1: Find the slope of the line segment OP.

The formula for the slope ($m'$) of a line segment passing through two points $(x_1, y_1)$ and $(x_2, y_2)$ is $m' = \frac{y_2 - y_1}{x_2 - x_1}$.

Using the coordinates of the origin $O(0, 0)$ and the point $P(-1, 2)$:

Slope of OP ($m_{OP}$) $= \frac{2 - 0}{-1 - 0} = \frac{2}{-1} = -2$

Step 2: Use the perpendicularity condition to find the slope of the line $y = mx + c$.

The line $y = mx + c$ has a slope of $m$. Since the line segment OP is perpendicular to the line $y = mx + c$, the product of their slopes is $-1$ (provided neither line is vertical or horizontal. The slope of OP is -2, which is neither undefined nor 0, so this condition applies).

$m_{OP} \times m = -1$

Substitute the value of $m_{OP}$:

$-2 \times m = -1$

Divide both sides by -2:

Step 3: Use the fact that P(-1, 2) lies on the line $y = mx + c$ to find the value of c.

Since the point $P(-1, 2)$ is on the line $y = mx + c$, its coordinates must satisfy the equation of the line. Substitute $x = -1$, $y = 2$, and the value of $m = \frac{1}{2}$ into the equation:

$2 = \left(\frac{1}{2}\right)(-1) + c$

$2 = -\frac{1}{2} + c$

Add $\frac{1}{2}$ to both sides to solve for $c$:

$c = 2 + \frac{1}{2}$

$c = \frac{4}{2} + \frac{1}{2}$

We have found the values of $m$ and $c$.

Answer:

The values are $\mathbf{m = \frac{1}{2}}$ and $\mathbf{c = \frac{5}{2}}$.

Given:

Line 1: $x \cos \theta - y \sin \theta = k \cos 2\theta$

Line 2: $x \sec \theta + y \text{cosec} \theta = k$

p is the length of the perpendicular from the origin (0, 0) to Line 1.

q is the length of the perpendicular from the origin (0, 0) to Line 2.

To Prove:

$p^2 + 4q^2 = k^2$

Solution:

The formula for the distance $d$ of a point $(x_0, y_0)$ from a line $Ax + By + C = 0$ is $d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$.

The origin is the point $(x_0, y_0) = (0, 0)$.

For the perpendicular length p from the origin to Line 1:

Line 1 is $x \cos \theta - y \sin \theta = k \cos 2\theta$.

Rewrite in the form $Ax + By + C = 0$:

Here, $A = \cos \theta$, $B = -\sin \theta$, $C = -k \cos 2\theta$, and $(x_0, y_0) = (0, 0)$.

Using the distance formula:

$p = \frac{|(\cos \theta)(0) + (-\sin \theta)(0) + (-k \cos 2\theta)|}{\sqrt{(\cos \theta)^2 + (-\sin \theta)^2}}$

$p = \frac{|-k \cos 2\theta|}{\sqrt{\cos^2 \theta + \sin^2 \theta}}$

Using the identity $\cos^2 \theta + \sin^2 \theta = 1$:

$p = \frac{|-k \cos 2\theta|}{\sqrt{1}} = |-k \cos 2\theta| = |k \cos 2\theta|$

Square both sides to find $p^2$:

For the perpendicular length q from the origin to Line 2:

Line 2 is $x \sec \theta + y \text{cosec} \theta = k$.

Rewrite in the form $Ax + By + C = 0$:

Here, $A = \sec \theta$, $B = \text{cosec} \theta$, $C = -k$, and $(x_0, y_0) = (0, 0)$.

Using the distance formula:

$q = \frac{|(\sec \theta)(0) + (\text{cosec} \theta)(0) + (-k)|}{\sqrt{(\sec \theta)^2 + (\text{cosec} \theta)^2}}$

$q = \frac{|-k|}{\sqrt{\sec^2 \theta + \text{cosec}^2 \theta}} = \frac{|k|}{\sqrt{\sec^2 \theta + \text{cosec}^2 \theta}}$

Square both sides to find $q^2$:

$q^2 = \frac{k^2}{\sec^2 \theta + \text{cosec}^2 \theta}$

Now, simplify the denominator using reciprocal identities ($\sec \theta = \frac{1}{\cos \theta}$, $\text{cosec} \theta = \frac{1}{\sin \theta}$) and the Pythagorean identity ($\sin^2 \theta + \cos^2 \theta = 1$):

$\sec^2 \theta + \text{cosec}^2 \theta = \frac{1}{\cos^2 \theta} + \frac{1}{\sin^2 \theta} = \frac{\sin^2 \theta + \cos^2 \theta}{\cos^2 \theta \sin^2 \theta} = \frac{1}{\cos^2 \theta \sin^2 \theta}$

Substitute this back into the expression for $q^2$:

$q^2 = \frac{k^2}{\frac{1}{\cos^2 \theta \sin^2 \theta}} = k^2 \cos^2 \theta \sin^2 \theta$

Using the double angle identity for sine, $\sin 2\theta = 2 \sin \theta \cos \theta$, we have $\sin^2 2\theta = 4 \sin^2 \theta \cos^2 \theta$.

So, $\sin^2 \theta \cos^2 \theta = \frac{1}{4} \sin^2 2\theta$.

Substitute this into the expression for $q^2$:

$q^2 = k^2 \left(\frac{1}{4} \sin^2 2\theta\right) = \frac{k^2}{4} \sin^2 2\theta$

Now, calculate $4q^2$:

Evaluate p$^2$ + 4q$^2$:

Add the expressions for $p^2$ from (i) and $4q^2$ from (ii):

$p^2 + 4q^2 = k^2 \cos^2 2\theta + k^2 \sin^2 2\theta$

Factor out $k^2$:

$p^2 + 4q^2 = k^2 (\cos^2 2\theta + \sin^2 2\theta)$

Using the Pythagorean identity $\cos^2 A + \sin^2 A = 1$ (with $A = 2\theta$):

$p^2 + 4q^2 = k^2 (1)$

$p^2 + 4q^2 = k^2$

This matches the required result.

Conclusion:

Based on the lengths of the perpendiculars from the origin to the given lines, we have proven that $p^2 + 4q^2 = k^2$.

Given:

The vertices of triangle ABC are $A(2, 3)$, $B(4, -1)$, and $C(1, 2)$.

To Find:

1. The equation of the altitude from vertex A.

2. The length of the altitude from vertex A.

Solution:

An altitude from a vertex of a triangle is the perpendicular line segment from the vertex to the opposite side. The altitude from vertex A is perpendicular to the side BC.

Part 1: Equation of the altitude from vertex A

The altitude from A is a line passing through A(2, 3) and is perpendicular to the line segment BC.

First, find the slope of the line segment BC. The formula for the slope ($m$) of a line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$ is $m = \frac{y_2 - y_1}{x_2 - x_1}$.

Using the coordinates of B(4, -1) and C(1, 2):

Slope of BC ($m_{BC}$) $= \frac{2 - (-1)}{1 - 4} = \frac{2 + 1}{-3} = \frac{3}{-3} = -1$

The altitude from A is perpendicular to BC. If two non-vertical lines are perpendicular, the product of their slopes is $-1$. Let the slope of the altitude from A be $m_{alt}$.

$m_{BC} \times m_{alt} = -1$

Substitute the value of $m_{BC}$:

$(-1) \times m_{alt} = -1$

$m_{alt} = 1$

The slope of the altitude from A is 1.

Now we have the slope of the altitude ($m = 1$) and a point it passes through ($A(2, 3)$). We can use the point-slope form of the equation of a line, $y - y_1 = m(x - x_1)$.

Substitute the values:

$y - 3 = 1(x - 2)$

$y - 3 = x - 2$

Rearrange the equation into the general form $Ax + By + C = 0$:

$0 = x - y - 2 + 3$

Equation of the altitude from A: $\mathbf{x - y + 1 = 0}$.

Part 2: Length of the altitude from vertex A

The length of the altitude from vertex A is the perpendicular distance from point A(2, 3) to the line containing side BC.

First, we need the equation of the line containing side BC in the general form $Ax + By + C = 0$. We found the slope of BC is $m_{BC} = -1$. The line BC passes through B(4, -1). Using the point-slope form $y - y_1 = m(x - x_1)$:

$y - (-1) = -1(x - 4)$

$y + 1 = -x + 4$

Rearrange into the general form:

$x + y + 1 - 4 = 0$

Equation of the line BC: $x + y - 3 = 0$.

This is in the form $Ax + By + C = 0$, where $A = 1$, $B = 1$, and $C = -3$.

The point is $A(x_0, y_0) = (2, 3)$.

The distance $d$ of a point $(x_0, y_0)$ from a line $Ax + By + C = 0$ is given by the formula:

$d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$

Substitute the values of the point A(2, 3) and the coefficients of the line BC ($A=1, B=1, C=-3$):

Length of altitude from A $= \frac{|(1)(2) + (1)(3) + (-3)|}{\sqrt{(1)^2 + (1)^2}}$

$= \frac{|2 + 3 - 3|}{\sqrt{1 + 1}}$

$= \frac{|2|}{\sqrt{2}} = \frac{2}{\sqrt{2}}$

Rationalize the denominator:

$\frac{2}{\sqrt{2}} = \frac{2}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{2\sqrt{2}}{2} = \sqrt{2}$

The length of the altitude from vertex A is $\sqrt{2}$ units.

Answer:

The equation of the altitude from vertex A is $\mathbf{x - y + 1 = 0}$.

The length of the altitude from vertex A is $\mathbf{\sqrt{2}}$.

Given:

A line has intercepts $a$ and $b$ on the x-axis and y-axis, respectively.

p is the length of the perpendicular from the origin (0, 0) to this line.

We assume that the line does not pass through the origin, which implies $a \neq 0$ and $b \neq 0$.

To Show:

$\frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2}$

Solution:

The equation of a line with x-intercept $a$ and y-intercept $b$ is given by the intercept form:

To find the distance from the origin, we need to rewrite this equation in the general form $Ax + By + C = 0$.

Find a common denominator on the left side:

$\frac{bx + ay}{ab} = 1$

Multiply both sides by $ab$:

$bx + ay = ab$

Move the constant term to the left side:

Comparing this with the general form $Ax + By + C = 0$, we have $A = b$, $B = a$, and $C = -ab$.

The distance $p$ of the point $(x_0, y_0) = (0, 0)$ (the origin) from the line $Ax + By + C = 0$ is given by the formula:

$p = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$

Substitute the values $A=b$, $B=a$, $C=-ab$, $x_0=0$, and $y_0=0$ into the distance formula:

$p = \frac{|(b)(0) + (a)(0) + (-ab)|}{\sqrt{(b)^2 + (a)^2}}$

$p = \frac{|0 + 0 - ab|}{\sqrt{b^2 + a^2}}$

$p = \frac{|-ab|}{\sqrt{a^2 + b^2}}$

Since $|-ab| = |ab|$, we have:

To prove the desired relationship, let's square both sides of the equation for $p$:

$p^2 = \left(\frac{|ab|}{\sqrt{a^2 + b^2}}\right)^2$

$p^2 = \frac{(ab)^2}{(\sqrt{a^2 + b^2})^2}$

Now, take the reciprocal of $p^2$:

$\frac{1}{p^2} = \frac{1}{\frac{a^2 b^2}{a^2 + b^2}}$

$\frac{1}{p^2} = \frac{a^2 + b^2}{a^2 b^2}$

Separate the terms in the numerator:

$\frac{1}{p^2} = \frac{a^2}{a^2 b^2} + \frac{b^2}{a^2 b^2}$

Simplify each fraction by cancelling common factors (since $a \neq 0$ and $b \neq 0$):

$\frac{1}{p^2} = \frac{\cancel{a^2}}{\cancel{a^2} b^2} + \frac{\cancel{b^2}}{a^2 \cancel{b^2}}$

$\frac{1}{p^2} = \frac{1}{b^2} + \frac{1}{a^2}$

Rearranging the terms on the right side:

$\frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2}$

This matches the required relationship.

Conclusion:

By expressing the equation of the line in general form and applying the distance formula from the origin, we have shown that $\frac{1}{p^{2}}$ = $\frac{1}{a^{2}}$ + $\frac{1}{b^{2}}$.

Given:

Three lines are given by the equations:

The lines are concurrent, meaning they intersect at a single point.

To Find:

The value of $k$.

Solution:

Since the three lines are concurrent, they all pass through the same point. This point is the intersection of any two of the lines. Let's find the intersection point of Line 1 and Line 3.

Rewrite the equations in a standard form, where the constant term is on the right side for solving the system:

For Line 3:

Now, let's solve the system of equations (i) and (iii) using the elimination method. Add equation (i) and equation (iii):

Divide both sides by 5:

Now substitute the value of $x = 1$ into equation (i) to find the value of $y$:

Subtract 2 from both sides:

So, the point of intersection of Line 1 and Line 3 is $(1, 1)$.

