Chapter 10 Straight Lines

Given:

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

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

To Find:

The equation of the line.

Solution:

Let the given point be $(x_1, y_1) = (2, 3)$.

Let the angle made by the line with the positive x-axis be $\theta = 30^\circ$.

The slope ($m$) of a line is given by the tangent of the angle it makes with the positive direction of the x-axis.

$m = \tan(\theta)$

Substituting the given value of $\theta$:

We know that the value of $\tan(30^\circ)$ is $\frac{1}{\sqrt{3}}$.

From equation (i):

The equation of a line passing through a point $(x_1, y_1)$ with slope $m$ is given by the point-slope form:

Substituting the point $(x_1, y_1) = (2, 3)$ and the slope m = $\frac{1}{\sqrt{3}}$ into equation (iii):

Now, we simplify equation (iv) to get the equation of the line.

Multiply both sides of equation (iv) by $\sqrt{3}$:

Expanding both sides of equation (v):

Rearrange the terms of equation (vi) to bring them to one side (standard form $Ax + By + C = 0$):

Alternate Solution (Slope-Intercept Form):

Alternatively, we can write the equation in slope-intercept form ($y = mx + c$).

From equation (vi):

Divide both sides of equation (viii) by $\sqrt{3}$:

Simplifying equation (ix):

Equation (vii) or equation (x) represents the required equation of the line.

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

Given:

Length of the perpendicular from the origin to the line (p) = 4 units.

Inclination of the perpendicular segment with the positive direction of x-axis ($\alpha$) = $30^\circ$.

To Find:

The equation of the line.

Solution:

The equation of a line in the Normal Form is given by:

where $p$ is the length of the perpendicular from the origin to the line and $\alpha$ is the angle the perpendicular makes with the positive x-axis.

Substitute the given values of $p = 4$ and $\alpha = 30^\circ$ into equation (i):

We know that $\cos 30^\circ = \frac{\sqrt{3}}{2}$ and $\sin 30^\circ = \frac{1}{2}$.

Substitute these trigonometric values into equation (ii):

Multiply both sides of equation (iii) by 2 to eliminate the denominators:

Simplifying equation (iv):

Rearrange equation (v) to write it in the standard form $Ax + By + C = 0$:

The equation of the line is $\sqrt{3}\text{x} + \text{y} - 8 = 0$.

To Prove:

Every straight line has an equation of the form $Ax + By + C = 0$, where A, B, and C are constants.

Proof:

Consider an arbitrary straight line in the Cartesian coordinate system. We can categorize straight lines into three types based on their orientation:

Case 1: The line is parallel to the y-axis.

A line parallel to the y-axis is a vertical line. The equation of a vertical line is of the form:

where $k$ is a constant (the x-intercept).

We can rewrite equation (i) as:

This equation is of the form $Ax + By + C = 0$ where $A=1$, $B=0$, and $C=-k$. Since $k$ is a constant, $A, B,$ and $C$ are constants.

Case 2: The line is parallel to the x-axis.

A line parallel to the x-axis is a horizontal line. The equation of a horizontal line is of the form:

where $k'$ is a constant (the y-intercept).

We can rewrite equation (iii) as:

This equation is of the form $Ax + By + C = 0$ where $A=0$, $B=1$, and $C=-k'$. Since $k'$ is a constant, $A, B,$ and $C$ are constants.

Case 3: The line is not parallel to either axis.

A line that is not parallel to either axis has a defined slope. Let the slope of the line be $m$. Let the line pass through a point $(x_1, y_1)$. The equation of such a line can be written using the point-slope form:

Expand equation (v):

Rearrange the terms of equation (vi) to bring them to one side:

This equation is of the form $Ax + By + C = 0$ where $A=m$, $B=-1$, and $C=y_1 - mx_1$. Since $m$, $x_1$, and $y_1$ are constants for a specific line, $A, B,$ and $C$ are constants.

Alternatively, if the line has slope $m$ and y-intercept $c$ (i.e., passes through $(0, c)$), its equation in slope-intercept form is:

Rearrange equation (viii):

This equation is of the form $Ax + By + C = 0$ where $A=m$, $B=-1$, and $C=c$. Since $m$ and $c$ are constants for a specific line, $A, B,$ and $C$ are constants.

In all three cases, we have shown that the equation of a straight line can be expressed in the form $Ax + By + C = 0$, where A, B, and C are constants.

Also, it is important to note that for the equation $Ax + By + C = 0$ to represent a straight line, at least one of A or B must be non-zero ($A \neq 0$ or $B \neq 0$). If both $A=0$ and $B=0$, the equation becomes $C=0$, which is either $0=0$ (true for all points, representing the entire plane, not a line) or $C \neq 0$ (false for all points, representing no points). The cases above inherently satisfy $A \neq 0$ or $B \neq 0$. In Case 1, $A=1, B=0$. In Case 2, $A=0, B=1$. In Case 3, $B=-1$, so $B \neq 0$.

Thus, every straight line has an equation of the form $Ax + By + C = 0$.

Conclusion:

Based on the analysis of all possible orientations of a straight line, we conclude that the general equation of a straight line is of the form $Ax + By + C = 0$, where A, B, and C are constants with at least one of A or B being non-zero.

Given:

The required line passes through the point $(1, 2)$.

The required line is perpendicular to the line $x + y + 7 = 0$.

To Find:

The equation of the required straight line.

Solution:

The equation of the given line is:

The general form of a linear equation is $Ax + By + C = 0$. Comparing equation (i) with the general form, we have $A=1$, $B=1$, and $C=7$.

The slope of a line in the form $Ax + By + C = 0$ is given by $m = -\frac{A}{B}$.

Let the slope of the given line (equation (i)) be $m_1$.

From equation (ii):

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 $m_2$.

Substitute the value of $m_1$ from equation (iii) into equation (iv):

Solving for $m_2$ from equation (v):

From equation (vi):

The required line passes through the point $(x_1, y_1) = (1, 2)$ and has a slope $m = 1$ (from equation (vii)).

Using the point-slope form of the equation of a line: $y - y_1 = m(x - x_1)$.

Substitute the values of $(x_1, y_1)$ and $m$ into the point-slope form:

Simplify equation (viii):

Rearrange the terms of equation (ix) to bring them to one side to get the standard form:

From equation (x):

The equation of the straight line passing through $(1, 2)$ and perpendicular to $x + y + 7 = 0$ is $x - y + 1 = 0$.

Given:

The equations of two lines are $3x + 4y = 9$ and $6x + 8y = 15$.

To Find:

The distance between the given lines.

Solution:

First, let's write the equations of the lines in the standard form $Ax + By + C = 0$.

Line 1:

Line 2:

To determine if the lines are parallel, we compare the ratios of the coefficients of x and y.

For line (i), $A_1 = 3$, $B_1 = 4$, $C_1 = -9$.

For line (ii), $A_2 = 6$, $B_2 = 8$, $C_2 = -15$.

Ratio of x coefficients: $\frac{A_1}{A_2} = \frac{3}{6} = \frac{1}{2}$.

Ratio of y coefficients: $\frac{B_1}{B_2} = \frac{4}{8} = \frac{1}{2}$.

Since $\frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{1}{2}$, the lines are parallel.

Now, we check the ratio of the constant terms: $\frac{C_1}{C_2} = \frac{-9}{-15} = \frac{9}{15} = \frac{3}{5}$.

Since $\frac{A_1}{A_2} = \frac{B_1}{B_2} \neq \frac{C_1}{C_2}$, the lines are distinct parallel lines.

To find the distance between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$, we need the coefficients of x and y to be the same in both equations.

Multiply equation (i) by 2:

Simplifying equation (iii):

Now we have the two parallel lines in the form $Ax + By + C = 0$ with the same A and B values:

Line (iv): $6x + 8y - 18 = 0$ ($A=6$, $B=8$, $C_1=-18$)

Line (ii): $6x + 8y - 15 = 0$ ($A=6$, $B=8$, $C_2=-15$)

The formula for the distance ($d$) between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is:

Substitute the values from lines (iv) and (ii) into equation (v):

Simplify equation (vi):

The distance between the lines $3x + 4y = 9$ and $6x + 8y = 15$ is $\frac{3}{10}$ units.

Given:

The equation of a variable line is $x \cos \alpha + y \sin \alpha = p$, where $p$ is a constant.

To Show:

The locus of the mid-point of the distance between the axes of the variable line is $\frac{1}{x^2} + \frac{1}{y^2} = \frac{4}{p^2}$.

Solution:

The equation of the variable line is:

To find the points where the line intersects the coordinate axes, we find the x-intercept and the y-intercept.

x-intercept: Set y = 0 in equation (i).

Assuming $\cos \alpha \neq 0$, we get:

The point where the line intersects the x-axis is A($\frac{p}{\cos \alpha}$, 0).

y-intercept: Set x = 0 in equation (i).

Assuming $\sin \alpha \neq 0$, we get:

The point where the line intersects the y-axis is B(0, $\frac{p}{\sin \alpha}$).

The segment of the line between the axes is the segment AB, connecting A($\frac{p}{\cos \alpha}$, 0) and B(0, $\frac{p}{\sin \alpha}$).

Let M($h$, $k$) be the mid-point of the segment AB. Using the midpoint formula:

From equation (viii):

From equation (ix):

We need to eliminate the parameter $\alpha$ from equations (x) and (xi) to find the locus of M($h$, $k$).

From equation (x), $\cos \alpha = \frac{p}{2h}$.

From equation (xi), $\sin \alpha = \frac{p}{2k}$.

Using the trigonometric identity $\sin^2 \alpha + \cos^2 \alpha = 1$:

Simplifying equation (xii):

Divide equation (xiii) by $p^2$ (assuming $p \neq 0$, otherwise the original line would pass through the origin, and the segment between axes would be just the origin, whose midpoint is the origin):

Multiply equation (xiv) by 4:

To find the equation of the locus of the mid-point M, we replace $h$ with $x$ and $k$ with $y$ in equation (xv).

Rearranging equation (xvi) gives the required form:

This is the equation of the locus of the mid-point of the segment intercepted between the axes by the variable line $x \cos \alpha + y \sin \alpha = p$.

Thus, the locus of the mid-point is $\frac{1}{x^2} + \frac{1}{y^2} = \frac{4}{p^2}$.

Given:

Point A: (2, 0)

Point B on the original line: (3, 1)

Angle of rotation about A: $15^\circ$ in the anticlockwise direction.

To Find:

The equation of the line in the new position.