Since the three lines are concurrent, this point $(1, 1)$ must also lie on Line 2, whose equation is $5x + ky - 3 = 0$. Substitute the coordinates of this point $(x=1, y=1)$ into the equation of Line 2:

Perform the multiplication:

Combine the constant terms:

Subtract 2 from both sides:

Answer:

The value of $k$ is $\mathbf{-2}$.

Given:

Line 1: $4x - y = 0$.

Point P $(4, 1)$.

A line (Line 2) passes through P(4, 1) and makes an angle of $135^\circ$ with the positive x-axis.

To Find:

The distance of Line 1 from point P, measured along Line 2.

This distance is the length of the segment of Line 2 from point P to the point where Line 2 intersects Line 1.

Solution:

First, find the equation of Line 2.

Line 2 passes through the point $(x_1, y_1) = (4, 1)$.

The inclination of Line 2 is $\theta = 135^\circ$.

The slope of Line 2 is $m_2 = \tan(\theta) = \tan(135^\circ)$.

We know that $\tan(135^\circ) = -1$.

So, the slope of Line 2 is $m_2 = -1$.

Using the point-slope form of the equation of a line, $y - y_1 = m(x - x_1)$:

$y - 1 = -1(x - 4)$

$y - 1 = -x + 4$

$x + y - 5 = 0$

This is the equation of Line 2.

Next, find the point of intersection of Line 1 ($4x - y = 0$) and Line 2 ($x + y - 5 = 0$). Let the point of intersection be $Q(x, y)$.

We have the system of equations:

We can solve this system by elimination. Add equation (i) and equation (ii):

$(4x - y) + (x + y) = 0 + 5$

$5x = 5$

Divide by 5:

$x = 1$

Substitute the value of $x=1$ into equation (ii) to find $y$:

$1 + y = 5$

Subtract 1 from both sides:

$y = 5 - 1 = 4$

The point of intersection is $Q(1, 4)$.

Finally, calculate the distance between point P(4, 1) and point Q(1, 4) using the distance formula:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Using points $P(4, 1)$ and $Q(1, 4)$: $(x_1, y_1) = (4, 1)$, $(x_2, y_2) = (1, 4)$.

$d = \sqrt{(1 - 4)^2 + (4 - 1)^2}$

$d = \sqrt{(-3)^2 + (3)^2}$

$d = \sqrt{9 + 9}$

$d = \sqrt{18}$

Simplify the square root:

$\sqrt{18} = \sqrt{9 \times 2} = \sqrt{9} \times \sqrt{2} = 3\sqrt{2}$

The distance is $3\sqrt{2}$ units.

Answer:

The distance of the line 4x – y = 0 from the point P (4, 1) measured along the line making an angle of 135° with the positive x-axis is $\mathbf{3\sqrt{2}}$.

Given:

Original point $P(1, 2)$.

Line (mirror) $L: x - 3y + 4 = 0$.

To Find:

The image of point P in the line L. Let the image point be $P'(x', y')$.

Solution:

The line L acts as a plane mirror. The image $P'(x', y')$ of a point $P(x, y)$ in a line L has the following properties:

1. The line segment PP' is perpendicular to the line L.

2. The midpoint of the line segment PP' lies on the line L.

Step 1: Use the perpendicularity condition.

First, find the slope of the given line $L: x - 3y + 4 = 0$. We can rewrite this equation in the slope-intercept form $y = mx + c$ to find its slope.

$x + 4 = 3y$

$y = \frac{1}{3}x + \frac{4}{3}$

The slope of the line L is $m_L = \frac{1}{3}$.

The slope of the line segment PP' passing through $P(1, 2)$ and $P'(x', y')$ is $m_{PP'} = \frac{y' - 2}{x' - 1}$.

Since PP' is perpendicular to L, the product of their slopes is $-1$ (as both lines are non-vertical and non-horizontal):

$m_{PP'} \times m_L = -1$

$\frac{y' - 2}{x' - 1} \times \frac{1}{3} = -1$

$\frac{y' - 2}{3(x' - 1)} = -1$

$y' - 2 = -3(x' - 1)$

$y' - 2 = -3x' + 3$

Rearrange to form an equation:

Step 2: Use the midpoint condition.

The midpoint M of the line segment PP' with endpoints $P(1, 2)$ and $P'(x', y')$ has coordinates:

$M\left(\frac{1 + x'}{2}, \frac{2 + y'}{2}\right)$

Since the midpoint M lies on the line $L: x - 3y + 4 = 0$, its coordinates must satisfy the equation of L.

Substitute the coordinates of M into the equation of L:

$\left(\frac{1 + x'}{2}\right) - 3\left(\frac{2 + y'}{2}\right) + 4 = 0$

Multiply the entire equation by 2 to clear the denominators:

$1 + x' - 3(2 + y') + 2(4) = 0$

$1 + x' - 6 - 3y' + 8 = 0$

Combine the constant terms:

$x' - 3y' + 3 = 0$

Rearrange to form an equation:

Step 3: Solve the system of equations.

We have a system of two linear equations with two variables $x'$ and $y'$:

(1) $3x' + y' = 5$

(2) $x' - 3y' = -3$}

From equation (1), express $y'$ in terms of $x'$: $y' = 5 - 3x'$.

Substitute this expression for $y'$ into equation (2):

$x' - 3(5 - 3x') = -3$

$x' - 15 + 9x' = -3$

Combine terms involving $x'$:

$10x' - 15 = -3$

Add 15 to both sides:

$10x' = -3 + 15$

$10x' = 12$

$x' = \frac{12}{10} = \frac{6}{5}$

Now substitute the value of $x' = \frac{6}{5}$ back into the expression for $y'$ ($y' = 5 - 3x'$):

$y' = 5 - 3\left(\frac{6}{5}\right)$

$y' = 5 - \frac{18}{5}$

$y' = \frac{25}{5} - \frac{18}{5}$

$y' = \frac{7}{5}$

The coordinates of the image point are $\left(\frac{6}{5}, \frac{7}{5}\right)$.

Answer:

The image of the point (1, 2) in the line x – 3y + 4 = 0 is $\mathbf{\left(\frac{6}{5}, \frac{7}{5}\right)}$.

Given:

Three lines with equations:

Line 1: $y = m_1x + c_1$

Line 2: $y = m_2x + c_2$

Line 3: $x = 0$ (the y-axis)

We assume $m_1 \neq m_2$ and $c_1 \neq c_2$, so the lines intersect at distinct points and form a triangle.

To Show:

The area of the triangle formed by these lines is $\frac{\left( c_1 \;-\; c_2 \right)^2}{2\left| m_1 \;-\; m_2 \right|}$.

Solution:

To find the area of the triangle, we first find the coordinates of its vertices, which are the points of intersection of the given lines.

Vertex 1: Intersection of Line 1 ($y = m_1x + c_1$) and Line 3 ($x = 0$).

Substitute $x = 0$ into the equation of Line 1:

$y = m_1(0) + c_1$

$y = c_1$

The first vertex is $V_1(0, c_1)$. This is the y-intercept of Line 1.

Vertex 2: Intersection of Line 2 ($y = m_2x + c_2$) and Line 3 ($x = 0$).

Substitute $x = 0$ into the equation of Line 2:

$y = m_2(0) + c_2$

$y = c_2$

The second vertex is $V_2(0, c_2)$. This is the y-intercept of Line 2.

Vertex 3: Intersection of Line 1 ($y = m_1x + c_1$) and Line 2 ($y = m_2x + c_2$).

Equate the expressions for $y$ from both equations:

$m_1x + c_1 = m_2x + c_2$

Rearrange the terms to solve for $x$:

$m_1x - m_2x = c_2 - c_1$

$x(m_1 - m_2) = c_2 - c_1$

Assuming $m_1 \neq m_2$, divide by $(m_1 - m_2)$:

$x = \frac{c_2 - c_1}{m_1 - m_2}$

Substitute this value of $x$ into the equation for Line 1 (or Line 2) to find the y-coordinate:

$y = m_1 \left(\frac{c_2 - c_1}{m_1 - m_2}\right) + c_1$

$y = \frac{m_1(c_2 - c_1) + c_1(m_1 - m_2)}{m_1 - m_2}$

$y = \frac{m_1c_2 - m_1c_1 + c_1m_1 - c_1m_2}{m_1 - m_2}$

$y = \frac{m_1c_2 - c_1m_2}{m_1 - m_2}$

The third vertex is $V_3\left(\frac{c_2 - c_1}{m_1 - m_2}, \frac{m_1c_2 - c_1m_2}{m_1 - m_2}\right)$.

The three vertices of the triangle are $V_1(0, c_1)$, $V_2(0, c_2)$, and $V_3\left(\frac{c_2 - c_1}{m_1 - m_2}, \frac{m_1c_2 - c_1m_2}{m_1 - m_2}\right)$.

The side $V_1V_2$ of the triangle lies on the y-axis ($x=0$). The length of this base is the distance between $(0, c_1)$ and $(0, c_2)$, which is $|c_2 - c_1|$ or $|c_1 - c_2|$.

The height of the triangle corresponding to this base is the perpendicular distance from the third vertex $V_3$ to the line $x=0$ (the y-axis).

The distance from a point $(x_0, y_0)$ to the line $x=0$ is $|x_0|$.

The x-coordinate of $V_3$ is $\frac{c_2 - c_1}{m_1 - m_2}$.

The height $h$ is $\left|\frac{c_2 - c_1}{m_1 - m_2}\right| = \frac{|c_2 - c_1|}{|m_1 - m_2|}$. Since $|c_2 - c_1| = |c_1 - c_2|$, the height is $\frac{|c_1 - c_2|}{|m_1 - m_2|}$.

The area of a triangle is given by $\frac{1}{2} \times \text{base} \times \text{height}$.

Area $= \frac{1}{2} \times |c_1 - c_2| \times \frac{|c_1 - c_2|}{|m_1 - m_2|}$

Area $= \frac{1}{2} \frac{|c_1 - c_2| \times |c_1 - c_2|}{|m_1 - m_2|}$

Area $= \frac{(c_1 - c_2)^2}{2|m_1 - m_2|}$

This is the required formula for the area of the triangle formed by the three lines.

Conclusion:

The area of the triangle formed by the lines y = m₁x + c₁ , y = m₂x + c₂ and x = 0 is shown to be $\mathbf{\frac{\left( c_1 \;-\; c_2 \right)^2}{2\left| m_1 \;-\; m_2 \right|}}$.

Given:

Line 1: $5x - y + 4 = 0$

Line 2: $3x + 4y - 4 = 0$

A line (let's call it L) intersects Line 1 at point A and Line 2 at point B.

The point $P(1, 5)$ is the midpoint of the line segment AB.

To Find:

The equation of line L.

Solution:

Let the points of intersection be $A(x_1, y_1)$ on Line 1 and $B(x_2, y_2)$ on Line 2.

Since $A(x_1, y_1)$ lies on Line 1 ($5x - y + 4 = 0$), its coordinates satisfy the equation:

Since $B(x_2, y_2)$ lies on Line 2 ($3x + 4y - 4 = 0$), its coordinates satisfy the equation:

We are given that $P(1, 5)$ is the midpoint of the line segment AB. Using the midpoint formula, the coordinates of P are:

$1 = \frac{x_1 + x_2}{2} \implies x_1 + x_2 = 2$

$5 = \frac{y_1 + y_2}{2} \implies y_1 + y_2 = 10$

Substitute the expressions for $y_1$ from (1) and $y_2$ from (4) into (4):

$(5x_1 + 4) + y_2 = 10 \implies 5x_1 + y_2 = 6$

Substitute the expressions for $x_2$ from (3) and $y_2$ from (4) into equation (2):

$3(2 - x_1) + 4(10 - y_1) - 4 = 0$

$6 - 3x_1 + 40 - 4y_1 - 4 = 0$

$42 - 3x_1 - 4y_1 = 0$

Now we have a system of two equations for $x_1$ and $y_1$ using (1) and (5):

y$_1$ = 5x$_1$ + 4

3x$_1$ + 4y$_1$ = 42

Substitute the first equation into the second:

$3x_1 + 4(5x_1 + 4) = 42$

$3x_1 + 20x_1 + 16 = 42$

$23x_1 = 42 - 16$

$23x_1 = 26$

$x_1 = \frac{26}{23}$

Now find $y_1$ using equation (1):

$y_1 = 5\left(\frac{26}{23}\right) + 4 = \frac{130}{23} + \frac{4 \times 23}{23} = \frac{130 + 92}{23} = \frac{222}{23}$

So, the point A is $\left(\frac{26}{23}, \frac{222}{23}\right)$.