Solution:

First, find the slope of the original line joining points A(2, 0) and B(3, 1).

The slope ($m_{AB}$) is given by the formula $m = \frac{y_2 - y_1}{x_2 - x_1}$.

From equation (i):

Let $\theta_{AB}$ be the angle the original line AB makes with the positive direction of the x-axis. Then $m_{AB} = \tan(\theta_{AB})$.

From equation (iv), the angle $\theta_{AB}$ is:

The line AB is rotated anticlockwise through an angle of $15^\circ$ about point A. Let the new line be AL, and let its angle with the positive x-axis be $\theta_{AL}$.

Substitute the value of $\theta_{AB}$ from equation (v) into equation (vi):

The slope of the new line AL ($m_{AL}$) is given by $m_{AL} = \tan(\theta_{AL})$.

From equation (ix):

The new line AL passes through the point A(2, 0) and has a slope $m_{AL} = \sqrt{3}$ (from equation (x)).

Using the point-slope form of the equation of a line $y - y_1 = m(x - x_1)$ with $(x_1, y_1) = (2, 0)$ and $m = \sqrt{3}$:

Simplify equation (xi):

Rearrange the terms of equation (xii) to express the equation in the standard form $Ax + By + C = 0$:

The equation of the line in the new position is $\sqrt{3}\text{x} - \text{y} - 2\sqrt{3} = 0$.

Given:

Point A: (3, 2)

Slope of the line passing through A: $m = \frac{3}{4}$

Distance from point A to the required points: $r = 5$ units.

To Find:

The coordinates of the points on the line which are 5 units away from point A.

Solution:

Let the point A be $(x_1, y_1) = (3, 2)$.

Let the slope of the line be $m = \frac{3}{4}$.

The slope of a line is given by $m = \tan \theta$, where $\theta$ is the angle the line makes with the positive direction of the x-axis.

From $\tan \theta = \frac{3}{4}$, we can find the values of $\sin \theta$ and $\cos \theta$. We can construct a right-angled triangle where the opposite side is 3 and the adjacent side is 4. The hypotenuse will be $\sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.

Since the slope is positive, the angle $\theta$ lies in the first or third quadrant. Thus, $\sin \theta$ and $\cos \theta$ must have the same sign.

We can take $\cos \theta = \frac{4}{5}$ and $\sin \theta = \frac{3}{5}$ (for the first quadrant) or $\cos \theta = -\frac{4}{5}$ and $\sin \theta = -\frac{3}{5}$ (for the third quadrant).

The coordinates of any point (x, y) on the line at a distance $r$ from the point $(x_1, y_1)$ are given by the parametric form (distance form) of the line equation:

In this problem, $(x_1, y_1) = (3, 2)$ and the distance is $r = \pm 5$ (since the points can be 5 units away from A in either direction along the line).

Using $\cos \theta = \frac{4}{5}$ and $\sin \theta = \frac{3}{5}$, and substituting the values into equations (ii) and (iii) for $r = 5$ and $r = -5$:

Case 1: For r = 5

From equation (iv):

From equation (v):

So, one point is (7, 5).

Case 2: For r = -5

From equation (viii):

From equation (ix):

So, the other point is (-1, -1).

The two points on the line which are 5 units away from the point A(3, 2) are (7, 5) and (-1, -1).

Given:

Equation of Line 1: $5x - 6y - 1 = 0$

Equation of Line 2: $3x + 2y + 5 = 0$

Equation of Line 3: $3x - 5y + 11 = 0$

To Find:

The equation of the straight line passing through the point of intersection of Line 1 and Line 2, and perpendicular to Line 3.

Solution:

Step 1: Find the point of intersection of the lines $5x - 6y - 1 = 0$ and $3x + 2y + 5 = 0$.

The given equations are:

Multiply equation (ii) by 3 to make the coefficient of y equal to that in equation (i):

Equation (iii) becomes:

Add equation (i) and equation (iv) to eliminate y:

From equation (vi), solve for x:

Substitute the value of x = -1 into equation (ii):

From equation (ix), solve for y:

The point of intersection of the two lines is (-1, -1).

Step 2: Find the slope of the line $3x - 5y + 11 = 0$.

The equation of Line 3 is:

The slope ($m_3$) of a line in the form $Ax + By + C = 0$ is given by $m = -\frac{A}{B}$.

Step 3: Find the slope of the required line.

The required line is perpendicular to Line 3. If $m_{req}$ is the slope of the required line, then the product of the slopes is -1:

Substitute the value of $m_3$ from equation (xv) into equation (xvi):

Solve for $m_{req}$ from equation (xvii):

Step 4: Find the equation of the required line.

The required line passes through the point of intersection (-1, -1) (from Step 1) and has a slope $m_{req} = -\frac{5}{3}$ (from Step 3).

Using the point-slope form of the equation of a line: $y - y_1 = m(x - x_1)$.

Substitute $(x_1, y_1) = (-1, -1)$ and $m = -\frac{5}{3}$:

Multiply both sides of equation (xx) by 3 to eliminate the denominator:

Expand both sides of equation (xxi):

Rearrange equation (xxii) to bring all terms to one side in the standard form $Ax + By + C = 0$:

The equation of the required straight line is $5x + 3y + 8 = 0$.

Given:

Starting point of the ray: P(1, 2).

Point of reflection on the x-axis: A.

Point the reflected ray passes through: Q(5, 3).

To Find:

The coordinates of the point A on the x-axis.

Solution:

Let the coordinates of the point A on the x-axis be $(x, 0)$.

When a ray of light is reflected from the x-axis, the incident ray and the reflected ray satisfy the law of reflection: the angle of incidence equals the angle of reflection.

An important property related to reflection is that the reflected ray, after reflection from a line (in this case, the x-axis), appears to come from the image of the source point in that line.

Let P' be the image of the point P(1, 2) in the x-axis.

The image of a point $(a, b)$ in the x-axis is $(a, -b)$.

So, the coordinates of P' are (1, -2).

According to the property of reflection, the reflected ray passing through A and Q must be collinear with the image point P'.

Therefore, the points P'(1, -2), A(x, 0), and Q(5, 3) are collinear.

For three points to be collinear, the slope between any two pairs of points must be equal.

We will use the condition that the slope of the segment P'A is equal to the slope of the segment P'Q.

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

Calculate the slope of the segment P'Q using P'(1, -2) and Q(5, 3):

Calculate the slope of the segment P'A using P'(1, -2) and A(x, 0):

Since P', A, and Q are collinear, their slopes must be equal:

Substitute the values from equations (iii) and (v) into equation (vi):

Cross-multiply equation (vii):

Add 5 to both sides of equation (ix):

Solve for x from equation (xi):

Since point A is on the x-axis, its coordinates are $(\frac{13}{5}, 0)$.

The coordinates of the point A where the ray is reflected from the x-axis are ($\frac{13}{5}$, 0).

Given:

Equation of one diagonal of the square: $8x - 15y = 0$.

Coordinates of one vertex of the square: (1, 2).

To Find:

The equations of the sides of the square passing through the given vertex.

Solution:

Let the given vertex be A(1, 2).

Let the equation of the given diagonal be $D_1: 8x - 15y = 0$.

First, let's check if the vertex A(1, 2) lies on the diagonal $D_1$. Substitute the coordinates of A into the equation of $D_1$:

Since -22 $\neq$ 0, the vertex A(1, 2) does not lie on the diagonal $D_1$. This means $D_1$ is the diagonal that does not pass through A. Let this diagonal be BD.

The other diagonal, AC, must pass through the vertex A(1, 2).

In a square, the diagonals are perpendicular to each other. Therefore, the diagonal AC is perpendicular to the diagonal BD.

The slope of the diagonal BD ($m_{BD}$) is found from the equation $8x - 15y = 0$. Rewrite the equation as $15y = 8x$, so $y = \frac{8}{15}x$.

Since AC is perpendicular to BD, the product of their slopes is -1. Let $m_{AC}$ be the slope of the diagonal AC.

Substitute $m_{BD}$ from equation (ii) into equation (iii):

From equation (iv):

The sides of the square passing through vertex A are AB and AD. The diagonal AC passes through A and bisects the angle $\angle BAD$, which is $90^\circ$. Thus, the angle between the diagonal AC and each of the sides AB and AD is $45^\circ$.

Let $m_s$ be the slope of a side of the square passing through A. The angle between the diagonal AC (with slope $m_{AC} = -\frac{15}{8}$) and the side (with slope $m_s$) is $45^\circ$.

Using the formula for the angle between two lines with slopes $m_1$ and $m_2$ is $\tan \phi = |\frac{m_1 - m_2}{1 + m_1 m_2}|$. Let $m_1 = m_{AC} = -\frac{15}{8}$ and $m_2 = m_s$, and $\phi = 45^\circ$ ($\tan 45^\circ = 1$).

From equation (viii), we have two possibilities:

Case 1: $\frac{-15 - 8m_s}{8 - 15m_s} = 1$

Case 2: $\frac{-15 - 8m_s}{8 - 15m_s} = -1$

The two slopes are $\frac{23}{7}$ and $-\frac{7}{23}$. These are the slopes of the two sides passing through the vertex A(1, 2).

Now, we use the point-slope form $y - y_1 = m(x - x_1)$ with $(x_1, y_1) = (1, 2)$ to find the equations of the sides.

Equation of the first side (with slope $\frac{23}{7}$):

Rearranging equation (xx) into the form $Ax + By + C = 0$:

Equation of the second side (with slope $-\frac{7}{23}$):

Rearranging equation (xxv) into the form $Ax + By + C = 0$:

The equations of the sides of the square passing through the vertex (1, 2) are $23x - 7y - 9 = 0$ and $7x + 23y - 53 = 0$.

Given:

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

To Find:

The inclination of the line with the positive direction of the x-axis.

Solution:

The equation of the line is:

We can rewrite the equation in the slope-intercept form, $y = mx + c$, where $m$ is the slope and $c$ is the y-intercept.

From equation (i):

Comparing equation (ii) with $y = mx + c$, the slope of the line is $m = 1$.

The inclination ($\theta$) of a line with the positive direction of the x-axis is related to its slope ($m$) by the formula:

Substitute the value of $m = 1$ into equation (iii):

For the principal value of the inclination, which lies in the range $[0^\circ, 180^\circ)$, the angle $\theta$ such that $\tan \theta = 1$ is $45^\circ$.

The inclination of the line $x - y + 3 = 0$ with the positive direction of the x-axis is $45^\circ$.

This corresponds to option (A).

Answer:

The correct option is (A) $45^\circ$.

Given:

Equation of Line 1: $ax + by = c$

Equation of Line 2: $a'x + b'y = c'$

To Find:

The condition for the two lines to be perpendicular.

Solution:

The slope of a line in the form $Ax + By + C = 0$ is given by $m = -\frac{A}{B}$.

Line 1: $ax + by - c = 0$. Let its slope be $m_1$.

Line 2: $a'x + b'y - c' = 0$. Let its slope be $m_2$.

Two lines are perpendicular if the product of their slopes is -1 (provided their slopes are defined and non-zero). If one slope is 0, the other must be undefined.

Case 1: $b \neq 0$ and $b' \neq 0$.

For the lines to be perpendicular, $m_1 \times m_2 = -1$.

Cross-multiplying equation (iv):

Rearranging equation (v):

Case 2: $b = 0$.

If $b = 0$, the first equation is $ax = c$. For this to be a line, $a \neq 0$. This is a vertical line (parallel to the y-axis). Its slope is undefined.

For a vertical line to be perpendicular to the second line $a'x + b'y = c'$, the second line must be horizontal (parallel to the x-axis).

A horizontal line has an equation of the form $y = k$, which in the form $a'x + b'y = c'$ implies $a' = 0$ (and $b' \neq 0$).

Let's check if the condition $aa' + bb' = 0$ holds for $b=0$ and $a'=0$ (with $a \neq 0$ and $b' \neq 0$):

The condition $aa' + bb' = 0$ holds.

Case 3: $b' = 0$.

If $b' = 0$, the second equation is $a'x = c'$. For this to be a line, $a' \neq 0$. This is a vertical line.

For a vertical line to be perpendicular to the first line $ax + by = c$, the first line must be horizontal.

A horizontal line implies $a = 0$ (and $b \neq 0$).

Let's check if the condition $aa' + bb' = 0$ holds for $b'=0$ and $a=0$ (with $a' \neq 0$ and $b \neq 0$):

The condition $aa' + bb' = 0$ holds.

Thus, the condition $aa' + bb' = 0$ covers all cases where the two lines $ax + by = c$ and $a'x + b'y = c'$ are perpendicular, provided that each equation represents a line (i.e., $a$ and $b$ are not both zero, and $a'$ and $b'$ are not both zero).

The condition for the two lines to be perpendicular is $aa' + bb' = 0$.

This corresponds to option (A).

Answer:

The correct option is (A) aa′ + bb′ = 0.

Given:

The line passes through the point (1, 2).

The required line is perpendicular to the line $x + y + 7 = 0$.

To Find:

The equation of the required line.

Solution:

The equation of the given line is:

The general form of a linear equation is $Ax + By + C = 0$. Comparing equation (i) with the general form, we have $A=1$, $B=1$, and $C=7$.

The slope of a line in the form $Ax + By + C = 0$ is given by $m = -\frac{A}{B}$.

Let the slope of the given line (equation (i)) be $m_1$.

From equation (ii):

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 $m_2$.

Substitute the value of $m_1$ from equation (iii) into equation (iv):

Solving for $m_2$ from equation (v):

From equation (vi):

The required line passes through the point $(x_1, y_1) = (1, 2)$ and has a slope $m = 1$ (from equation (vii)).

Using the point-slope form of the equation of a line: $y - y_1 = m(x - x_1)$.

Substitute the values of $(x_1, y_1)$ and $m$ into the point-slope form:

Simplify equation (viii):

Rearrange the terms of equation (ix) to bring them to one side in the form $Ax + By + C = 0$:

Now, we check the given options to see which one matches equation (xi).

(A) y – x + 1 = 0 $\implies$ -x + y + 1 = 0

(B) y – x – 1 = 0 $\implies$ -x + y - 1 = 0 or x - y + 1 = 0 (by multiplying by -1)

(C) y – x + 2 = 0 $\implies$ -x + y + 2 = 0

(D) y – x – 2 = 0 $\implies$ -x + y - 2 = 0

The equation $x - y + 1 = 0$ matches option (B) when rearranged or multiplied by -1.

The equation of the line passing through (1, 2) and perpendicular to $x + y + 7 = 0$ is $y - x - 1 = 0$ (or $x - y + 1 = 0$).

The correct option is (B) y – x – 1 = 0.

Answer:

The correct option is (B) y – x – 1 = 0.

Given:

Point P: (1, -3)

Equation of the line: $2y - 3x = 4$

To Find:

The distance of the point P from the line.

Solution:

The equation of the line is $2y - 3x = 4$. We need to rewrite it in the standard form $Ax + By + C = 0$:

Comparing equation (i) with $Ax + By + C = 0$, we have $A = -3$, $B = 2$, and $C = -4$.

The given point is $(x_1, y_1) = (1, -3)$.

The distance ($d$) of a point $(x_1, y_1)$ from a line $Ax + By + C = 0$ is given by the formula:

Substitute the values of $A$, $B$, $C$, $x_1$, and $y_1$ into equation (ii):

Simplify the numerator and the denominator of equation (iii):

To rationalize the denominator, multiply the numerator and denominator by $\sqrt{13}$:

Cancel out the common factor 13:

The distance of the point P(1, -3) from the line $2y - 3x = 4$ is $\sqrt{13}$ units.

This corresponds to option (C).

Answer:

The correct option is (C) $\sqrt{13}$.

Given:

Point P: (2, 3)

Equation of the line L: $x + y - 11 = 0$

To Find:

The coordinates of the foot of the perpendicular from point P to line L.

Solution:

Let the foot of the perpendicular from the point P(2, 3) to the line $x + y - 11 = 0$ be the point Q($h, k$).

Since the point Q($h, k$) lies on the line $x + y - 11 = 0$, its coordinates must satisfy the equation of the line:

The line segment PQ is perpendicular to the line $x + y - 11 = 0$.

First, find the slope of the line $x + y - 11 = 0$. The equation is in the form $Ax + By + C = 0$, where $A=1$, $B=1$, and $C=-11$.

The slope ($m_L$) of the line is given by $m_L = -\frac{A}{B}$.

Next, find the slope of the line segment PQ joining P(2, 3) and Q($h, k$).

The slope ($m_{PQ}$) is given by $m_{PQ} = \frac{y_2 - y_1}{x_2 - x_1}$.

Since PQ is perpendicular to the line L, the product of their slopes is -1:

Substitute the slopes from equations (ii) and (iii) into equation (iv):

Simplify equation (v):

Cross-multiply equation (vi):

Rearrange equation (vii) to get a second equation relating $h$ and $k$:

Now we have a system of two linear equations with two variables $h$ and $k$:

Add equation (x) and equation (xi) to eliminate $h$:

Solve for $k$ from equation (xiii):

Substitute the value of $k = 6$ into equation (x):

Solve for $h$ from equation (xv):

The coordinates of the foot of the perpendicular Q are ($h, k$) = (5, 6).

The coordinates of the foot of the perpendicular from the point (2, 3) on the line $x + y - 11 = 0$ are (5, 6).

This corresponds to option (B).

Answer:

The correct option is (B) (5, 6).

Given:

The intercept cut off by the line from the y-axis is twice that from the x-axis.

The line passes through the point (1, 2).

To Find:

The equation of the line.

Solution:

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

According to the given information, the y-intercept is twice the x-intercept.

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

Substitute $b = 2a$ from equation (i) into equation (ii):

Multiply equation (iii) by the least common multiple of the denominators, which is 2a (assuming $a \neq 0$. If $a=0$, then $b=0$, and the line passes through the origin. A line passing through (1, 2) and the origin would have a slope of $\frac{2}{1} = 2$, and the equation $y=2x$ or $2x-y=0$. This does not match any option, so $a \neq 0$).

Simplify equation (iv):

This is the equation of the line in terms of 'a'. We are given that the line passes through the point (1, 2). Substitute the coordinates of this point into equation (v) to find the value of 'a'.

From equation (viii), solve for 'a':

Now substitute the value of $a = 2$ back into equation (v) to get the equation of the line.

This equation can also be written as $2x + y - 4 = 0$.

The equation of the line is $2x + y = 4$.

This corresponds to option (A).

Answer:

The correct option is (A) 2x + y = 4.

Given:

The line passes through the point P (1, 2).

The intercept of the line between the coordinate axes is bisected at point P.

To Find:

The equation of the line.

Solution:

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

Let the x-intercept be 'a'. The coordinates of point A on the x-axis are $(a, 0)$.

Let the y-intercept be 'b'. The coordinates of point B on the y-axis are $(0, b)$.

The line segment intercepted between the axes is AB.

We are given that the point P(1, 2) is the mid-point of the segment AB.

Using the mid-point formula, the coordinates of the mid-point of AB are $\left(\frac{a + 0}{2}, \frac{0 + b}{2}\right) = \left(\frac{a}{2}, \frac{b}{2}\right)$.

Since P(1, 2) is the mid-point, we equate the coordinates:

From equation (i):

From equation (ii):

Now we have the x-intercept $a = 2$ and the y-intercept $b = 4$.

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

Substitute the values of $a = 2$ and $b = 4$ from equations (iii) and (iv) into equation (v):

To eliminate the denominators, multiply both sides of equation (vi) by the least common multiple of 2 and 4, which is 4:

Simplify equation (vii):

Rearrange equation (ix) into the standard form $Ax + By + C = 0$:

Now, compare equation (x) with the given options:

(A) x + 2y = 5 $\implies$ x + 2y - 5 = 0

(B) x – y + 1 = 0

(C) x + y – 3 = 0

(D) 2x + y – 4 = 0

Equation (x) matches option (D).

The equation of the line is $2x + y - 4 = 0$.

This corresponds to option (D).

Answer:

The correct option is (D) 2x + y – 4 = 0.

Given:

The original point P: (4, -13)

The line of reflection L: $5x + y + 6 = 0$

To Find:

The coordinates of the reflection of point P about the line L.

Solution:

Let the reflection of the point P(4, -13) about the line $5x + y + 6 = 0$ be P'($x'$, $y'$).

There are two key properties of the reflection:

1. The line segment PP' is perpendicular to the line of reflection L.

2. The midpoint of the line segment PP' lies on the line of reflection L.

Step 1: Use the midpoint property.

The midpoint M of the segment PP' has coordinates $\left(\frac{4 + x'}{2}, \frac{-13 + y'}{2}\right)$.

Since M lies on the line $5x + y + 6 = 0$, its coordinates must satisfy the equation:

Multiply equation (i) by 2 to clear the denominators:

Expand and simplify equation (ii):

Step 2: Use the perpendicularity property.