Line L passes through point $P(1, 5)$ and point $A\left(\frac{26}{23}, \frac{222}{23}\right)$. We can find the equation of Line L using these two points.

Using the two-point form of the equation of a line $\frac{y - y_P}{y_A - y_P} = \frac{x - x_P}{x_A - x_P}$:

$\frac{y - 5}{\frac{222}{23} - 5} = \frac{x - 1}{\frac{26}{23} - 1}$

Simplify the denominators:

$\frac{222}{23} - 5 = \frac{222 - 5 \times 23}{23} = \frac{222 - 115}{23} = \frac{107}{23}$

$\frac{26}{23} - 1 = \frac{26 - 1 \times 23}{23} = \frac{26 - 23}{23} = \frac{3}{23}$

Substitute these simplified denominators back into the two-point formula:

$\frac{y - 5}{\frac{107}{23}} = \frac{x - 1}{\frac{3}{23}}$

Multiply both sides by $\frac{1}{1/23} = 23$:

$\frac{y - 5}{107} = \frac{x - 1}{3}$

Cross-multiply:

$3(y - 5) = 107(x - 1)$

$3y - 15 = 107x - 107$

Rearrange the equation into the general form $Ax + By + C = 0$:

$0 = 107x - 3y - 107 + 15$

$107x - 3y - 92 = 0$

Answer:

The equation of the line is $\mathbf{107x - 3y - 92 = 0}$.

Given:

Line 1: $3x - 2y = 5$, which can be written in the general form as $3x - 2y - 5 = 0$.

Line 2: $3x + 2y = 5$, which can be written in the general form as $3x + 2y - 5 = 0$.

A moving point $P(x, y)$ such that its distance from Line 1 is equal to its distance from Line 2.

To Show:

The path (locus) of the moving point P is a straight line.

Solution:

Let $P(x, y)$ be a moving point such that its distance from Line 1 is equal to its distance from Line 2.

The distance $d$ of a point $(x_0, y_0)$ from a line $Ax + By + C = 0$ is given by the formula:

$d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$

First, find the distance of the point $P(x, y)$ from Line 1 ($3x - 2y - 5 = 0$).

Here, $(x_0, y_0) = (x, y)$, $A_1 = 3$, $B_1 = -2$, $C_1 = -5$.

Distance from P to Line 1 ($d_1$) $= \frac{|3x - 2y - 5|}{\sqrt{3^2 + (-2)^2}} = \frac{|3x - 2y - 5|}{\sqrt{9 + 4}} = \frac{|3x - 2y - 5|}{\sqrt{13}}$

Next, find the distance of the point $P(x, y)$ from Line 2 ($3x + 2y - 5 = 0$).

Here, $(x_0, y_0) = (x, y)$, $A_2 = 3$, $B_2 = 2$, $C_2 = -5$.

Distance from P to Line 2 ($d_2$) $= \frac{|3x + 2y - 5|}{\sqrt{3^2 + 2^2}} = \frac{|3x + 2y - 5|}{\sqrt{9 + 4}} = \frac{|3x + 2y - 5|}{\sqrt{13}}$

According to the problem condition, the distance from P to Line 1 is equal to the distance from P to Line 2 ($d_1 = d_2$).

$\frac{|3x - 2y - 5|}{\sqrt{13}} = \frac{|3x + 2y - 5|}{\sqrt{13}}$

Since the denominators are equal and positive, we can cancel them:

$|3x - 2y - 5| = |3x + 2y - 5|$

The equality of absolute values $|A| = |B|$ implies either $A = B$ or $A = -B$.

Case 1: $3x - 2y - 5 = 3x + 2y - 5$

Subtract $3x$ from both sides: $-2y - 5 = 2y - 5$

Add $5$ to both sides: $-2y = 2y$

Add $2y$ to both sides: $0 = 4y$

Divide by 4:

This is the equation of the x-axis, which is a straight line.

Case 2: $3x - 2y - 5 = -(3x + 2y - 5)$

$3x - 2y - 5 = -3x - 2y + 5$

Add $3x$ to both sides: $6x - 2y - 5 = -2y + 5$

Add $2y$ to both sides: $6x - 5 = 5$

Add $5$ to both sides:

$6x = 10$

Divide by 6:

This is the equation of a vertical line, $x = \frac{5}{3}$, which is also a straight line.

The locus of the moving point P satisfying the given condition is the set of points $(x, y)$ such that $y=0$ or $x=\frac{5}{3}$. Both $y=0$ and $x=\frac{5}{3}$ are equations of straight lines.

The two given lines $3x - 2y - 5 = 0$ and $3x + 2y - 5 = 0$ are not parallel (their slopes are $\frac{3}{2}$ and $-\frac{3}{2}$). They intersect at the point $(\frac{5}{3}, 0)$. The derived lines $y=0$ and $x=\frac{5}{3}$ are the angle bisectors of the angles formed by the intersecting lines. The locus of points equidistant from two intersecting lines is indeed the pair of lines bisecting the angles between them.

Conclusion:

The equations derived for the locus of the point P are $y=0$ and $x=\frac{5}{3}$. Since both of these equations represent straight lines, the path of the moving point is a straight line (or a pair of straight lines, which is still a set of straight lines).

Given the equation of the line is:

$(k - 3) x - (4 - k^2) y + k^2 - 7k + 6 = 0$

This equation is in the form $Ax + By + C = 0$, where:

$A = k - 3$

$B = -(4 - k^2) = k^2 - 4$

$C = k^2 - 7k + 6$

(a) Parallel to the x-axis

A line $Ax + By + C = 0$ is parallel to the x-axis if $A = 0$ and $B \neq 0$.

Setting $A = 0$, we get:

$k - 3 = 0$

Solving for $k$:

$k = 3$

Now we check if $B \neq 0$ for $k = 3$:

$B = k^2 - 4 = (3)^2 - 4 = 9 - 4 = 5$

Since $5 \neq 0$, the condition $B \neq 0$ is satisfied for $k = 3$.

Thus, the line is parallel to the x-axis when $k = 3$.

(b) Parallel to the y-axis

A line $Ax + By + C = 0$ is parallel to the y-axis if $B = 0$ and $A \neq 0$.

Setting $B = 0$, we get:

$k^2 - 4 = 0$

Factoring the quadratic equation:

$(k - 2)(k + 2) = 0$

Solving for $k$:

$k = 2$ or $k = -2$

Now we check if $A \neq 0$ for these values of $k$.

For $k = 2$:

$A = k - 3 = 2 - 3 = -1$

Since $-1 \neq 0$, the condition $A \neq 0$ is satisfied for $k = 2$.

For $k = -2$:

$A = k - 3 = -2 - 3 = -5$

Since $-5 \neq 0$, the condition $A \neq 0$ is satisfied for $k = -2$.

Thus, the line is parallel to the y-axis when $k = 2$ or $k = -2$.

(c) Passing through the origin

A line $Ax + By + C = 0$ passes through the origin $(0, 0)$ if substituting $x = 0$ and $y = 0$ satisfies the equation.

Substituting $(0, 0)$ into the given equation:

$(k - 3) (0) - (4 - k^2) (0) + k^2 - 7k + 6 = 0$

This simplifies to:

$k^2 - 7k + 6 = 0$

Factoring the quadratic equation:

$(k - 1)(k - 6) = 0$

Solving for $k$:

$k = 1$ or $k = 6$

Thus, the line passes through the origin when $k = 1$ or $k = 6$.

Let the x-intercept be $a$ and the y-intercept be $b$.

The equation of a line in intercept form is given by:

$\frac{x}{a} + \frac{y}{b} = 1$

According to the problem, the sum of the intercepts is 1:

$a + b = 1$

... (i)

And the product of the intercepts is -6:

$ab = -6$

... (ii)

From equation (i), we can express $b$ in terms of $a$:

$b = 1 - a$

Substitute this into equation (ii):

$a(1 - a) = -6$

$a - a^2 = -6$

Rearrange the terms to form a quadratic equation:

$a^2 - a - 6 = 0$

Factor the quadratic equation:

$(a - 3)(a + 2) = 0$

This gives two possible values for $a$:

$a = 3$ or $a = -2$

Now, we find the corresponding values for $b$ using $b = 1 - a$ for each value of $a$.

Case 1: When $a = 3$

$b = 1 - 3 = -2$

The intercepts are $a = 3$ and $b = -2$.

Substitute these values into the intercept form $\frac{x}{a} + \frac{y}{b} = 1$:

$\frac{x}{3} + \frac{y}{-2} = 1$

To eliminate the denominators, multiply the entire equation by the LCM of 3 and -2, which is 6:

$6 \left( \frac{x}{3} \right) + 6 \left( \frac{y}{-2} \right) = 6(1)$

$2x - 3y = 6$

Rearrange into the general form $Ax + By + C = 0$:

$2x - 3y - 6 = 0$

Case 2: When $a = -2$

$b = 1 - (-2) = 1 + 2 = 3$

The intercepts are $a = -2$ and $b = 3$.

Substitute these values into the intercept form $\frac{x}{a} + \frac{y}{b} = 1$:

$\frac{x}{-2} + \frac{y}{3} = 1$

To eliminate the denominators, multiply the entire equation by the LCM of -2 and 3, which is 6:

$6 \left( \frac{x}{-2} \right) + 6 \left( \frac{y}{3} \right) = 6(1)$

$-3x + 2y = 6$

Rearrange into the general form $Ax + By + C = 0$:

$-3x + 2y - 6 = 0$

Alternatively, multiply by -1:

$3x - 2y + 6 = 0$

Therefore, the equations of the lines are $2x - 3y - 6 = 0$ and $3x - 2y + 6 = 0$.

Let the equation of the line be $\frac{x}{3} + \frac{y}{4} = 1$.

To rewrite this in the general form $Ax + By + C = 0$, we find a common denominator, which is 12:

$\frac{4x + 3y}{12} = 1$

$4x + 3y = 12$

$4x + 3y - 12 = 0$

Here, $A = 4$, $B = 3$, and $C = -12$.

A point on the y-axis has coordinates of the form $(0, y)$.

The distance $d$ from a point $(x_0, y_0)$ to a line $Ax + By + C = 0$ is given by the formula:

$d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$

We are given that the distance is 4 units, and the point is $(0, y)$. So, $d = 4$, $x_0 = 0$, and $y_0 = y$.

Substitute these values into the distance formula:

$4 = \frac{|4(0) + 3(y) - 12|}{\sqrt{4^2 + 3^2}}$

$4 = \frac{|3y - 12|}{\sqrt{16 + 9}}$

$4 = \frac{|3y - 12|}{\sqrt{25}}$

$4 = \frac{|3y - 12|}{5}$

Multiply both sides by 5:

$20 = |3y - 12|$

This absolute value equation means either $3y - 12 = 20$ or $3y - 12 = -20$.

Case 1: $3y - 12 = 20$

$3y = 20 + 12$

$3y = 32$

$y = \frac{32}{3}$

The point on the y-axis is $\left(0, \frac{32}{3}\right)$.

Case 2: $3y - 12 = -20$

$3y = -20 + 12$

$3y = -8$

$y = -\frac{8}{3}$

The point on the y-axis is $\left(0, -\frac{8}{3}\right)$.

Therefore, the points on the y-axis whose distance from the given line is 4 units are $\left(0, \frac{32}{3}\right)$ and $\left(0, -\frac{8}{3}\right)$.

Let the two given points be $P_1(\cos \theta, \sin \theta)$ and $P_2(\cos \phi, \sin \phi)$.

The equation of a line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by:

$\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}$

Substitute the coordinates of $P_1$ and $P_2$ into the formula:

$\frac{y - \sin \theta}{\sin \phi - \sin \theta} = \frac{x - \cos \theta}{\cos \phi - \cos \theta}$

Rearrange the equation to the general form $Ax + By + C = 0$:

$(y - \sin \theta)(\cos \phi - \cos \theta) = (x - \cos \theta)(\sin \phi - \sin \theta)$

$y(\cos \phi - \cos \theta) - \sin \theta (\cos \phi - \cos \theta) = x(\sin \phi - \sin \theta) - \cos \theta (\sin \phi - \sin \theta)$

$x(\sin \phi - \sin \theta) - y(\cos \phi - \cos \theta) + \sin \theta (\cos \phi - \cos \theta) - \cos \theta (\sin \phi - \sin \theta) = 0$

Let's simplify the constant term:

$C = \sin \theta \cos \phi - \sin \theta \cos \theta - \cos \theta \sin \phi + \cos \theta \sin \theta$

$C = \sin \theta \cos \phi - \cos \theta \sin \phi$

Using the trigonometric identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$, we have:

$C = \sin(\theta - \phi)$

So the equation of the line is:

$(\sin \phi - \sin \theta) x - (\cos \phi - \cos \theta) y + \sin(\theta - \phi) = 0$

This is in the form $Ax + By + C = 0$, where:

$A = \sin \phi - \sin \theta$

$B = -(\cos \phi - \cos \theta) = \cos \theta - \cos \phi$

$C = \sin(\theta - \phi)$

The perpendicular distance $d$ from the origin $(0, 0)$ to the line $Ax + By + C = 0$ is given by the formula:

$d = \frac{|A(0) + B(0) + C|}{\sqrt{A^2 + B^2}} = \frac{|C|}{\sqrt{A^2 + B^2}}$

Let's calculate the denominator $\sqrt{A^2 + B^2}$:

$A^2 + B^2 = (\sin \phi - \sin \theta)^2 + (\cos \theta - \cos \phi)^2$

$A^2 + B^2 = (\sin^2 \phi - 2 \sin \phi \sin \theta + \sin^2 \theta) + (\cos^2 \theta - 2 \cos \theta \cos \phi + \cos^2 \phi)$

Rearrange the terms:

$A^2 + B^2 = (\sin^2 \phi + \cos^2 \phi) + (\sin^2 \theta + \cos^2 \theta) - 2 (\sin \phi \sin \theta + \cos \phi \cos \theta)$

Using the identities $\sin^2 x + \cos^2 x = 1$ and $\cos(X - Y) = \cos X \cos Y + \sin X \sin Y$:

$A^2 + B^2 = 1 + 1 - 2 \cos(\phi - \theta)$

$A^2 + B^2 = 2 - 2 \cos(\phi - \theta)$

Using the half-angle identity $1 - \cos(2x) = 2 \sin^2 x$, we have $1 - \cos(\phi - \theta) = 2 \sin^2 \left(\frac{\phi - \theta}{2}\right)$:

$A^2 + B^2 = 2 \left(1 - \cos(\phi - \theta)\right) = 2 \left(2 \sin^2 \left(\frac{\phi - \theta}{2}\right)\right)$

$A^2 + B^2 = 4 \sin^2 \left(\frac{\phi - \theta}{2}\right)$

So, $\sqrt{A^2 + B^2} = \sqrt{4 \sin^2 \left(\frac{\phi - \theta}{2}\right)} = |2 \sin \left(\frac{\phi - \theta}{2}\right)|$.

Now, let's consider the numerator $|C| = |\sin(\theta - \phi)|$.

Using the identity $\sin(2x) = 2 \sin x \cos x$, we have $\sin(\theta - \phi) = 2 \sin \left(\frac{\theta - \phi}{2}\right) \cos \left(\frac{\theta - \phi}{2}\right)$.

Also, $\sin \left(\frac{\theta - \phi}{2}\right) = -\sin \left(\frac{\phi - \theta}{2}\right)$.

So, $\sin(\theta - \phi) = -2 \sin \left(\frac{\phi - \theta}{2}\right) \cos \left(\frac{\phi - \theta}{2}\right)$.

Thus, $|C| = |-2 \sin \left(\frac{\phi - \theta}{2}\right) \cos \left(\frac{\phi - \theta}{2}\right)| = |2 \sin \left(\frac{\phi - \theta}{2}\right) \cos \left(\frac{\phi - \theta}{2}\right)|$.

Substitute the values of $|C|$ and $\sqrt{A^2 + B^2}$ into the distance formula:

$d = \frac{|2 \sin \left(\frac{\phi - \theta}{2}\right) \cos \left(\frac{\phi - \theta}{2}\right)|}{|2 \sin \left(\frac{\phi - \theta}{2}\right)|}$

Assuming the two points are distinct, $\phi - \theta \neq 2n\pi$ for any integer $n$. This implies $\frac{\phi - \theta}{2} \neq n\pi$, so $\sin \left(\frac{\phi - \theta}{2}\right) \neq 0$. We can cancel the $|2 \sin \left(\frac{\phi - \theta}{2}\right)|$ terms.

$d = |\cos \left(\frac{\phi - \theta}{2}\right)|$

The perpendicular distance from the origin to the line is $|\cos \left(\frac{\phi - \theta}{2}\right)|$.

We are given two lines:

Line 1: $x - 7y + 5 = 0$

Line 2: $3x + y = 0$

We need to find the equation of a line that passes through the point of intersection of these two lines and is parallel to the y-axis.

First, let's find the point of intersection of the two lines by solving the system of equations:

$x - 7y = -5$ (1)

$3x + y = 0$ (2)

From equation (2), we can express $y$ in terms of $x$:

$y = -3x$

Substitute this expression for $y$ into equation (1):

$x - 7(-3x) = -5$

$x + 21x = -5$

$22x = -5$

$x = -\frac{5}{22}$

Now substitute the value of $x$ back into $y = -3x$ to find $y$:

$y = -3 \left(-\frac{5}{22}\right)$

$y = \frac{15}{22}$

The point of intersection of the two lines is $\left(-\frac{5}{22}, \frac{15}{22}\right)$.

A line parallel to the y-axis has an equation of the form $x = c$, where $c$ is a constant.

Since the required line passes through the point of intersection $\left(-\frac{5}{22}, \frac{15}{22}\right)$, the x-coordinate of this point must satisfy the equation of the line.

Therefore, $c = -\frac{5}{22}$.

The equation of the line parallel to the y-axis and passing through the point of intersection is:

$x = -\frac{5}{22}$

This can also be written as:

$x + \frac{5}{22} = 0$

Multiplying by 22 to remove the fraction, we get:

$22x + 5 = 0$

The equation of the line is $22x + 5 = 0$.

The given equation of the line is $\frac{x}{4} + \frac{y}{6} = 1$.

To find the point where this line meets the y-axis, we set $x = 0$ in the equation.

$\frac{0}{4} + \frac{y}{6} = 1$

$0 + \frac{y}{6} = 1$

$\frac{y}{6} = 1$

$y = 6$

The point where the line meets the y-axis is $(0, 6)$.

Now, let's find the slope of the given line.

The equation $\frac{x}{4} + \frac{y}{6} = 1$ can be rewritten in the slope-intercept form $y = mx + c$.

Multiply the equation by the LCM of 4 and 6, which is 12:

$12 \left(\frac{x}{4} + \frac{y}{6}\right) = 12(1)$

$3x + 2y = 12$

Solve for $y$:

$2y = -3x + 12$

$y = -\frac{3}{2}x + \frac{12}{2}$

$y = -\frac{3}{2}x + 6$

The slope of the given line is $m_1 = -\frac{3}{2}$.

The required line is perpendicular to the given line.

If two lines are perpendicular, the product of their slopes is -1.

Let the slope of the required line be $m_2$.

$m_1 \times m_2 = -1$

$-\frac{3}{2} \times m_2 = -1$

$m_2 = \frac{-1}{-\frac{3}{2}} = \frac{2}{3}$

The slope of the required line is $m_2 = \frac{2}{3}$.

The required line passes through the point $(0, 6)$ and has a slope of $\frac{2}{3}$.

Using the point-slope form of a linear equation $y - y_1 = m(x - x_1)$:

Substitute $(x_1, y_1) = (0, 6)$ and $m = \frac{2}{3}$:

$y - 6 = \frac{2}{3}(x - 0)$

$y - 6 = \frac{2}{3}x$

To eliminate the fraction, multiply both sides by 3:

$3(y - 6) = 2x$

$3y - 18 = 2x$

Rearrange the terms to get the equation in general form $Ax + By + C = 0$:

$2x - 3y + 18 = 0$

The equation of the line perpendicular to $\frac{x}{4} + \frac{y}{6} = 1$ and passing through the point where it meets the y-axis is $2x - 3y + 18 = 0$.

The equations of the three given lines are:

Line 1: $y - x = 0 \implies y = x$

Line 2: $x + y = 0 \implies y = -x$

Line 3: $x - k = 0 \implies x = k$

To find the vertices of the triangle, we find the points of intersection of these lines.

Intersection of Line 1 ($y = x$) and Line 2 ($y = -x$):

Set $x = -x$

$2x = 0$

$x = 0$

Since $y = x$, $y = 0$.

The first vertex is $V_1 = (0, 0)$.

Intersection of Line 1 ($y = x$) and Line 3 ($x = k$):

Substitute $x = k$ into $y = x$.

$y = k$

The second vertex is $V_2 = (k, k)$.

Intersection of Line 2 ($y = -x$) and Line 3 ($x = k$):

Substitute $x = k$ into $y = -x$.

$y = -k$

The third vertex is $V_3 = (k, -k)$.

The vertices of the triangle are $(0, 0)$, $(k, k)$, and $(k, -k)$.

We can calculate the area of the triangle using the coordinates of its vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ using the formula:

Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$

Let $(x_1, y_1) = (0, 0)$, $(x_2, y_2) = (k, k)$, and $(x_3, y_3) = (k, -k)$.

Area $= \frac{1}{2} |0(k - (-k)) + k(-k - 0) + k(0 - k)|$

Area $= \frac{1}{2} |0(2k) + k(-k) + k(-k)|$

Area $= \frac{1}{2} |0 - k^2 - k^2|$

Area $= \frac{1}{2} |-2k^2|$

Area $= \frac{1}{2} |2k^2|$

Since $k^2 \geq 0$, $|2k^2| = 2k^2$.

Area $= \frac{1}{2} \times 2k^2$

Area $= k^2$

Alternatively, we can find the base and height of the triangle.

Let the base of the triangle be the segment connecting $V_2(k, k)$ and $V_3(k, -k)$. Both points have the same x-coordinate $k$, so the base is a vertical segment.

The length of the base is the distance between $(k, k)$ and $(k, -k)$: $|k - (-k)| = |2k|$.

The height of the triangle is the perpendicular distance from the third vertex $V_1(0, 0)$ to the line containing the base, which is the line $x = k$ (or $x - k = 0$).

The distance from a point $(x_0, y_0)$ to the line $Ax + By + C = 0$ is $\frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$.

For the line $x - k = 0$, $A=1$, $B=0$, $C=-k$. The point is $(0, 0)$.

Height $= \frac{|1(0) + 0(0) - k|}{\sqrt{1^2 + 0^2}} = \frac{|-k|}{\sqrt{1}} = |-k| = |k|$.

Area $= \frac{1}{2} \times \text{base} \times \text{height}$

Area $= \frac{1}{2} \times |2k| \times |k|$

Area $= \frac{1}{2} \times 2|k| \times |k|$

Area $= |k|^2 = k^2$

The area of the triangle formed by the given lines is $k^2$ square units.

Let the three given lines be:

$L_1: 3x + y - 2 = 0$

$L_2: px + 2y - 3 = 0$

$L_3: 2x - y - 3 = 0$

For the three lines to intersect at one point, the point of intersection of any two lines must lie on the third line.

Let's find the point of intersection of lines $L_1$ and $L_3$. We can write the equations as:

Add equation (1) and equation (2):

$(3x + y) + (2x - y) = 2 + 3$

$5x = 5$

Solving for $x$:

$x = \frac{5}{5}$

$x = 1$

Substitute the value of $x = 1$ into equation (1):

$3(1) + y = 2$

$3 + y = 2$

Solving for $y$:

$y = 2 - 3$

$y = -1$

The point of intersection of lines $L_1$ and $L_3$ is $(1, -1)$.

For the three lines to intersect at one point, the point $(1, -1)$ must satisfy the equation of line $L_2$ ($px + 2y - 3 = 0$).

Substitute $x = 1$ and $y = -1$ into the equation of $L_2$:

$p(1) + 2(-1) - 3 = 0$

$p - 2 - 3 = 0$

$p - 5 = 0$

Solving for $p$:

$p = 5$

Thus, the value of $p$ for which the three lines intersect at one point is $p = 5$.

Given the equations of the three lines:

$L_1: y = m_1x + c_1$

$L_2: y = m_2x + c_2$

$L_3: y = m_3x + c_3$

We can rewrite these equations in the general form $Ax + By + C = 0$:

$L_1: m_1x - y + c_1 = 0$

$L_2: m_2x - y + c_2 = 0$

$L_3: m_3x - y + c_3 = 0$

Three lines $A_1x + B_1y + C_1 = 0$, $A_2x + B_2y + C_2 = 0$, and $A_3x + B_3y + C_3 = 0$ are concurrent if and only if the determinant of their coefficients is zero.