The slope of the line of reflection L ($5x + y + 6 = 0$) is $m_L = -\frac{A}{B} = -\frac{5}{1} = -5$.

The slope of the segment PP' joining P(4, -13) and P'($x'$, $y'$) is $m_{PP'} = \frac{y' - (-13)}{x' - 4} = \frac{y' + 13}{x' - 4}$.

Since PP' is perpendicular to L, the product of their slopes is -1:

Substitute the slopes into equation (v):

Simplify equation (vi):

Cross-multiply equation (vii):

Rearrange equation (ix):

Step 3: Solve the system of linear equations for x' and y'.

We have the system:

Multiply equation (A) by 5:

Add equation (B) and equation (C):

Solve for x' from equation (xiv):

Substitute the value of $x' = -1$ into equation (A):

Solve for y' from equation (xvii):

The coordinates of the reflection P' are (-1, -14).

The coordinates of the reflection of the point (4, -13) about the line $5x + y + 6 = 0$ are (-1, -14).

This corresponds to option (A).

Answer:

The correct option is (A) (– 1, – 14).

Given:

Fixed point F: (4, 0)

Fixed line L (Directrix): $x = 16$, which can be written as $x - 16 = 0$.

The distance of a moving point P from the fixed point is half its distance from the fixed line.

To Find:

The locus of the moving point.

Solution:

Let the moving point be P(x, y).

The distance of point P(x, y) from the fixed point F(4, 0) is given by the distance formula:

The distance of point P(x, y) from the line $x - 16 = 0$ is given by the perpendicular distance formula from a point to a line $Ax + By + C = 0$, which is $\frac{|Ax + By + C|}{\sqrt{A^2 + B^2}}$.

For the line $x - 16 = 0$, we have $A=1$, $B=0$, and $C=-16$.

According to the given condition, the distance from P to F is half the distance from P to L:

Substitute the expressions for the distances from equations (ii) and (v) into equation (vi):

To find the locus, we need to remove the square root and absolute value. Square both sides of equation (vii):

Expand the squared terms in equation (ix):

Multiply equation (x) by 4 to clear the fraction:

Move all terms to one side of equation (xii):

Combine like terms in equation (xiii):

Rearrange equation (xiv) to match the options:

This is the equation of the locus of the point P(x, y).

The equation of the locus is $3x^2 + 4y^2 = 192$.

This corresponds to option (A).

Answer:

The correct option is (A) 3x² + 4y² = 192.

Given:

The line passes through the point (1, -2).

The line cuts off equal intercepts from the axes.

To Find:

The equation of the straight line.

Solution:

Let the equation of the straight line be in the intercept form:

where 'a' is the x-intercept and 'b' is the y-intercept.

According to the problem, the line cuts off equal intercepts from the axes. This means:

Substitute $b = a$ from equation (ii) into equation (i):

We can simplify equation (iii) by multiplying both sides by 'a' (assuming $a \neq 0$ since the point (1, -2) is not the origin, a line passing through a non-origin point and having zero intercepts ($a=b=0$) would imply the line passes through the origin, which is not the case here for non-zero intercepts).

The line passes through the point (1, -2). Substitute the coordinates of this point into equation (iv):

From equation (v):

Since $a = -1 \neq 0$, our assumption was valid. Substitute the value of $a = -1$ from equation (vi) back into equation (iv):

Rearrange equation (vii) into the standard form $Ax + By + C = 0$:

This line $x+y+1=0$ has x-intercept (-1) and y-intercept (-1), which are equal. It passes through (1, -2) since $1 + (-2) + 1 = 0$.

The equation of the straight line is $x + y + 1 = 0$.

Given:

The required line passes through the point P (5, 2).

The required line is perpendicular to the line joining points A (2, 3) and B (3, -1).

To Find:

The equation of the line.

Solution:

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

Let the coordinates of A be $(x_1, y_1) = (2, 3)$ and the coordinates of B be $(x_2, y_2) = (3, -1)$.

The slope of the line AB ($m_{AB}$) is given by the formula $m = \frac{y_2 - y_1}{x_2 - x_1}$.

From equation (i):

The required line is perpendicular to the line AB. If two lines are perpendicular, the product of their slopes is -1.

Let the slope of the required line be $m_{req}$.

Substitute the value of $m_{AB}$ from equation (iii) into equation (iv):

Solve for $m_{req}$ from equation (v):

The required line passes through the point P(5, 2) and has a slope $m = \frac{1}{4}$ (from equation (vii)).

Using the point-slope form of the equation of a line: $y - y_1 = m(x - x_1)$.

Substitute the coordinates of P $(x_1, y_1) = (5, 2)$ and the slope $m = \frac{1}{4}$ into the point-slope form:

Multiply both sides of equation (viii) by 4 to eliminate the denominator:

Expand both sides of equation (ix):

Rearrange equation (x) to bring all terms to one side in the standard form $Ax + By + C = 0$:

The equation of the line passing through (5, 2) and perpendicular to the line joining (2, 3) and (3, -1) is $x - 4y + 3 = 0$.

Given:

Equation of Line 1: $y = (2 – \sqrt{3}) (x + 5)$

Equation of Line 2: $y = (2 + \sqrt{3}) (x – 7)$

To Find:

The angle between the two lines.

Solution:

The equations of the lines are given in the slope-intercept form $y = mx + c$, where $m$ is the slope.

For Line 1, the equation is $y = (2 - \sqrt{3})(x + 5)$. The slope of Line 1 ($m_1$) is the coefficient of x.

For Line 2, the equation is $y = (2 + \sqrt{3})(x - 7)$. The slope of Line 2 ($m_2$) is the coefficient of x.

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

First, calculate the product $m_1 m_2$:

Using the difference of squares formula $(a - b)(a + b) = a^2 - b^2$:

Next, calculate the difference $m_1 - m_2$:

Now substitute the values of $(m_1 - m_2)$ and $(1 + m_1 m_2)$ into the formula for $\tan \theta$ (equation (iii)):

We need to find the angle $\theta$ whose tangent is $\sqrt{3}$. The principal value of such an angle is $60^\circ$ or $\frac{\pi}{3}$ radians.

The angle between the two lines is typically taken as the acute angle unless otherwise specified. Since $60^\circ$ is an acute angle, this is the required angle.

The angle between the lines $y = (2 – \sqrt{3}) (x + 5)$ and $y = (2 + \sqrt{3}) (x – 7)$ is $60^\circ$.

Given:

The line passes through the point (3, 4).

The sum of the x-intercept and y-intercept is 14.

To Find:

The equation of the straight line.

Solution:

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

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

According to the problem statement, the sum of the intercepts is 14:

From equation (ii), we can express $b$ in terms of $a$:

Substitute $b = 14 - a$ from equation (iii) into the intercept form (equation (i)):

The line passes through the point (3, 4). Substitute $x = 3$ and $y = 4$ into equation (iv):

To solve for $a$, find a common denominator for the fractions in equation (v):

Multiply both sides of equation (vi) by $a(14 - a)$ (assuming $a \neq 0$ and $a \neq 14$):

Expand and simplify equation (vii):

Rearrange equation (ix) into a quadratic equation in $a$:

Solve the quadratic equation (xi) for $a$ by factoring:

This gives two possible values for $a$:

Now find the corresponding values for $b$ using $b = 14 - a$ (equation (iii)).

Case 1: If $a = 6$

The intercepts are $a=6$ and $b=8$. Substitute these values into the intercept form (equation (i)):

Multiply equation (xv) by the least common multiple of 6 and 8, which is 24:

Rearrange equation (xvii) into the standard form:

Case 2: If $a = 7$

The intercepts are $a=7$ and $b=7$. Substitute these values into the intercept form (equation (i)):

Multiply equation (xx) by 7:

Rearrange equation (xxii) into the standard form:

There are two such lines that satisfy the given conditions.

The equations of the lines are $4x + 3y - 24 = 0$ and $x + y - 7 = 0$.

Given:

Equation of the first line (L1): $x + y = 4$.

Equation of the second line (L2): $4x + 3y = 10$.

The distance from the points on L1 to L2 is 1 unit.

To Find:

The coordinates of the points on line L1 that are at a unit distance from line L2.

Solution:

Let the coordinates of a point on the line $x + y = 4$ be $(x_1, y_1)$. Since the point lies on the line, we have $y_1 = 4 - x_1$. So, the coordinates of any point on L1 can be represented as $(x_1, 4 - x_1)$.

The equation of the second line is $4x + 3y = 10$. We rewrite this in the standard form $Ax + By + C = 0$:

Comparing equation (i) with $Ax + By + C = 0$, we have $A=4$, $B=3$, and $C=-10$.

The distance ($d$) of a point $(x_1, y_1)$ from a line $Ax + By + C = 0$ is given by the formula:

We are given that the distance is 1 unit. Substitute $d=1$, $A=4$, $B=3$, $C=-10$, $x_1=x_1$, and $y_1=4-x_1$ into equation (ii):

Simplify the numerator and the denominator of equation (iii):

From equation (vi), we have:

The absolute value equation (vii) gives two possibilities for $x_1 + 2$:

Case 1: $x_1 + 2 = 5$

If $x_1 = 3$, the corresponding $y_1$ is $y_1 = 4 - x_1 = 4 - 3 = 1$.

The first point is (3, 1).

Case 2: $x_1 + 2 = -5$

If $x_1 = -7$, the corresponding $y_1$ is $y_1 = 4 - x_1 = 4 - (-7) = 4 + 7 = 11$.

The second point is (-7, 11).

The two points on the line $x + y = 4$ that lie at a unit distance from the line $4x + 3y = 10$ are (3, 1) and (-7, 11).

Given:

Equation of Line 1: $\frac{x}{a} + \frac{y}{b} = 1$

Equation of Line 2: $\frac{x}{a} - \frac{y}{b} = 1$

To Show:

The tangent of the angle between the lines is $\frac{2ab}{a^2 - b^2}$.

Solution:

First, find the slopes of the given lines. We can rewrite the equations in the slope-intercept form $y = mx + c$ or the standard form $Ax + By + C = 0$.

Line 1: $\frac{x}{a} + \frac{y}{b} = 1$. Multiply by $ab$ (assuming $a \neq 0$ and $b \neq 0$) to eliminate denominators:

Rearrange equation (i) into the standard form $Ax + By + C = 0$:

The slope ($m_1$) of Line 1 is $m_1 = -\frac{A}{B}$.

Line 2: $\frac{x}{a} - \frac{y}{b} = 1$. Multiply by $ab$ (assuming $a \neq 0$ and $b \neq 0$) to eliminate denominators:

Rearrange equation (iv) into the standard form $Ax + By + C = 0$:

The slope ($m_2$) of Line 2 is $m_2 = -\frac{A}{B}$.

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

Substitute the values of $m_1 = -\frac{b}{a}$ and $m_2 = \frac{b}{a}$ into equation (vii):

Simplify the numerator and the denominator of equation (viii):

Find a common denominator in the denominator:

Invert the denominator and multiply:

Simplify equation (xi):

The tangent of the angle between the lines is usually taken as the positive value (the acute angle), so we remove the absolute value.

If $a^2 \neq b^2$, then $\frac{2ab}{a^2 - b^2}$ can be positive or negative depending on the values of a and b. The magnitude of the tangent is $\frac{2|ab|}{|a^2 - b^2|}$. However, the problem asks to show that the tangent is $\frac{2ab}{a^2 - b^2}$, which implies considering the directional angle or the specific angle $\theta$ such that $\tan \theta = \frac{m_1 - m_2}{1 + m_1 m_2}$.

Let's reconsider equation (ix) without the absolute value for one of the angles between the lines:

The angles between two lines are $\theta$ and $180^\circ - \theta$. If $\tan \theta = K$, then $\tan(180^\circ - \theta) = -\tan \theta = -K$. So, the tangents of the angles between the lines are $\frac{-2ab}{a^2 - b^2}$ and $\frac{2ab}{a^2 - b^2}$.

The question asks to show that "the tangent of an angle" is $\frac{2ab}{a^2 - b^2}$, implying that this value is one of the possible tangents of the angles between the lines.

We have found $\tan \theta = -\frac{2ab}{a^2 - b^2}$. The other angle will have a tangent of $\frac{2ab}{a^2 - b^2}$.

This holds true provided $a^2 \neq b^2$ (so the denominator is not zero) and $a \neq 0, b \neq 0$ (so the slopes are defined).

If $a=0$, the first line is $y/b = 1 \implies y = b$ (horizontal), $m_1 = 0$. The second line is $-y/b = 1 \implies y = -b$ (horizontal), $m_2 = 0$. If $b \neq 0$, these are parallel distinct lines, angle is 0, $\tan 0 = 0$. Formula gives $\frac{2(0)b}{0^2 - b^2} = 0$. Holds.

If $b=0$, the first line is $x/a = 1 \implies x = a$ (vertical), $m_1$ is undefined. The second line is $x/a = 1 \implies x = a$ (vertical), $m_2$ is undefined. If $a \neq 0$, these are the same line, angle is 0, $\tan 0 = 0$. Formula gives $\frac{2a(0)}{a^2 - 0^2} = 0$. Holds.

If $a^2 = b^2$, then $a = \pm b$. If $a=b$, the slopes are $m_1 = -\frac{b}{b} = -1$ and $m_2 = \frac{b}{b} = 1$. $m_1 m_2 = (-1)(1) = -1$, so the lines are perpendicular. The angle is $90^\circ$, and the tangent is undefined. The denominator $a^2 - b^2 = a^2 - a^2 = 0$. The formula is undefined, which is consistent.

If $a=-b$, the slopes are $m_1 = -\frac{-a}{a} = 1$ and $m_2 = \frac{-a}{a} = -1$. $m_1 m_2 = (1)(-1) = -1$, so the lines are perpendicular. The angle is $90^\circ$, tangent is undefined. The denominator $a^2 - b^2 = a^2 - (-a)^2 = a^2 - a^2 = 0$. The formula is undefined, consistent.

Therefore, one of the tangents of the angle between the lines is indeed $\frac{2ab}{a^2 - b^2}$.

The tangent of the angle between the lines $\frac{x}{a} + \frac{y}{b} = 1$ and $\frac{x}{a} - \frac{y}{b} = 1$ is $\frac{2ab}{a^2 - b^2}$.

Given:

The line passes through the point (1, 2).

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

To Find:

The equation of the line(s).

Solution:

Let the angle that the line makes with the positive direction of the x-axis be $\theta$.

The angle made by the line with the y-axis is given as $30^\circ$. A line making an angle $\phi$ with the y-axis makes an angle $\theta$ with the x-axis such that $\theta = 90^\circ \pm \phi$.

In this case, $\phi = 30^\circ$. So, the possible angles with the positive x-axis are:

There are two such lines passing through the point (1, 2).

The slope ($m$) of a line with inclination $\theta$ is given by $m = \tan \theta$.

Case 1: Inclination $\theta_1 = 60^\circ$

The slope $m_1$ is:

The line passes through $(x_1, y_1) = (1, 2)$. Using the point-slope form $y - y_1 = m(x - x_1)$:

Simplify and rearrange equation (v):

Case 2: Inclination $\theta_2 = 120^\circ$

The slope $m_2$ is:

We know that $\tan(120^\circ) = \tan(180^\circ - 60^\circ) = -\tan(60^\circ)$.

The line passes through $(x_1, y_1) = (1, 2)$. Using the point-slope form $y - y_1 = m(x - x_1)$:

Simplify and rearrange equation (x):

The equations of the lines passing through (1, 2) and making an angle of $30^\circ$ with the y-axis are $\sqrt{3}\text{x} - \text{y} + (2 - \sqrt{3}) = 0$ and $\sqrt{3}\text{x} + \text{y} - (2 + \sqrt{3}) = 0$.

Given:

Line 1: $2x + y = 5$

Line 2: $x + 3y + 8 = 0$

Line 3: $3x + 4y = 7$

To Find:

The equation of the line passing through the point of intersection of Line 1 and Line 2, and parallel to Line 3.

Solution:

Step 1: Find the point of intersection of the lines $2x + y = 5$ and $x + 3y + 8 = 0$.

The given equations are:

From equation (i), express y in terms of x:

Substitute the expression for y from equation (iii) into equation (ii):

Simplify and solve for x from equation (iv):

Substitute the value of x from equation (viii) into equation (iii) to find y:

The point of intersection is P($\frac{23}{5}, -\frac{21}{5}$).

Step 2: Find the slope of the line $3x + 4y = 7$.

The equation of Line 3 is $3x + 4y = 7$. Rewrite it in the standard form $Ax + By + C = 0$:

The slope ($m_3$) of a line in the form $Ax + By + C = 0$ is given by $m = -\frac{A}{B}$.

Step 3: Determine the slope of the required line.

The required line is parallel to Line 3. If two lines are parallel, their slopes are equal.

Let the slope of the required line be $m_{req}$.

Step 4: Find the equation of the required line.

The required line passes through the point P($\frac{23}{5}, -\frac{21}{5}$) (from Step 1) and has a slope $m = -\frac{3}{4}$ (from equation (xiv)).

Using the point-slope form of the equation of a line: $y - y_1 = m(x - x_1)$.

Substitute $(x_1, y_1) = (\frac{23}{5}, -\frac{21}{5})$ and $m = -\frac{3}{4}$:

Multiply both sides of equation (xvi) by 20 (the LCM of 5 and 4) to eliminate denominators:

Rearrange equation (xxi) to bring all terms to one side in the standard form $Ax + By + C = 0$:

Divide the entire equation (xxiii) by the common factor 5:

The equation of the line passing through the point of intersection of $2x + y = 5$ and $x + 3y + 8 = 0$ and parallel to the line $3x + 4y = 7$ is $3x + 4y + 3 = 0$.

Given:

Equation of the first line (L1): $ax + by + 8 = 0$.

Equation of the second line (L2): $2x - 3y + 6 = 0$.

The intercepts cut off on the coordinate axes by L1 are equal in length but opposite in signs to those cut off by L2.

To Find:

The values of $a$ and $b$.

Solution:

First, find the intercepts cut off by the line $2x - 3y + 6 = 0$ on the coordinate axes.

The equation of Line 2 is:

To find the x-intercept of Line 2, set $y = 0$ in equation (i):

To find the y-intercept of Line 2, set $x = 0$ in equation (i):

So, the x-intercept of L2 is -3 and the y-intercept of L2 is 2.

Next, find the intercepts cut off by the line $ax + by + 8 = 0$ on the coordinate axes.

The equation of Line 1 is:

To find the x-intercept of Line 1, set $y = 0$ in equation (x):

Assuming $a \neq 0$, the x-intercept of L1 is $\frac{-8}{a}$.

To find the y-intercept of Line 1, set $x = 0$ in equation (x):

Assuming $b \neq 0$, the y-intercept of L1 is $\frac{-8}{b}$.

According to the given condition, the intercepts of L1 are equal in length but opposite in signs to those of L2.

Let the x-intercept of L1 be $X_1$ and the y-intercept of L1 be $Y_1$.

Let the x-intercept of L2 be $X_2 = -3$ and the y-intercept of L2 be $Y_2 = 2$.

The condition means:

Substitute the intercept values into equations (xv) and (xvi):

Simplify equation (xvii):

Solve for $a$ from equation (xix):

Simplify equation (xviii):

Solve for $b$ from equation (xxii):

Our assumed conditions $a \neq 0$ and $b \neq 0$ are satisfied by the calculated values $a = -\frac{8}{3}$ and $b = 4$.

The values of $a$ and $b$ are $a = -\frac{8}{3}$ and $b = 4$.

Given:

The line cuts off an intercept between the coordinate axes.

The point P(-5, 4) divides this intercept in the ratio 1:2.

To Find:

The equation of the line.

Solution:

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

Let the coordinates of the x-intercept be A($a, 0$) and the coordinates of the y-intercept be B($0, b$).

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

The point P(-5, 4) divides the line segment AB in the ratio 1:2. The phrasing does not specify the order of division (from A to B, or from B to A). We consider both cases.

Case 1: P(-5, 4) divides the segment AB in the ratio 1:2 (AP : PB = 1 : 2).

Using the section formula, the coordinates of a point dividing a segment joining ($x_1, y_1$) and ($x_2, y_2$) in the ratio $m:n$ are $\left(\frac{nx_1 + mx_2}{m + n}, \frac{ny_1 + my_2}{m + n}\right)$.

Here, $(x_1, y_1) = (a, 0)$, $(x_2, y_2) = (0, b)$, $(x, y) = (-5, 4)$, $m = 1$, $n = 2$.