The condition for the concurrency of the three given lines is:

$\begin{vmatrix} m_1 & -1 & c_1 \\ m_2 & -1 & c_2 \\ m_3 & -1 & c_3 \end{vmatrix} = 0$

Expand the determinant along the first row:

$m_1 \begin{vmatrix} -1 & c_2 \\ -1 & c_3 \end{vmatrix} - (-1) \begin{vmatrix} m_2 & c_2 \\ m_3 & c_3 \end{vmatrix} + c_1 \begin{vmatrix} m_2 & -1 \\ m_3 & -1 \end{vmatrix} = 0$

Evaluate the $2 \times 2$ determinants:

$m_1((-1)(c_3) - (c_2)(-1)) + 1((m_2)(c_3) - (c_2)(m_3)) + c_1((m_2)(-1) - (-1)(m_3)) = 0$

$m_1(-c_3 + c_2) + (m_2c_3 - m_3c_2) + c_1(-m_2 + m_3) = 0$

Rearrange the terms to match the required expression:

$m_1(c_2 - c_3) + m_2c_3 - m_3c_2 - c_1m_2 + c_1m_3 = 0$

$m_1(c_2 - c_3) + (m_2c_3 - c_1m_2) + (c_1m_3 - m_3c_2) = 0$

$m_1(c_2 - c_3) + m_2(c_3 - c_1) + m_3(c_1 - c_2) = 0$

Thus, if the three lines are concurrent, then the condition $m_1(c_2 – c_3) + m_2(c_3 – c_1) + m_3(c_1 – c_2) = 0$ holds.

The given point is $(x_0, y_0) = (3, 2)$.

The given line is $x - 2y = 3$.

First, find the slope of the given line. Rewrite the equation in slope-intercept form $y = mx + c$:

$x - 2y = 3$

$-2y = -x + 3$

$y = \frac{-x}{-2} + \frac{3}{-2}$

$y = \frac{1}{2}x - \frac{3}{2}$

The slope of the given line is $m_1 = \frac{1}{2}$.

Let $m_2$ be the slope of the required line.

The angle between the required line and the given line is $\theta = 45^\circ$.

The formula for the tangent of the angle between two lines with slopes $m_1$ and $m_2$ is:

$\tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right|$

Substitute $\theta = 45^\circ$ and $m_1 = \frac{1}{2}$:

$\tan 45^\circ = \left| \frac{m_2 - \frac{1}{2}}{1 + \frac{1}{2} m_2} \right|$

$1 = \left| \frac{m_2 - \frac{1}{2}}{1 + \frac{1}{2} m_2} \right|$

This gives two cases:

Case 1: $\frac{m_2 - \frac{1}{2}}{1 + \frac{1}{2} m_2} = 1$

$m_2 - \frac{1}{2} = 1 + \frac{1}{2} m_2$

Multiply by 2 to clear fractions:

$2m_2 - 1 = 2 + m_2$

$2m_2 - m_2 = 2 + 1$

$m_2 = 3$

Using the point-slope form $y - y_0 = m(x - x_0)$ with $(x_0, y_0) = (3, 2)$ and $m_2 = 3$:

$y - 2 = 3(x - 3)$

$y - 2 = 3x - 9$

Rearrange to the general form:

$3x - y - 9 + 2 = 0$

Equation of the first line: $3x - y - 7 = 0$

Case 2: $\frac{m_2 - \frac{1}{2}}{1 + \frac{1}{2} m_2} = -1$

$m_2 - \frac{1}{2} = -1 \left(1 + \frac{1}{2} m_2\right)$

$m_2 - \frac{1}{2} = -1 - \frac{1}{2} m_2$

Multiply by 2 to clear fractions:

$2m_2 - 1 = -2 - m_2$

$2m_2 + m_2 = -2 + 1$

$3m_2 = -1$

$m_2 = -\frac{1}{3}$

Using the point-slope form $y - y_0 = m(x - x_0)$ with $(x_0, y_0) = (3, 2)$ and $m_2 = -\frac{1}{3}$:

$y - 2 = -\frac{1}{3}(x - 3)$

Multiply by 3 to clear the fraction:

$3(y - 2) = -(x - 3)$

$3y - 6 = -x + 3$

Rearrange to the general form:

$x + 3y - 6 - 3 = 0$

Equation of the second line: $x + 3y - 9 = 0$

The equations of the required lines are $3x - y - 7 = 0$ and $x + 3y - 9 = 0$.

Let the two given lines be:

To find the point of intersection, we solve this system of equations.

Multiply equation (2) by 2:

Subtract equation (3) from equation (1):

$(4x + 7y - 3) - (4x - 6y + 2) = 0 - 0$

$4x + 7y - 3 - 4x + 6y - 2 = 0$

$13y - 5 = 0$

$13y = 5$

$y = \frac{5}{13}$

Substitute the value of $y = \frac{5}{13}$ into equation (2):

$2x - 3\left(\frac{5}{13}\right) + 1 = 0$

$2x - \frac{15}{13} + 1 = 0$

$2x = \frac{15}{13} - 1$

$2x = \frac{15 - 13}{13}$

$2x = \frac{2}{13}$

$x = \frac{1}{13}$

The point of intersection of the two lines is $\left(\frac{1}{13}, \frac{5}{13}\right)$.

Let the equation of the required line have equal intercepts on the axes. This means the x-intercept and the y-intercept are equal.

Let the equal intercept be $a$.

If the intercept is non-zero ($a \neq 0$), the equation of the line in intercept form is:

$\frac{x}{a} + \frac{y}{a} = 1$

Multiplying by $a$, we get:

$x + y = a$

Since the required line passes through the point of intersection $\left(\frac{1}{13}, \frac{5}{13}\right)$, these coordinates must satisfy the equation $x + y = a$.

Substitute $x = \frac{1}{13}$ and $y = \frac{5}{13}$:

$\frac{1}{13} + \frac{5}{13} = a$

$a = \frac{1 + 5}{13}$

$a = \frac{6}{13}$

Since the point of intersection $\left(\frac{1}{13}, \frac{5}{13}\right)$ is not the origin $(0,0)$, the equal intercept $a$ must be non-zero ($a = \frac{6}{13} \neq 0$).

The equation of the line is $x + y = \frac{6}{13}$.

To eliminate the fraction, multiply both sides by 13:

$13(x + y) = 13 \left(\frac{6}{13}\right)$

$13x + 13y = 6$

Rearrange to the general form:

$13x + 13y - 6 = 0$

The equation of the line is $13x + 13y - 6 = 0$.

Let the equation of the given line be:

$y = mx + c$

The slope of this line is $m_1 = m$.

Let the equation of the required line passing through the origin $(0, 0)$ be:

$y = m'x$

The slope of this line is $m'$.

The angle between the given line and the required line is $\theta$.

The formula for the tangent of the angle between two lines with slopes $m_1$ and $m_2$ is given by:

$\tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right|$

Substituting $m_1 = m$ and $m_2 = m'$ into the formula:

$\tan \theta = \left| \frac{m' - m}{1 + m m'} \right|$

This absolute value equation yields two possibilities:

Case 1: $\tan \theta = \frac{m' - m}{1 + m m'}$

Multiply both sides by $(1 + mm')$:

$\tan \theta (1 + mm') = m' - m$

$\tan \theta + mm' \tan \theta = m' - m$

Rearrange to group terms with $m'$:

$m + \tan \theta = m' - mm' \tan \theta$

$m + \tan \theta = m'(1 - m \tan \theta)$

Solving for $m'$:

$m' = \frac{m + \tan \theta}{1 - m \tan \theta}$

Case 2: $\tan \theta = -\left(\frac{m' - m}{1 + m m'}\right) = \frac{m - m'}{1 + m m'}$

Multiply both sides by $(1 + mm')$:

$\tan \theta (1 + mm') = m - m'$

$\tan \theta + mm' \tan \theta = m - m'$

Rearrange to group terms with $m'$:

$m' + mm' \tan \theta = m - \tan \theta$

$m'(1 + m \tan \theta) = m - \tan \theta$

Solving for $m'$:

$m' = \frac{m - \tan \theta}{1 + m \tan \theta}$

Combining both cases, the slope of the required line $m'$ is given by:

$m' = \frac{m \pm \tan \theta}{1 \mp m \tan \theta}$

Note: The $\mp$ symbol in the denominator corresponds to the $\pm$ symbol in the numerator. If the numerator is $(m + \tan \theta)$, the denominator is $(1 - m \tan \theta)$. If the numerator is $(m - \tan \theta)$, the denominator is $(1 + m \tan \theta)$.

The equation of the line passing through the origin $(0, 0)$ with slope $m'$ is $y = m'x$.

Substitute the expression for $m'$:

$y = \left(\frac{m \;\pm\; \tan \theta}{1 \;\mp\; m \tan \theta}\right) x$

Assuming $x \neq 0$, we can divide both sides by $x$:

$\frac{y}{x} = \frac{m \;\pm\; \tan \theta}{1 \;\mp\; m \tan \theta}$

This shows the required equation of the line passing through the origin and making an angle $\theta$ with the line $y = mx + c$ is $\frac{y}{x} = \frac{m \;\pm\; tan \;θ}{1 \;\mp\; m tan \;θ}$.

Let the two given points be $\mathbf{P(-1, 1)}$ and $\mathbf{Q(5, 7)}$.

Let the equation of the given line be $\mathbf{x + y = 4}$.

Assume that the line $x + y = 4$ divides the line segment joining $P(-1, 1)$ and $Q(5, 7)$ at a point $\mathbf{R}$ in the ratio $\mathbf{\lambda : 1}$.

Using the section formula, the coordinates of the point R are given by:

$R = \left(\frac{\lambda x_2 + 1 x_1}{\lambda + 1}, \frac{\lambda y_2 + 1 y_1}{\lambda + 1}\right)$

Here, $(x_1, y_1) = (-1, 1)$ and $(x_2, y_2) = (5, 7)$.

So, the coordinates of R are:

$R = \left(\frac{\lambda (5) + 1 (-1)}{\lambda + 1}, \frac{\lambda (7) + 1 (1)}{\lambda + 1}\right)$

$R = \left(\frac{5\lambda - 1}{\lambda + 1}, \frac{7\lambda + 1}{\lambda + 1}\right)$

Since the point R lies on the line $x + y = 4$, its coordinates must satisfy the equation of the line.

Substitute the coordinates of R into the equation $x + y = 4$:

$\left(\frac{5\lambda - 1}{\lambda + 1}\right) + \left(\frac{7\lambda + 1}{\lambda + 1}\right) = 4$

Combine the terms on the left side:

$\frac{(5\lambda - 1) + (7\lambda + 1)}{\lambda + 1} = 4$

$\frac{5\lambda - 1 + 7\lambda + 1}{\lambda + 1} = 4$

$\frac{12\lambda}{\lambda + 1} = 4$

Multiply both sides by $(\lambda + 1)$ (assuming $\lambda + 1 \neq 0$, which means $\lambda \neq -1$, the ratio cannot be -1:1 for a finite point of division):

$12\lambda = 4(\lambda + 1)$

$12\lambda = 4\lambda + 4$

Subtract $4\lambda$ from both sides:

$12\lambda - 4\lambda = 4$

$8\lambda = 4$

Solve for $\lambda$:

$\lambda = \frac{4}{8}$

$\lambda = \frac{1}{2}$

The ratio in which the line divides the line segment is $\lambda : 1 = \frac{1}{2} : 1$.

To express this ratio in integers, multiply both parts by 2:

$\left(\frac{1}{2} \times 2\right) : (1 \times 2) = 1 : 2$

Since $\lambda$ is positive, the point of division R lies internally on the line segment PQ.

The line $x + y = 4$ divides the line segment joining (–1, 1) and (5, 7) in the ratio 1 : 2 internally.

We are asked to find the distance from the point $\mathbf{(1, 2)}$ to the line $\mathbf{4x + 7y + 5 = 0}$, measured along the line $\mathbf{2x - y = 0}$.

This means we need to find the point where the line $\mathbf{2x - y = 0}$ intersects the line $\mathbf{4x + 7y + 5 = 0}$. The required distance is the distance between the point $(1, 2)$ and this point of intersection.

Let's find the point of intersection of the two lines:

From equation (2), we can express $y$ in terms of $x$:

$y = 2x$

Substitute this expression for $y$ into equation (1):

$4x + 7(2x) + 5 = 0$

$4x + 14x + 5 = 0$

$18x + 5 = 0$

$18x = -5$

$x = -\frac{5}{18}$

Now substitute the value of $x$ back into the equation $y = 2x$ to find the corresponding $y$-coordinate:

$y = 2 \left(-\frac{5}{18}\right)$

$y = -\frac{10}{18}$

$y = -\frac{5}{9}$

The point of intersection, let's call it Q, is $\mathbf{Q\left(-\frac{5}{18}, -\frac{5}{9}\right)}$.

The given point is $\mathbf{P(1, 2)}$.

The distance we need to find is the distance between P and Q.