Equating the coordinates of P:

Substitute the values of $a = -\frac{15}{2}$ and $b = 12$ into the intercept form (equation (i)):

Multiply equation (ix) by the least common multiple of 15 and 12, which is 60:

Rearrange equation (xii) into the standard form:

Case 2: P(-5, 4) divides the segment BA in the ratio 1:2 (BP : PA = 1 : 2).

Here, $(x_1, y_1) = (0, b)$, $(x_2, y_2) = (a, 0)$, $(x, y) = (-5, 4)$, $m = 1$, $n = 2$.

Equating the coordinates of P:

Substitute the values of $a = -15$ and $b = 6$ into the intercept form (equation (i)):

Multiply equation (xx) by the least common multiple of 15 and 6, which is 30:

Rearrange equation (xxiii) into the standard form:

Both equations (xiii) and (xxiv) are valid lines satisfying the given condition based on the direction of the ratio.

The equations of the lines are $8x - 5y + 60 = 0$ and $2x - 5y + 30 = 0$.

Given:

Length of the perpendicular from the origin to the line ($p$) = 4 units.

The line makes an angle ($\theta$) of $120^\circ$ with the positive direction of the x-axis.

To Find:

The equation of the straight line.

Solution:

The equation of a straight line can be expressed in the Normal Form, which is $x \cos \omega + y \sin \omega = p$.

In this form, $p$ is the length of the perpendicular from the origin to the line, and $\omega$ is the angle which the perpendicular from the origin to the line makes with the positive direction of the x-axis.

We are given $p = 4$ units.

We are given that the line itself makes an angle $\theta = 120^\circ$ with the positive x-axis.

The perpendicular from the origin to the line makes an angle $\omega$ with the positive x-axis, where $\omega$ is related to $\theta$ by $\omega = \theta \pm 90^\circ$.

In this case, $\theta = 120^\circ$.

Possible values for $\omega$ are $120^\circ - 90^\circ = 30^\circ$ or $120^\circ + 90^\circ = 210^\circ$.

The hint suggests using $\omega = 30^\circ$. Let's use this value.

The equation of the line in Normal Form is:

Substitute $p = 4$ and $\omega = 30^\circ$ into equation (ii):

We know that $\cos 30^\circ = \frac{\sqrt{3}}{2}$ and $\sin 30^\circ = \frac{1}{2}$.

Substitute these trigonometric values into equation (iii):

Multiply both sides of equation (iv) by 2 to eliminate the denominators:

Simplifying equation (v):

Rearrange equation (vi) to write it in the standard form $Ax + By + C = 0$:

Using $\omega = 210^\circ$ would give the parallel line $\sqrt{3}x + y + 8 = 0$, which is also 4 units away from the origin, but the hint guides us to use $\omega = 30^\circ$.

The equation of the straight line is $\sqrt{3}\text{x} + \text{y} - 8 = 0$.

Given:

Equation of the hypotenuse of an isosceles right-angled triangle: $3x + 4y = 4$.

Coordinates of the vertex opposite the hypotenuse (the right-angle vertex): A(2, 2).

To Find:

The equation of one of the sides of the triangle passing through the vertex (2, 2).

Solution:

Let the vertex opposite the hypotenuse be A(2, 2).

The sides passing through vertex A are the two equal legs of the isosceles right-angled triangle. These legs are perpendicular to each other, forming the right angle at A.

In an isosceles right-angled triangle, the angles opposite the equal sides are equal, and their sum is $180^\circ - 90^\circ = 90^\circ$. Thus, each of these angles is $45^\circ$. These are the angles between the hypotenuse and the two legs.

Let the equation of the hypotenuse be $H: 3x + 4y = 4$. We rewrite this equation in the slope-intercept form $y = mx + c$ to find its slope.

The slope of the hypotenuse ($m_h$) is the coefficient of x in equation (ii).

Let $m_s$ be the slope of one of the sides (legs) passing through A(2, 2). The angle between this side and the hypotenuse is $45^\circ$.

The tangent of the angle ($\phi$) between two lines with slopes $m_1$ and $m_2$ is given by $\tan \phi = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right|$.

Here, $\phi = 45^\circ$, $m_1 = m_s$, and $m_2 = m_h = -\frac{3}{4}$. $\tan 45^\circ = 1$.

Equation (vii) gives two possibilities:

Case 1: $\frac{4m_s + 3}{4 - 3m_s} = 1$

The first side passes through (2, 2) and has slope $\frac{1}{7}$. Using the point-slope form $y - y_1 = m(x - x_1)$:

Case 2: $\frac{4m_s + 3}{4 - 3m_s} = -1$

The second side passes through (2, 2) and has slope -7. Using the point-slope form $y - y_1 = m(x - x_1)$:

The equations of the two sides of the isosceles right-angled triangle passing through the vertex (2, 2) are $x - 7y + 12 = 0$ and $7x + y - 16 = 0$. The question asks for the equation of one of the sides.

The equation of one of the sides is $x - 7y + 12 = 0$ or $7x + y - 16 = 0$.

Given:

Equation of the base of an equilateral triangle is $x + y = 2$.

Vertex of the triangle is $(2, -1)$.

To Find:

Length of the side of the triangle.

Solution:

The equation of the base is $x + y = 2$. We can rewrite this in the general form $Ax + By + C = 0$ as $x + y - 2 = 0$.

Comparing this with the standard form, we have $A = 1$, $B = 1$, and $C = -2$.

The given vertex (a point not on the base) is $(x_1, y_1) = (2, -1)$.

The length of the perpendicular ($p$) from a point $(x_1, y_1)$ to the line $Ax + By + C = 0$ is given by the formula:

$p = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$

Substituting the values of the point and the line coefficients into the formula, we get:

$p = \frac{|-1|}{\sqrt{2}}$

$p = \frac{1}{\sqrt{2}}$

This perpendicular distance $p$ represents the altitude of the equilateral triangle from the given vertex to the base.

Let $l$ be the length of the side of the equilateral triangle.

In an equilateral triangle, the relationship between the altitude ($p$) and the side length ($l$) is given by:

We know that the value of $\sin 60^\circ$ is $\frac{\sqrt{3}}{2}$.

Substitute the value of $p$ we calculated and the value of $\sin 60^\circ$ into equation (iii):

$\frac{1}{\sqrt{2}} = l \times \frac{\sqrt{3}}{2}$

Now, we solve for $l$ by isolating it:

$l = \frac{1}{\sqrt{2}} \times \frac{2}{\sqrt{3}}$

$l = \frac{2}{\sqrt{2}\sqrt{3}}$

$l = \frac{2}{\sqrt{6}}$

To rationalize the denominator, multiply both the numerator and the denominator by $\sqrt{6}$:

$l = \frac{2}{\sqrt{6}} \times \frac{\sqrt{6}}{\sqrt{6}}$

$l = \frac{2\sqrt{6}}{6}$

We can simplify the fraction by cancelling the common factor of 2 in the numerator and the denominator:

$l = \frac{\cancel{2}^1 \sqrt{6}}{\cancel{6}_3}$

$l = \frac{\sqrt{6}}{3}$

Final Answer:

The length of the side of the equilateral triangle is $\frac{\sqrt{6}}{3}$.

Given:

Three points are (2, 0), (0, 2), and (1, 1).

A variable line passes through a fixed point P.

The algebraic sum of the perpendiculars drawn from the three given points to the line is zero.

To Find:

The coordinates of the fixed point P.

Solution:

Let the fixed point through which the variable line passes be $P(x_1, y_1)$.

Let the slope of the variable line be $m$.

The equation of the line passing through $P(x_1, y_1)$ with slope $m$ is given by:

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

Rearranging this equation into the general form $Ax + By + C = 0$, we get:

$mx - y + (y_1 - mx_1) = 0$

Here, $A = m$, $B = -1$, and $C = y_1 - mx_1$.

The algebraic perpendicular distance from a point $(x_0, y_0)$ to the line $Ax + By + C = 0$ is given by $\frac{Ax_0 + By_0 + C}{\sqrt{A^2 + B^2}}$. Note that we do not take the absolute value for the algebraic sum.

Let $d_1$, $d_2$, and $d_3$ be the algebraic perpendicular distances from the points (2, 0), (0, 2), and (1, 1) respectively to the line $mx - y + (y_1 - mx_1) = 0$.

For the point (2, 0):

$d_1 = \frac{m(2) - (0) + (y_1 - mx_1)}{\sqrt{m^2 + (-1)^2}} = \frac{2m + y_1 - mx_1}{\sqrt{m^2 + 1}}$

For the point (0, 2):

$d_2 = \frac{m(0) - (2) + (y_1 - mx_1)}{\sqrt{m^2 + 1}} = \frac{-2 + y_1 - mx_1}{\sqrt{m^2 + 1}}$

For the point (1, 1):

$d_3 = \frac{m(1) - (1) + (y_1 - mx_1)}{\sqrt{m^2 + 1}} = \frac{m - 1 + y_1 - mx_1}{\sqrt{m^2 + 1}}$

According to the problem, the algebraic sum of these distances is zero:

Substituting the expressions for $d_1$, $d_2$, and $d_3$ into equation (i):

$\frac{2m + y_1 - mx_1}{\sqrt{m^2 + 1}} + \frac{-2 + y_1 - mx_1}{\sqrt{m^2 + 1}} + \frac{m - 1 + y_1 - mx_1}{\sqrt{m^2 + 1}} = 0$

Since the denominator $\sqrt{m^2 + 1}$ is non-zero for any real value of $m$, we can multiply the entire equation by $\sqrt{m^2 + 1}$:

$(2m + y_1 - mx_1) + (-2 + y_1 - mx_1) + (m - 1 + y_1 - mx_1) = 0$

Combine like terms:

$(2m + m) + (y_1 + y_1 + y_1) + (-mx_1 - mx_1 - mx_1) + (-2 - 1) = 0$

$3m + 3y_1 - 3mx_1 - 3 = 0$

Divide the entire equation by 3:

$m + y_1 - mx_1 - 1 = 0$

Rearrange the terms to group those involving $m$:

This equation must hold true for any variable line passing through $P(x_1, y_1)$. This means equation (ii) must be true for any possible slope $m$ of the variable line (as long as the line is not vertical, a case we can check separately or handle by considering the form $x=k$).

For a linear equation in $m$, $(1 - x_1)m + (y_1 - 1) = 0$, to be true for all values of $m$, the coefficient of $m$ must be zero, and the constant term must also be zero.

From $1 - x_1 = 0$, we get $x_1 = 1$.

From $y_1 - 1 = 0$, we get $y_1 = 1$.

Thus, the coordinates of the fixed point P are $(1, 1)$.