Using the distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$, which is $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:

$d = \sqrt{\left(-\frac{5}{18} - 1\right)^2 + \left(-\frac{5}{9} - 2\right)^2}$

Calculate the differences in coordinates:

$x_2 - x_1 = -\frac{5}{18} - \frac{18}{18} = -\frac{23}{18}$

$y_2 - y_1 = -\frac{5}{9} - \frac{18}{9} = -\frac{23}{9}$

Calculate the squares of the differences:

$(x_2 - x_1)^2 = \left(-\frac{23}{18}\right)^2 = \frac{(-23)^2}{18^2} = \frac{529}{324}$

$(y_2 - y_1)^2 = \left(-\frac{23}{9}\right)^2 = \frac{(-23)^2}{9^2} = \frac{529}{81}$

To add these, find a common denominator for the squares (324):

$\frac{529}{81} = \frac{529 \times 4}{81 \times 4} = \frac{2116}{324}$

Now substitute these back into the distance formula:

$d = \sqrt{\frac{529}{324} + \frac{2116}{324}}$

$d = \sqrt{\frac{529 + 2116}{324}}$

$d = \sqrt{\frac{2645}{324}}$

Simplify the square root:

$d = \frac{\sqrt{2645}}{\sqrt{324}}$

$d = \frac{\sqrt{2645}}{18}$

The distance of the line $4x + 7y + 5 = 0$ from the point $(1, 2)$ along the line $2x - y = 0$ is $\frac{\sqrt{2645}}{18}$ units.

Given point P: $(-1, 2)$

Given line L1: $x + y = 4$

Let the required line passing through P make an angle $\alpha$ with the positive x-axis. Any point Q on this line at a distance $r$ from P($x_0, y_0$) has coordinates $(x, y)$ given by the parametric equations:

$x = x_0 + r \cos \alpha$

$y = y_0 + r \sin \alpha$

In this case, the starting point is $P(-1, 2)$, so $(x_0, y_0) = (-1, 2)$. The point of intersection Q is at a distance $r=3$ units from P. Therefore, the coordinates of the point of intersection Q are:

$x = -1 + 3 \cos \alpha$

$y = 2 + 3 \sin \alpha$

Since the point of intersection Q lies on the line $x + y = 4$, its coordinates must satisfy the equation of L1.

Substitute the coordinates of Q into the equation $x + y = 4$:

$(-1 + 3 \cos \alpha) + (2 + 3 \sin \alpha) = 4$

$1 + 3 \cos \alpha + 3 \sin \alpha = 4$

Rearranging the terms:

Divide equation (i) by 3:

To solve for $\alpha$, we can rearrange equation (ii) and square both sides:

$\cos \alpha = 1 - \sin \alpha$

Squaring both sides:

$(\cos \alpha)^2 = (1 - \sin \alpha)^2$

$\cos^2 \alpha = 1 - 2 \sin \alpha + \sin^2 \alpha$

Using the trigonometric identity $\cos^2 \alpha = 1 - \sin^2 \alpha$, substitute into the equation:

$1 - \sin^2 \alpha = 1 - 2 \sin \alpha + \sin^2 \alpha$

Rearranging the terms to form a quadratic equation in $\sin \alpha$:

$0 = \sin^2 \alpha + \sin^2 \alpha - 2 \sin \alpha + 1 - 1$

$2 \sin^2 \alpha - 2 \sin \alpha = 0$

Factor out $2 \sin \alpha$:

From equation (iii), we have two possible cases for the value of $\sin \alpha$:

Case A: $2 \sin \alpha = 0 \implies \sin \alpha = 0$

Case B: $\sin \alpha - 1 = 0 \implies \sin \alpha = 1$

Since we squared equation (ii), we must check these potential solutions for $\sin \alpha$ in the original equation (ii): $\cos \alpha + \sin \alpha = 1$, to eliminate any extraneous solutions.

Check Case A: $\sin \alpha = 0$

If $\sin \alpha = 0$, then $\cos \alpha = \pm \sqrt{1 - \sin^2 \alpha} = \pm \sqrt{1 - 0^2} = \pm 1$.

Substitute $\sin \alpha = 0$ and $\cos \alpha = 1$ into equation (ii):

$1 + 0 = 1$. This is true.

Substitute $\sin \alpha = 0$ and $\cos \alpha = -1$ into equation (ii):

$-1 + 0 = -1 \neq 1$. This is false.

So, the solution $\sin \alpha = 0$ is valid only when $\cos \alpha = 1$. This occurs when $\alpha = 0$ (or $2n\pi$ for integer $n$).

Check Case B: $\sin \alpha = 1$

If $\sin \alpha = 1$, then $\cos \alpha = \pm \sqrt{1 - \sin^2 \alpha} = \pm \sqrt{1 - 1^2} = 0$.

Substitute $\sin \alpha = 1$ and $\cos \alpha = 0$ into equation (ii):

$0 + 1 = 1$. This is true.

So, the solution $\sin \alpha = 1$ is valid when $\cos \alpha = 0$. This occurs when $\alpha = \frac{\pi}{2}$ (or $\frac{\pi}{2} + 2n\pi$ for integer $n$).

The valid values for the angle $\alpha$ representing the direction of the line are $0$ and $\frac{\pi}{2}$. These are the angles the required line must make with the positive x-axis.

An angle of $\alpha = 0$ radians ($0^\circ$) corresponds to a line parallel to the x-axis (horizontal direction).

An angle of $\alpha = \frac{\pi}{2}$ radians ($90^\circ$) corresponds to a line parallel to the y-axis (vertical direction).

The direction in which the straight line must be drawn through the point $(-1, 2)$ is such that it makes an angle of $0^\circ$ (or 0 radians) or $90^\circ$ (or $\frac{\pi}{2}$ radians) with the positive x-axis.

This means the line must be drawn either horizontally or vertically from the point $(-1, 2)$.

Let the endpoints of the hypotenuse be $A(1, 3)$ and $B(-4, 1)$.

In a right-angled triangle, the legs are perpendicular to each other. If the legs are parallel to the axes, one leg must be parallel to the x-axis (horizontal) and the other must be parallel to the y-axis (vertical).

The vertex where the right angle is formed must have coordinates $(x, y)$ such that one coordinate is from point A and the other is from point B. There are two such possibilities for this vertex.

Case 1: The right angle vertex has the x-coordinate of A and the y-coordinate of B.

The vertex is $C_1(1, 1)$.

One leg passes through $C_1(1, 1)$ and is parallel to the y-axis. The equation of a line parallel to the y-axis is of the form $x = \text{constant}$. Since it passes through $(1, 1)$, the constant is 1.

The equation of this vertical leg is $x = 1$. This leg connects $C_1(1, 1)$ and $A(1, 3)$.

The other leg passes through $C_1(1, 1)$ and is parallel to the x-axis. The equation of a line parallel to the x-axis is of the form $y = \text{constant}$. Since it passes through $(1, 1)$, the constant is 1.

The equation of this horizontal leg is $y = 1$. This leg connects $C_1(1, 1)$ and $B(-4, 1)$.

The triangle formed has vertices $(1, 3)$, $(-4, 1)$, and $(1, 1)$. The legs are $x=1$ and $y=1$. These are perpendicular to each other.

In this case, the equations of the legs are $x = 1$ and $y = 1$.

Case 2: The right angle vertex has the x-coordinate of B and the y-coordinate of A.

The vertex is $C_2(-4, 3)$.

One leg passes through $C_2(-4, 3)$ and is parallel to the y-axis. The equation is $x = \text{constant}$. Since it passes through $(-4, 3)$, the constant is -4.

The equation of this vertical leg is $x = -4$. This leg connects $C_2(-4, 3)$ and $B(-4, 1)$.

The other leg passes through $C_2(-4, 3)$ and is parallel to the x-axis. The equation is $y = \text{constant}$. Since it passes through $(-4, 3)$, the constant is 3.

The equation of this horizontal leg is $y = 3$. This leg connects $C_2(-4, 3)$ and $A(1, 3)$.

The triangle formed has vertices $(1, 3)$, $(-4, 1)$, and $(-4, 3)$. The legs are $x=-4$ and $y=3$. These are perpendicular to each other.

In this case, the equations of the legs are $x = -4$ and $y = 3$.

There are two possible sets of equations for the legs of the triangle:

Set 1: $x = 1$ and $y = 1$

Set 2: $x = -4$ and $y = 3$

Let the given point be $\mathbf{P(3, 8)}$.

Let the equation of the given line (plane mirror) be $\mathbf{L: x + 3y = 7}$, which can be written as $x + 3y - 7 = 0$.

Let the image of point P with respect to the line L be $\mathbf{P'(x', y')}$.

The line segment PP' is perpendicular to the line L, and the midpoint of PP' lies on the line L.

Method 1: Using properties of image

The equation of the line L is $x + 3y - 7 = 0$. The slope of this line is $m_L$. To find the slope, rewrite the equation in slope-intercept form $y = mx + c$:

$3y = -x + 7$

$y = -\frac{1}{3}x + \frac{7}{3}$

So, the slope of line L is $m_L = -\frac{1}{3}$.

The line segment PP' is perpendicular to L. The slope of PP', let's call it $m_{PP'}$, must satisfy $m_{PP'} \times m_L = -1$.

$m_{PP'} \times \left(-\frac{1}{3}\right) = -1$

$m_{PP'} = 3$

The line passing through P(3, 8) and P'(x', y') has a slope of 3. Using the point-slope form $y - y_1 = m(x - x_1)$ with P(3, 8):

$y - 8 = 3(x - 3)$

$y - 8 = 3x - 9$

$3x - y - 1 = 0$

Since P'(x', y') lies on this line, its coordinates satisfy the equation:

The midpoint M of PP' has coordinates $\left(\frac{3 + x'}{2}, \frac{8 + y'}{2}\right)$.

Since the midpoint M lies on the line L ($x + 3y = 7$), its coordinates must satisfy the equation of L:

$\frac{3 + x'}{2} + 3\left(\frac{8 + y'}{2}\right) = 7$

Multiply by 2 to clear denominators:

$3 + x' + 3(8 + y') = 14$

$3 + x' + 24 + 3y' = 14$

$x' + 3y' + 27 = 14$

$x' + 3y' = 14 - 27$

Now we solve the system of linear equations (i) and (ii) for $x'$ and $y'$.

From (i): $y' = 3x' - 1$

Substitute this into (ii):

$x' + 3(3x' - 1) = -13$

$x' + 9x' - 3 = -13$

$10x' = -13 + 3$

$10x' = -10$

$x' = -1$

Substitute $x' = -1$ back into the equation for $y'$:

$y' = 3(-1) - 1$

$y' = -3 - 1$

$y' = -4$

The image of the point (3, 8) is $(-1, -4)$.

Method 2: Using direct formula

The formula for the image $(x', y')$ of a point $(x_1, y_1)$ with respect to the line $Ax + By + C = 0$ is given by:

$\frac{x' - x_1}{A} = \frac{y' - y_1}{B} = -2 \frac{Ax_1 + By_1 + C}{A^2 + B^2}$

Given point $(x_1, y_1) = (3, 8)$.

Given line $x + 3y - 7 = 0$. Here, $A = 1$, $B = 3$, and $C = -7$.

Substitute these values into the formula:

$\frac{x' - 3}{1} = \frac{y' - 8}{3} = -2 \frac{1(3) + 3(8) - 7}{1^2 + 3^2}$

$\frac{x' - 3}{1} = \frac{y' - 8}{3} = -2 \frac{3 + 24 - 7}{1 + 9}$

$\frac{x' - 3}{1} = \frac{y' - 8}{3} = -2 \frac{20}{10}$

$\frac{x' - 3}{1} = \frac{y' - 8}{3} = -2(2)$

From equation (iii), equate the first part to -4:

$\frac{x' - 3}{1} = -4$

$x' - 3 = -4$

$x' = -4 + 3$

$x' = -1$

From equation (iii), equate the second part to -4:

$\frac{y' - 8}{3} = -4$

$y' - 8 = 3(-4)$

$y' - 8 = -12$

$y' = -12 + 8$

$y' = -4$

The image of the point (3, 8) is $(-1, -4)$.

The image of the point (3, 8) with respect to the line x + 3y = 7 is (-1, -4).

Let the slopes of the three lines be $m_1, m_2$, and $m_3$.

The first line is $y = 3x + 1$. Its slope is $m_1 = 3$.

The second line is $2y = x + 3$. Dividing by 2, we get $y = \frac{1}{2}x + \frac{3}{2}$. Its slope is $m_2 = \frac{1}{2}$.

The third line is $y = mx + 4$. Its slope is $m_3 = m$.

If the line $y = mx + 4$ is equally inclined to the lines $y = 3x + 1$ and $2y = x + 3$, it means the angle between the third line and the first line is equal to the angle between the third line and the second line.

Let $\theta_1$ be the angle between the line with slope $m_3$ and the line with slope $m_1$.

Let $\theta_2$ be the angle between the line with slope $m_3$ and the line with slope $m_2$.

The condition for equally inclined lines implies $\tan \theta_1 = \tan \theta_2$.

Using the formula for the tangent of the angle between two lines with slopes $m_a$ and $m_b$, which is $\tan \theta = \left| \frac{m_a - m_b}{1 + m_a m_b} \right|$:

$\tan \theta_1 = \left| \frac{m_3 - m_1}{1 + m_3 m_1} \right| = \left| \frac{m - 3}{1 + m(3)} \right| = \left| \frac{m - 3}{1 + 3m} \right|$

$\tan \theta_2 = \left| \frac{m_3 - m_2}{1 + m_3 m_2} \right| = \left| \frac{m - \frac{1}{2}}{1 + m\left(\frac{1}{2}\right)} \right| = \left| \frac{\frac{2m - 1}{2}}{\frac{2 + m}{2}} \right| = \left| \frac{2m - 1}{2 + m} \right|$

Setting $\tan \theta_1 = \tan \theta_2$:

$\left| \frac{m - 3}{1 + 3m} \right| = \left| \frac{2m - 1}{2 + m} \right|$

This equation holds if $\frac{m - 3}{1 + 3m} = \frac{2m - 1}{2 + m}$ or $\frac{m - 3}{1 + 3m} = -\frac{2m - 1}{2 + m}$.

Case 1: $\frac{m - 3}{1 + 3m} = \frac{2m - 1}{2 + m}$

Cross-multiply:

$(m - 3)(2 + m) = (2m - 1)(1 + 3m)$

$m(2) + m(m) - 3(2) - 3(m) = 2m(1) + 2m(3m) - 1(1) - 1(3m)$

$2m + m^2 - 6 - 3m = 2m + 6m^2 - 1 - 3m$

$m^2 - m - 6 = 6m^2 - m - 1$

Rearrange the terms:

$m^2 - 6m^2 - m + m - 6 + 1 = 0$

$-5m^2 - 5 = 0$

Divide by -5:

$m^2 + 1 = 0$

$m^2 = -1$

This equation has no real solutions for $m$.

Case 2: $\frac{m - 3}{1 + 3m} = -\frac{2m - 1}{2 + m} = \frac{-(2m - 1)}{2 + m} = \frac{1 - 2m}{2 + m}$

Cross-multiply:

$(m - 3)(2 + m) = (1 - 2m)(1 + 3m)$

$m^2 - m - 6 = 1(1) + 1(3m) - 2m(1) - 2m(3m)$

$m^2 - m - 6 = 1 + 3m - 2m - 6m^2$

$m^2 - m - 6 = -6m^2 + m + 1$

Rearrange the terms to form a quadratic equation:

$m^2 + 6m^2 - m - m - 6 - 1 = 0$

$7m^2 - 2m - 7 = 0$

Use the quadratic formula $m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ to solve for $m$, where $a=7$, $b=-2$, $c=-7$:

$m = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(7)(-7)}}{2(7)}$

$m = \frac{2 \pm \sqrt{4 + 196}}{14}$

$m = \frac{2 \pm \sqrt{200}}{14}$

$m = \frac{2 \pm \sqrt{100 \times 2}}{14}$

$m = \frac{2 \pm 10\sqrt{2}}{14}$

Factor out 2 from the numerator and simplify:

$m = \frac{2(1 \pm 5\sqrt{2})}{14}$

$m = \frac{1 \pm 5\sqrt{2}}{7}$

The values of $m$ for which the line $y = mx + 4$ is equally inclined to the other two lines are $\frac{1 + 5\sqrt{2}}{7}$ and $\frac{1 - 5\sqrt{2}}{7}$.

Let the given lines be:

$L_1: x + y - 5 = 0$

$L_2: 3x - 2y + 7 = 0$

Let P(x, y) be the variable point.

The formula for the perpendicular distance from a point $(x_0, y_0)$ to a line $Ax + By + C = 0$ is given by:

$d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$

The perpendicular distance from P(x, y) to line $L_1$ ($x + y - 5 = 0$) is $d_1$.

Here, $A_1 = 1$, $B_1 = 1$, $C_1 = -5$.

$d_1 = \frac{|x + y - 5|}{\sqrt{1^2 + 1^2}} = \frac{|x + y - 5|}{\sqrt{2}}$

The perpendicular distance from P(x, y) to line $L_2$ ($3x - 2y + 7 = 0$) is $d_2$.

Here, $A_2 = 3$, $B_2 = -2$, $C_2 = 7$.

$d_2 = \frac{|3x - 2y + 7|}{\sqrt{3^2 + (-2)^2}} = \frac{|3x - 2y + 7|}{\sqrt{9 + 4}} = \frac{|3x - 2y + 7|}{\sqrt{13}}$

According to the problem, the sum of the perpendicular distances is always 10.

$d_1 + d_2 = 10$

$\frac{|x + y - 5|}{\sqrt{2}} + \frac{|3x - 2y + 7|}{\sqrt{13}} = 10$

The absolute value $|E|$ of an expression E means that $E$ can be positive or negative. Thus, the equation involves terms with absolute values. For any point P(x, y) in the plane, the expressions $(x + y - 5)$ and $(3x - 2y + 7)$ have specific signs. The lines $x + y - 5 = 0$ and $3x - 2y + 7 = 0$ divide the plane into regions where the signs of these expressions are constant.

In each region, the equation $\frac{|x + y - 5|}{\sqrt{2}} + \frac{|3x - 2y + 7|}{\sqrt{13}} = 10$ can be written without the absolute value signs by considering the specific signs of $(x + y - 5)$ and $(3x - 2y + 7)$. This leads to four possible cases:

Case 1: $x + y - 5 \geq 0$ and $3x - 2y + 7 \geq 0$

$\frac{(x + y - 5)}{\sqrt{2}} + \frac{(3x - 2y + 7)}{\sqrt{13}} = 10$

Multiply by $\sqrt{26}$ (LCM of $\sqrt{2}$ and $\sqrt{13}$):

$\sqrt{13}(x + y - 5) + \sqrt{2}(3x - 2y + 7) = 10\sqrt{26}$

$\sqrt{13}x + \sqrt{13}y - 5\sqrt{13} + 3\sqrt{2}x - 2\sqrt{2}y + 7\sqrt{2} = 10\sqrt{26}$

Group x and y terms:

$(\sqrt{13} + 3\sqrt{2})x + (\sqrt{13} - 2\sqrt{2})y + (-5\sqrt{13} + 7\sqrt{2} - 10\sqrt{26}) = 0$

This is a linear equation in the form $Ax + By + C = 0$, where $A = \sqrt{13} + 3\sqrt{2}$, $B = \sqrt{13} - 2\sqrt{2}$, and $C = -5\sqrt{13} + 7\sqrt{2} - 10\sqrt{26}$. Since A and B are not both zero, this equation represents a straight line.

Case 2: $x + y - 5 \geq 0$ and $3x - 2y + 7 < 0$

$\frac{(x + y - 5)}{\sqrt{2}} - \frac{(3x - 2y + 7)}{\sqrt{13}} = 10$

Multiply by $\sqrt{26}$:

$\sqrt{13}(x + y - 5) - \sqrt{2}(3x - 2y + 7) = 10\sqrt{26}$

$\sqrt{13}x + \sqrt{13}y - 5\sqrt{13} - 3\sqrt{2}x + 2\sqrt{2}y - 7\sqrt{2} = 10\sqrt{26}$

Group x and y terms:

$(\sqrt{13} - 3\sqrt{2})x + (\sqrt{13} + 2\sqrt{2})y + (-5\sqrt{13} - 7\sqrt{2} - 10\sqrt{26}) = 0$

This is also a linear equation in the form $A'x + B'y + C' = 0$, representing a straight line.

Case 3: $x + y - 5 < 0$ and $3x - 2y + 7 \geq 0$

$-\frac{(x + y - 5)}{\sqrt{2}} + \frac{(3x - 2y + 7)}{\sqrt{13}} = 10$

Multiply by $\sqrt{26}$:

$-\sqrt{13}(x + y - 5) + \sqrt{2}(3x - 2y + 7) = 10\sqrt{26}$

$-\sqrt{13}x - \sqrt{13}y + 5\sqrt{13} + 3\sqrt{2}x - 2\sqrt{2}y + 7\sqrt{2} = 10\sqrt{26}$

Group x and y terms:

$(-\sqrt{13} + 3\sqrt{2})x + (-\sqrt{13} - 2\sqrt{2})y + (5\sqrt{13} + 7\sqrt{2} - 10\sqrt{26}) = 0$

This is a linear equation in the form $A''x + B''y + C'' = 0$, representing a straight line.

Case 4: $x + y - 5 < 0$ and $3x - 2y + 7 < 0$

$-\frac{(x + y - 5)}{\sqrt{2}} - \frac{(3x - 2y + 7)}{\sqrt{13}} = 10$

Multiply by $\sqrt{26}$:

$-\sqrt{13}(x + y - 5) - \sqrt{2}(3x - 2y + 7) = 10\sqrt{26}$

$-\sqrt{13}x - \sqrt{13}y + 5\sqrt{13} - 3\sqrt{2}x + 2\sqrt{2}y - 7\sqrt{2} = 10\sqrt{26}$

Group x and y terms:

$(-\sqrt{13} - 3\sqrt{2})x + (-\sqrt{13} + 2\sqrt{2})y + (5\sqrt{13} - 7\sqrt{2} - 10\sqrt{26}) = 0$

This is a linear equation in the form $A'''x + B'''y + C''' = 0$, representing a straight line.

In each of the four regions defined by the lines $x + y - 5 = 0$ and $3x - 2y + 7 = 0$, the locus of points P(x, y) satisfying the given condition lies on a specific straight line. The collection of these points across all regions forms the complete locus.

The equation $\frac{|L_1|}{\sqrt{A_1^2+B_1^2}} + \frac{|L_2|}{\sqrt{A_2^2+B_2^2}} = k$ (where $L_1=0$ and $L_2=0$ are intersecting lines) represents a locus that is a polygon whose sides are segments of the four lines derived above. However, the phrasing "P must move on a line" strongly implies that the locus itself is a single straight line.

This implies that the relationship between the coordinates $(x, y)$ of point P is always linear.

Since each of the derived equations is a linear equation of the form $Ax + By + C = 0$, this shows that for any point P satisfying the condition, its coordinates $(x, y)$ must satisfy one of these linear equations.

Therefore, the point P must move on a line (or a union of parts of lines that are themselves straight lines).

Thus, we have shown that the equation representing the locus of P is linear in x and y in each region of the plane, and hence P must move on a line.

Let the equations of the two given lines be:

First, let's verify that the lines are parallel by finding their slopes.

For $L_1: 9x + 6y - 7 = 0$, rewrite in slope-intercept form $y = mx + c$:

$6y = -9x + 7$

$y = -\frac{9}{6}x + \frac{7}{6}$

$y = -\frac{3}{2}x + \frac{7}{6}$

The slope of $L_1$ is $m_1 = -\frac{3}{2}$.

For $L_2: 3x + 2y + 6 = 0$, rewrite in slope-intercept form $y = mx + c$:

$2y = -3x - 6$

$y = -\frac{3}{2}x - 3$

The slope of $L_2$ is $m_2 = -\frac{3}{2}$.

Since $m_1 = m_2$, the lines are parallel.

The line that is equidistant from two parallel lines is also a parallel line located exactly in the middle of them.

Let the equation of the required line be $L: Ax + By + C = 0$. Since it is parallel to $L_1$ and $L_2$, the coefficients of x and y will be proportional. We can use the coefficients from $L_1$ or a multiple of the coefficients from $L_2$. It's helpful to make the coefficients of x and y the same for both lines when considering distance.

Multiply the equation of $L_2$ by 3:

$3 \times (3x + 2y + 6) = 3 \times 0$

$9x + 6y + 18 = 0$

So the two parallel lines can be written as:

$L_1: 9x + 6y - 7 = 0$ (where $A=9, B=6, C_1=-7$)

$L_2': 9x + 6y + 18 = 0$ (where $A=9, B=6, C_2=18$)

Let P(x, y) be any point on the equidistant line. The perpendicular distance from P(x, y) to $L_1$ is equal to the perpendicular distance from P(x, y) to $L_2'$.

The distance from $(x, y)$ to $Ax + By + C = 0$ is $\frac{|Ax + By + C|}{\sqrt{A^2 + B^2}}$.

Distance from P to $L_1$: $d_1 = \frac{|9x + 6y - 7|}{\sqrt{9^2 + 6^2}}$

Distance from P to $L_2'$: $d_2 = \frac{|9x + 6y + 18|}{\sqrt{9^2 + 6^2}}$

Since $d_1 = d_2$:

$\frac{|9x + 6y - 7|}{\sqrt{81 + 36}} = \frac{|9x + 6y + 18|}{\sqrt{81 + 36}}$

$|9x + 6y - 7| = |9x + 6y + 18|$

This equation implies two possibilities:

1) $9x + 6y - 7 = 9x + 6y + 18$

$-7 = 18$ (This is impossible, so no point satisfies this condition).