We can verify this by checking the case of a vertical line through (1, 1), which has the equation $x = 1$, or $x - 1 = 0$. The distances from (2, 0), (0, 2), and (1, 1) are $\frac{2-1}{\sqrt{1^2+0^2}} = 1$, $\frac{0-1}{\sqrt{1^2+0^2}} = -1$, and $\frac{1-1}{\sqrt{1^2+0^2}} = 0$. The sum is $1 + (-1) + 0 = 0$, which is correct.

Final Answer:

The coordinates of the fixed point P are $(1, 1)$.

Given:

Point P is $(1, 2)$.

Equation of the line is $x + y = 4$.

The distance from P to the point of intersection Q is $\frac{\sqrt{6}}{3}$.

To Find:

The direction (slope or angle with x-axis) of the line passing through P.

Solution:

Let the equation of the line passing through the point $P(1, 2)$ be $y - 2 = m(x - 1)$, where $m$ is the slope of the line.

We can rewrite this equation in the form $mx - y + (2 - m) = 0$.

Let the point of intersection of this line with the line $x + y = 4$ be $Q(x, y)$.

The coordinates of Q must satisfy both equations:

From equation (ii), we have $y = 4 - x$. Substitute this into equation (i):

$mx - (4 - x) + (2 - m) = 0$

$mx - 4 + x + 2 - m = 0$

Group the terms with $x$ and the constant terms:

$(m + 1)x - 2 - m = 0$

$(m + 1)x = m + 2$

If $m \neq -1$, we can solve for $x$:

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

Now, substitute this value of $x$ back into equation (ii) to find $y$:

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

So, the coordinates of the intersection point Q are $\left(\frac{m + 2}{m + 1}, \frac{3m + 2}{m + 1}\right)$.

The distance between the point P(1, 2) and Q$\left(\frac{m + 2}{m + 1}, \frac{3m + 2}{m + 1}\right)$ is given as $\frac{\sqrt{6}}{3}$. The square of the distance is $\left(\frac{\sqrt{6}}{3}\right)^2 = \frac{6}{9} = \frac{2}{3}$.

Using the distance formula, the square of the distance PQ is:

$(PQ)^2 = \left(\frac{m + 2}{m + 1} - 1\right)^2 + \left(\frac{3m + 2}{m + 1} - 2\right)^2$

Simplify the terms inside the parentheses:

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

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

Now, substitute these back into the distance formula:

$(PQ)^2 = \left(\frac{1}{m + 1}\right)^2 + \left(\frac{m}{m + 1}\right)^2 = \frac{1^2}{(m + 1)^2} + \frac{m^2}{(m + 1)^2} = \frac{1 + m^2}{(m + 1)^2}$

We are given that $(PQ)^2 = \frac{2}{3}$. So, we set up the equation:

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

Cross-multiply:

$3(1 + m^2) = 2(m + 1)^2$

$3 + 3m^2 = 2(m^2 + 2m + 1)$

$3 + 3m^2 = 2m^2 + 4m + 2$

Rearrange the terms to form a quadratic equation in $m$:

$3m^2 - 2m^2 - 4m + 3 - 2 = 0$

We can solve this quadratic equation for $m$ using the quadratic formula $m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a=1$, $b=-4$, and $c=1$.

$m = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(1)}}{2(1)}$

$m = \frac{4 \pm \sqrt{16 - 4}}{2}$

$m = \frac{4 \pm \sqrt{12}}{2}$

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

Divide the numerator and denominator by 2:

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

$m = 2 \pm \sqrt{3}$

Thus, there are two possible values for the slope $m$: $m_1 = 2 + \sqrt{3}$ and $m_2 = 2 - \sqrt{3}$. These slopes represent the directions in which the line can be drawn.

Alternatively, the direction can be specified by the angle $\theta$ the line makes with the positive x-axis, where $m = \tan \theta$.

If $m = 2 + \sqrt{3}$, then $\tan \theta = 2 + \sqrt{3}$. This corresponds to $\theta = 75^\circ$ or $\frac{5\pi}{12}$ radians.

If $m = 2 - \sqrt{3}$, then $\tan \theta = 2 - \sqrt{3}$. This corresponds to $\theta = 15^\circ$ or $\frac{\pi}{12}$ radians.

Final Answer:

The line should be drawn in a direction having a slope of $2 + \sqrt{3}$ or $2 - \sqrt{3}$.

Equivalently, the direction is at an angle of $\frac{\pi}{12}$ or $\frac{5\pi}{12}$ radians (or $15^\circ$ or $75^\circ$) with the positive x-axis.

Given:

A straight line has x-intercept $a$ and y-intercept $b$.

The sum of the reciprocals of its intercepts on the axes is constant, i.e., $\frac{1}{a} + \frac{1}{b} = \text{constant}$.

To Show:

The line passes through a fixed point.

Solution:

Let the equation of the straight line in the intercept form be:

where $a$ is the x-intercept and $b$ is the y-intercept of the line.

According to the given condition, the sum of the reciprocals of the intercepts is constant. Let this constant be denoted by $C$.

Since $C$ is a constant, we can write $C = \frac{1}{k}$ for some constant $k$ (assuming $C \neq 0$). If $C=0$, then $\frac{1}{a} = -\frac{1}{b}$, so $b = -a$. The line equation would be $\frac{x}{a} - \frac{y}{a} = 1$, or $x - y = a$. This line passes through the origin if $a=0$, which implies $1/a$ is undefined. If $a \neq 0$, the line passes through $(a, 0)$ and $(0, -a)$, e.g., $(1, 0)$ and $(0, -1)$ gives $x-y=1$, $(2,0)$ and $(0,-2)$ gives $x-y=2$. This form $x-y=a$ does not pass through a single fixed point unless $a$ is fixed, which it isn't since $a$ is a variable intercept. Thus, we must have $C \neq 0$.

So, let $C = \frac{1}{k}$ where $k$ is a non-zero constant.

Substituting $C = \frac{1}{k}$ into equation (ii), we get:

Now, multiply equation (iii) by $k$:

$k\left(\frac{1}{a} + \frac{1}{b}\right) = k\left(\frac{1}{k}\right)$

Compare equation (iv) with the equation of the line (i): $\frac{x}{a} + \frac{y}{b} = 1$.

If we substitute $x = k$ and $y = k$ into equation (i), we get $\frac{k}{a} + \frac{k}{b} = 1$.

This equation is identical to equation (iv), which is derived directly from the given condition $\frac{1}{a} + \frac{1}{b} = C = \frac{1}{k}$.

This means that the coordinates $(k, k)$ satisfy the equation of the line $\frac{x}{a} + \frac{y}{b} = 1$ for any values of $a$ and $b$ that satisfy the condition $\frac{1}{a} + \frac{1}{b} = \frac{1}{k}$.

Since $k$ is a constant determined by the given constant sum, the point $(k, k)$ is a fixed point.

Therefore, the line always passes through the fixed point $(k, k)$.

Conclusion:

Since the coordinates $(k, k)$ satisfy the equation of the line for all possible values of the intercepts $a$ and $b$ obeying the given condition, the line must pass through the fixed point $(k, k)$.

The fixed point is $(k, k)$, where $\frac{1}{k}$ is the constant sum of the reciprocals of the intercepts.

Final Answer:

The line passes through the fixed point $\left(\frac{1}{C}, \frac{1}{C}\right)$, where $C$ is the constant sum of the reciprocals of the intercepts.

Given:

The line passes through the point $P(-4, 3)$.

The portion of the line intercepted between the axes is divided internally in the ratio 5 : 3 by the point $P(-4, 3)$.

To Find:

The equation of the line.

Solution:

Let the equation of the line in the intercept form be:

where $a$ is the x-intercept and $b$ is the y-intercept.

The points where the line intercepts the axes are $A(a, 0)$ on the x-axis and $B(0, b)$ on the y-axis.

The point $P(-4, 3)$ lies on the line and divides the segment $AB$ internally in the ratio $5 : 3$.

Using the section formula, if a point $(x, y)$ divides the segment joining $(x_1, y_1)$ and $(x_2, y_2)$ internally in the ratio $m : n$, then:

$x = \frac{nx_1 + mx_2}{m+n}$ and $y = \frac{ny_1 + my_2}{m+n}$

Here, the point is $P(-4, 3)$, the segment is $AB$ with $A(a, 0)$ and $B(0, b)$, and the ratio is $5 : 3$ ($m=5$, $n=3$).

Applying the section formula for the x-coordinate:

$-4 = \frac{3a}{8}$

$3a = -4 \times 8$

$3a = -32$

Applying the section formula for the y-coordinate:

$3 = \frac{5b}{8}$

$5b = 3 \times 8$

$5b = 24$

Now substitute the values of $a$ and $b$ from equations (iii) and (v) into the intercept form equation (i):

$\frac{x}{-32/3} + \frac{y}{24/5} = 1$}

$\frac{3x}{-32} + \frac{5y}{24} = 1$}

To eliminate the denominators, we find the Least Common Multiple (LCM) of 32 and 24.

Prime factorization of 32: $2^5$

Prime factorization of 24: $2^3 \times 3$

LCM(32, 24) = $2^5 \times 3 = 32 \times 3 = 96$.

Multiply the equation by 96:

$96 \left(\frac{3x}{-32}\right) + 96 \left(\frac{5y}{24}\right) = 96(1)$

$\cancel{96}^{-3} \left(\frac{3x}{\cancel{-32}^1}\right) + \cancel{96}^{4} \left(\frac{5y}{\cancel{24}^1}\right) = 96$

$-3(3x) + 4(5y) = 96$

$-9x + 20y = 96$

Rearrange the terms to get the general form of the equation of the line:

$9x - 20y + 96 = 0$

This is the required equation of the line.

Final Answer:

The equation of the line is $9x - 20y + 96 = 0$.

Given:

Lines $L_1: x - y + 1 = 0$ and $L_2: 2x - 3y + 5 = 0$.

A point $P(3, 2)$.

Distance from P to the required line is $\frac{7}{5}$.

To Find:

The equations of the lines passing through the intersection of $L_1$ and $L_2$ and having a distance of $\frac{7}{5}$ from P(3, 2).

Solution:

First, we find the point of intersection of the lines $x - y + 1 = 0$ and $2x - 3y + 5 = 0$.

Consider the system of equations:

Multiply equation (i) by 2:

Subtract equation (iii) from equation (ii):

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

$2x - 3y - 2x + 2y = -5 + 2$

$-y = -3$

$y = 3$

Substitute $y = 3$ into equation (i):

$x - 3 = -1$

$x = -1 + 3$

$x = 2$}

The point of intersection of the two lines is $(2, 3)$.