2) $9x + 6y - 7 = -(9x + 6y + 18)$

$9x + 6y - 7 = -9x - 6y - 18$}

Bring all terms to one side:

$9x + 9x + 6y + 6y - 7 + 18 = 0$

$18x + 12y + 11 = 0$

This is the equation of the line equidistant from the two given lines.

Alternate Method: Using the property of the constant term

For two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$, the equation of the line equidistant from them is $Ax + By + \frac{C_1 + C_2}{2} = 0$.

We have the standardized equations:

$9x + 6y - 7 = 0$ ($C_1 = -7$)

$9x + 6y + 18 = 0$ ($C_2 = 18$)

The equation of the equidistant line is:

$9x + 6y + \frac{-7 + 18}{2} = 0$

$9x + 6y + \frac{11}{2} = 0$

Multiply by 2 to remove the fraction:

$18x + 12y + 11 = 0$

Both methods yield the same result.

The equation of the line which is equidistant from parallel lines $9x + 6y – 7 = 0$ and $3x + 2y + 6 = 0$ is $18x + 12y + 11 = 0$.

Let the initial point of the ray be $\mathbf{P(1, 2)}$.

Let the point on the x-axis where the reflection occurs be $\mathbf{A(x, 0)}$.

Let the point through which the reflected ray passes be $\mathbf{Q(5, 3)}$.

According to the laws of reflection, the angle of incidence equals the angle of reflection. For reflection on the x-axis, this means the incident ray PA and the reflected ray AQ make equal angles with the x-axis.

A useful property for reflection problems is that the reflected ray appears to come from the image of the initial point with respect to the reflecting surface.

Let $\mathbf{P'(x', y')}$ be the image of the point P(1, 2) with respect to the x-axis.

When a point $(x_0, y_0)$ is reflected across the x-axis, its image is $(x_0, -y_0)$.

So, the image of P(1, 2) is $\mathbf{P'(1, -2)}$.

The reflected ray AQ lies on the straight line passing through the image point P'(1, -2) and the point Q(5, 3).

The point of reflection A(x, 0) is the point where the line segment PQ touches the x-axis after reflection, which is the intersection point of the line P'Q with the x-axis.

We need to find the equation of the line passing through P'(1, -2) and Q(5, 3).

Using the two-point form of the equation of a line $\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}$ with $(x_1, y_1) = (1, -2)$ and $(x_2, y_2) = (5, 3)$:

$\frac{y - (-2)}{3 - (-2)} = \frac{x - 1}{5 - 1}$

$\frac{y + 2}{5} = \frac{x - 1}{4}$

Cross-multiply to find the equation of the line P'Q:

$4(y + 2) = 5(x - 1)$

$4y + 8 = 5x - 5$

Rearrange into the general form:

$5x - 4y - 5 - 8 = 0$

The point of reflection A(x, 0) lies on the x-axis, which is the line $y = 0$. It is also the intersection of the line P'Q (equation (i)) and the x-axis ($y=0$).

Substitute $y = 0$ into equation (i):

$5x - 4(0) - 13 = 0$

$5x - 13 = 0$

$5x = 13$

$x = \frac{13}{5}$

The coordinates of point A are $(x, 0)$, so $A = \left(\frac{13}{5}, 0\right)$.

The coordinates of A are $\left(\frac{13}{5}, 0\right)$.

Let the two given points be $P_1\left(\sqrt{a^2 - b^2}, 0\right)$ and $P_2\left(-\sqrt{a^2 - b^2}, 0\right)$. These points are the foci of an ellipse if $a > b$. Let $c = \sqrt{a^2 - b^2}$. The points are $(c, 0)$ and $(-c, 0)$.

Let the equation of the given line be $\frac{x}{a} \cos \theta + \frac{y}{b} \sin \theta = 1$.

To use the distance formula, rewrite the line equation in the general form $Ax + By + C = 0$:

$\frac{\cos \theta}{a}x + \frac{\sin \theta}{b}y - 1 = 0$

Here, $A = \frac{\cos \theta}{a}$, $B = \frac{\sin \theta}{b}$, and $C = -1$.

The formula for the perpendicular distance from a point $(x_0, y_0)$ to a line $Ax + By + C = 0$ is $d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$.

The square of the denominator is $A^2 + B^2 = \left(\frac{\cos \theta}{a}\right)^2 + \left(\frac{\sin \theta}{b}\right)^2 = \frac{\cos^2 \theta}{a^2} + \frac{\sin^2 \theta}{b^2}$.

$\sqrt{A^2 + B^2} = \sqrt{\frac{\cos^2 \theta}{a^2} + \frac{\sin^2 \theta}{b^2}}$

The perpendicular distance from $P_1(c, 0)$ to the line is $d_1$:

$d_1 = \frac{\left|\frac{\cos \theta}{a}(c) + \frac{\sin \theta}{b}(0) - 1\right|}{\sqrt{\frac{\cos^2 \theta}{a^2} + \frac{\sin^2 \theta}{b^2}}} = \frac{\left|\frac{c \cos \theta}{a} - 1\right|}{\sqrt{\frac{\cos^2 \theta}{a^2} + \frac{\sin^2 \theta}{b^2}}}$

The perpendicular distance from $P_2(-c, 0)$ to the line is $d_2$:

$d_2 = \frac{\left|\frac{\cos \theta}{a}(-c) + \frac{\sin \theta}{b}(0) - 1\right|}{\sqrt{\frac{\cos^2 \theta}{a^2} + \frac{\sin^2 \theta}{b^2}}} = \frac{\left|-\frac{c \cos \theta}{a} - 1\right|}{\sqrt{\frac{\cos^2 \theta}{a^2} + \frac{\sin^2 \theta}{b^2}}} = \frac{\left|\frac{c \cos \theta}{a} + 1\right|}{\sqrt{\frac{\cos^2 \theta}{a^2} + \frac{\sin^2 \theta}{b^2}}}$

Note that $|-E| = |E|$, so $\left|-\frac{c \cos \theta}{a} - 1\right| = \left|-\left(\frac{c \cos \theta}{a} + 1\right)\right| = \left|\frac{c \cos \theta}{a} + 1\right|$.

The product of the lengths of the perpendiculars is $d_1 d_2$:

$d_1 d_2 = \frac{\left|\frac{c \cos \theta}{a} - 1\right| \left|\frac{c \cos \theta}{a} + 1\right|}{\left(\sqrt{\frac{\cos^2 \theta}{a^2} + \frac{\sin^2 \theta}{b^2}}\right)^2}$

$d_1 d_2 = \frac{\left|\left(\frac{c \cos \theta}{a}\right)^2 - 1^2\right|}{\frac{\cos^2 \theta}{a^2} + \frac{\sin^2 \theta}{b^2}}$

$d_1 d_2 = \frac{\left|\frac{c^2 \cos^2 \theta}{a^2} - 1\right|}{\frac{b^2 \cos^2 \theta + a^2 \sin^2 \theta}{a^2 b^2}}$

Substitute $c^2 = a^2 - b^2$ in the numerator:

Numerator $= \left|\frac{(a^2 - b^2) \cos^2 \theta}{a^2} - 1\right|$

Numerator $= \left|\frac{a^2 \cos^2 \theta - b^2 \cos^2 \theta}{a^2} - \frac{a^2}{a^2}\right|$

Numerator $= \left|\frac{a^2 \cos^2 \theta - b^2 \cos^2 \theta - a^2}{a^2}\right|$

Numerator $= \left|\frac{a^2 (\cos^2 \theta - 1) - b^2 \cos^2 \theta}{a^2}\right|$

Using $\cos^2 \theta - 1 = -\sin^2 \theta$:

Numerator $= \left|\frac{-a^2 \sin^2 \theta - b^2 \cos^2 \theta}{a^2}\right|$

Numerator $= \left|\frac{-(a^2 \sin^2 \theta + b^2 \cos^2 \theta)}{a^2}\right|$

Numerator $= \frac{a^2 \sin^2 \theta + b^2 \cos^2 \theta}{a^2}$ (since $a^2, b^2, \sin^2 \theta, \cos^2 \theta$ are non-negative, their sum is non-negative)

Now substitute the numerator and the denominator back into the expression for $d_1 d_2$:

$d_1 d_2 = \frac{\frac{a^2 \sin^2 \theta + b^2 \cos^2 \theta}{a^2}}{\frac{b^2 \cos^2 \theta + a^2 \sin^2 \theta}{a^2 b^2}}$

$d_1 d_2 = \frac{a^2 \sin^2 \theta + b^2 \cos^2 \theta}{a^2} \times \frac{a^2 b^2}{b^2 \cos^2 \theta + a^2 \sin^2 \theta}$

Cancel the common term $(a^2 \sin^2 \theta + b^2 \cos^2 \theta)$ and $a^2$ (assuming $a^2 \sin^2 \theta + b^2 \cos^2 \theta \neq 0$ and $a^2 \neq 0$):

$d_1 d_2 = \frac{\cancel{(a^2 \sin^2 \theta + b^2 \cos^2 \theta)}}{\cancel{a^2}} \times \frac{\cancel{a^2} b^2}{\cancel{(b^2 \cos^2 \theta + a^2 \sin^2 \theta)}}$

$d_1 d_2 = b^2$

Thus, the product of the lengths of the perpendiculars drawn from the given points to the line is $b^2$.

To reach the path $6x - 7y + 8 = 0$ in the least time, the person must follow the path that is perpendicular to this line and passes through the person's starting point.

The starting point of the person is the point of intersection of the paths $2x – 3y + 4 = 0$ and $3x + 4y – 5 = 0$. Let's find this point of intersection.

We need to solve the system of equations:

Multiply equation (1) by 4 and equation (2) by 3 to eliminate $y$:

Add equation (3) and equation (4):

(8x - 12y) + (9x + 12y) = -16 + 15

8x - 12y + 9x + 12y = -1

17x = -1

Solving for $x$:

$x = -\frac{1}{17}$

Substitute the value of $x$ into equation (1):

$2\left(-\frac{1}{17}\right) - 3y = -4$

$-\frac{2}{17} - 3y = -4$}

$-3y = -4 + \frac{2}{17}$

$-3y = \frac{-4 \times 17 + 2}{17}$

$-3y = \frac{-68 + 2}{17}$

$-3y = \frac{-66}{17}$

Solving for $y$:

$y = \frac{-66}{17 \times -3} = \frac{22}{17}$

The point of intersection is $\left(-\frac{1}{17}, \frac{22}{17}\right)$. This is the starting point of the person.

The person wants to reach the path $6x - 7y + 8 = 0$ in the least time. The path of least time is perpendicular to this line.

Let the equation of the target path be $L_3: 6x - 7y + 8 = 0$.

Rewrite $L_3$ in slope-intercept form $y = mx + c$ to find its slope:

$-7y = -6x - 8$

$y = \frac{-6}{-7}x + \frac{-8}{-7}$

$y = \frac{6}{7}x + \frac{8}{7}$

The slope of $L_3$ is $m_3 = \frac{6}{7}$.

The required path is perpendicular to $L_3$. If the slope of the required path is $m_\perp$, then $m_\perp \times m_3 = -1$.

$m_\perp \times \frac{6}{7} = -1$

$m_\perp = -1 \times \frac{7}{6} = -\frac{7}{6}$

The slope of the path the person should follow is $-\frac{7}{6}$.

The required path passes through the point $\left(-\frac{1}{17}, \frac{22}{17}\right)$ and has a slope of $-\frac{7}{6}$. Using the point-slope form $y - y_1 = m(x - x_1)$:

$y - \frac{22}{17} = -\frac{7}{6}\left(x - \left(-\frac{1}{17}\right)\right)$

$y - \frac{22}{17} = -\frac{7}{6}\left(x + \frac{1}{17}\right)$

$y - \frac{22}{17} = -\frac{7}{6}x - \frac{7}{6 \times 17}$

$y - \frac{22}{17} = -\frac{7}{6}x - \frac{7}{102}$

To eliminate the denominators, multiply the entire equation by the LCM of 17, 6, and 102, which is 102:

$102 \left(y - \frac{22}{17}\right) = 102 \left(-\frac{7}{6}x - \frac{7}{102}\right)$

$102y - 102 \times \frac{22}{17} = 102 \times \left(-\frac{7}{6}x\right) - 102 \times \frac{7}{102}$

$102y - (6 \times 17) \times \frac{22}{17} = (17 \times 6) \times \left(-\frac{7}{6}x\right) - 7$

$102y - 6 \times 22 = 17 \times (-7x) - 7$

$102y - 132 = -119x - 7$

Rearrange the terms to put the equation in the form $Ax + By + C = 0$:

$119x + 102y - 132 + 7 = 0$

$119x + 102y - 125 = 0$

The equation of the path that the person should follow to reach the line $6x – 7y + 8 = 0$ in the least time is $119x + 102y - 125 = 0$.