Let the equation of the required line passing through $(2, 3)$ have slope $m$. The equation of the line in point-slope form is:

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

Rearranging this into the general form $Ax + By + C = 0$, we get:

The distance from the point $(x_0, y_0) = (3, 2)$ to the line $Ax + By + C = 0$ is given by the formula $d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$.

Here, $A = m$, $B = -1$, $C = 3 - 2m$, and the given distance is $d = \frac{7}{5}$.

Substituting these values into the distance formula:

$\frac{7}{5} = \frac{|m(3) + (-1)(2) + (3 - 2m)|}{\sqrt{m^2 + (-1)^2}}$

$\frac{7}{5} = \frac{|3m - 2 + 3 - 2m|}{\sqrt{m^2 + 1}}$

Square both sides of equation (v) to eliminate the absolute value and the square root:

$\left(\frac{7}{5}\right)^2 = \left(\frac{m + 1}{\sqrt{m^2 + 1}}\right)^2$

$\frac{49}{25} = \frac{(m + 1)^2}{m^2 + 1}$

$\frac{49}{25} = \frac{m^2 + 2m + 1}{m^2 + 1}$

Cross-multiply:

$49(m^2 + 1) = 25(m^2 + 2m + 1)$

$49m^2 + 49 = 25m^2 + 50m + 25$

Rearrange the terms to form a quadratic equation in $m$:

$49m^2 - 25m^2 - 50m + 49 - 25 = 0$}

$24m^2 - 50m + 24 = 0$}

Divide the entire equation by 2:

Solve the quadratic equation (vi) for $m$ by factoring. We look for two numbers that multiply to $12 \times 12 = 144$ and add up to $-25$. These numbers are $-9$ and $-16$.

$12m^2 - 16m - 9m + 12 = 0$}

Group the terms and factor by grouping:

$4m(3m - 4) - 3(3m - 4) = 0$}

$(4m - 3)(3m - 4) = 0$}

This gives two possible values for the slope $m$:

$4m - 3 = 0 \implies 4m = 3 \implies m = \frac{3}{4}$

$3m - 4 = 0 \implies 3m = 4 \implies m = \frac{4}{3}$

Now, we find the equation of the line for each value of $m$, using the point $(2, 3)$.

Case 1: $m = \frac{3}{4}$

Using point-slope form $y - y_1 = m(x - x_1)$ with $(x_1, y_1) = (2, 3)$:

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

Multiply by 4:

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

$4y - 12 = 3x - 6$}

Rearrange into the general form:

$3x - 4y - 6 + 12 = 0$}

$3x - 4y + 6 = 0$}

Case 2: $m = \frac{4}{3}$

Using point-slope form $y - y_1 = m(x - x_1)$ with $(x_1, y_1) = (2, 3)$:

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

Multiply by 3:

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

$3y - 9 = 4x - 8$}

Rearrange into the general form:

$4x - 3y - 8 + 9 = 0$}

$4x - 3y + 1 = 0$}

Thus, there are two lines that satisfy the given conditions.

Final Answer:

The equations of the required lines are $3x - 4y + 6 = 0$ and $4x - 3y + 1 = 0$.

Given:

A moving point in a plane, let's call its coordinates $(x, y)$.

The sum of the distances of the point from the axes is 1.

To Find:

The locus of the moving point.

Solution:

Let the moving point be $P(x, y)$.

The distance of the point $P(x, y)$ from the x-axis is the absolute value of the y-coordinate, which is $|y|$.

The distance of the point $P(x, y)$ from the y-axis is the absolute value of the x-coordinate, which is $|x|$.

According to the given condition, the sum of these distances is 1.

So, we have the equation:

$|x| + |y| = 1$

This equation represents the locus of the point $P(x, y)$. To understand the shape of this locus, we consider the equation in different quadrants:

• In the first quadrant ($x \ge 0, y \ge 0$), the equation becomes $x + y = 1$. This is the equation of a straight line segment connecting the points (1, 0) and (0, 1).
• In the second quadrant ($x \le 0, y \ge 0$), the equation becomes $-x + y = 1$. This is the equation of a straight line segment connecting the points (-1, 0) and (0, 1).
• In the third quadrant ($x \le 0, y \le 0$), the equation becomes $-x - y = 1$, which can be written as $x + y = -1$. This is the equation of a straight line segment connecting the points (-1, 0) and (0, -1).
• In the fourth quadrant ($x \ge 0, y \le 0$), the equation becomes $x - y = 1$. This is the equation of a straight line segment connecting the points (1, 0) and (0, -1).

When we combine these four line segments, they form a square with vertices at (1, 0), (0, 1), (-1, 0), and (0, -1).

Final Answer:

The locus of the point is given by the equation $|x| + |y| = 1$. This equation represents a square with vertices at $(1, 0)$, $(0, 1)$, $(-1, 0)$, and $(0, -1)$.

Given:

Two lines given by the equation $y - \sqrt{3}|x| = 2$.

Points P₁ and P₂ lie on these lines at a distance of 5 units from their point of intersection.

To Find:

The coordinates of the foot of the perpendiculars drawn from P₁ and P₂ on the bisector of the angle between the given lines.

Solution:

The given equation of the lines is $y - \sqrt{3}|x| = 2$, which can be written as $y = \sqrt{3}|x| + 2$.

This equation represents two lines:

• When $x \ge 0$, $y = \sqrt{3}x + 2$. Let's call this line $L_1$.
• When $x < 0$, $y = -\sqrt{3}x + 2$. Let's call this line $L_2$.

The point of intersection of these two lines is found by setting the equations equal for $x=0$: $y = \sqrt{3}(0) + 2 = 2$ and $y = -\sqrt{3}(0) + 2 = 2$. The point of intersection is $I(0, 2)$.

Let's determine the angle between the lines. The slope of $L_1$ is $m_1 = \sqrt{3}$. The angle $\theta_1$ that $L_1$ makes with the positive x-axis is $\tan \theta_1 = \sqrt{3}$, so $\theta_1 = 60^\circ$.

The slope of $L_2$ is $m_2 = -\sqrt{3}$. The angle $\theta_2$ that $L_2$ makes with the positive x-axis is $\tan \theta_2 = -\sqrt{3}$. Since the line passes through $(0, 2)$ and has a negative slope for $x < 0$, this corresponds to $\theta_2 = 120^\circ$.

The angle between the lines is $|\theta_2 - \theta_1| = |120^\circ - 60^\circ| = 60^\circ$.

The bisector of the angle between the lines $y = m_1x + c$ and $y = m_2x + c$ passing through the common y-intercept $(0, c)$ is the y-axis ($x=0$) if $m_1 = -m_2$, which is true here ($\sqrt{3}$ and $-\sqrt{3}$). Thus, the bisector of the angle between the lines is the y-axis, with equation $x = 0$.

Points P₁ and P₂ are on the lines $L_1$ and $L_2$ respectively, at a distance of 5 units from the intersection point $I(0, 2)$.

Consider point P₁ on $L_1: y = \sqrt{3}x + 2$ with $x \ge 0$. Let $P_1$ have coordinates $(x_1, y_1)$. The distance from $I(0, 2)$ to $P_1(x_1, y_1)$ is 5.

$\sqrt{(x_1 - 0)^2 + (y_1 - 2)^2} = 5$

$(x_1)^2 + (y_1 - 2)^2 = 25$

Since $P_1$ is on $L_1$, $y_1 = \sqrt{3}x_1 + 2$, so $y_1 - 2 = \sqrt{3}x_1$.

Substitute this into the distance equation:

$(x_1)^2 + (\sqrt{3}x_1)^2 = 25$

$x_1^2 + 3x_1^2 = 25$

$4x_1^2 = 25$

$x_1^2 = \frac{25}{4}$

$x_1 = \pm \frac{5}{2}$

Since $P_1$ is on the part of $L_1$ where $x \ge 0$, we take the positive value: $x_1 = \frac{5}{2}$.

Then $y_1 = \sqrt{3}\left(\frac{5}{2}\right) + 2 = \frac{5\sqrt{3}}{2} + 2$.

So, $P_1$ has coordinates $\left(\frac{5}{2}, \frac{5\sqrt{3}}{2} + 2\right)$.

Consider point P₂ on $L_2: y = -\sqrt{3}x + 2$ with $x < 0$. Let $P_2$ have coordinates $(x_2, y_2)$. The distance from $I(0, 2)$ to $P_2(x_2, y_2)$ is 5.

$\sqrt{(x_2 - 0)^2 + (y_2 - 2)^2} = 5$

$(x_2)^2 + (y_2 - 2)^2 = 25$

Since $P_2$ is on $L_2$, $y_2 = -\sqrt{3}x_2 + 2$, so $y_2 - 2 = -\sqrt{3}x_2$.

Substitute this into the distance equation:

$(x_2)^2 + (-\sqrt{3}x_2)^2 = 25$

$x_2^2 + 3x_2^2 = 25$

$4x_2^2 = 25$

$x_2^2 = \frac{25}{4}$

$x_2 = \pm \frac{5}{2}$

Since $P_2$ is on the part of $L_2$ where $x < 0$, we take the negative value: $x_2 = -\frac{5}{2}$.

Then $y_2 = -\sqrt{3}\left(-\frac{5}{2}\right) + 2 = \frac{5\sqrt{3}}{2} + 2$.

So, $P_2$ has coordinates $\left(-\frac{5}{2}, \frac{5\sqrt{3}}{2} + 2\right)$.

The bisector of the angle between the lines is the y-axis, which has the equation $x = 0$.

The foot of the perpendicular from a point $(x_0, y_0)$ to the line $x = 0$ is the point $(0, y_0)$.

The foot of the perpendicular from $P_1\left(\frac{5}{2}, \frac{5\sqrt{3}}{2} + 2\right)$ to the y-axis is $F_1\left(0, \frac{5\sqrt{3}}{2} + 2\right)$.

The foot of the perpendicular from $P_2\left(-\frac{5}{2}, \frac{5\sqrt{3}}{2} + 2\right)$ to the y-axis is $F_2\left(0, \frac{5\sqrt{3}}{2} + 2\right)$.

We see that the foot of the perpendicular is the same point for both P₁ and P₂.

The y-coordinate of the foot of the perpendicular is $2 + \frac{5\sqrt{3}}{2}$. Note that $\frac{\sqrt{3}}{2} = \cos 30^\circ$. So the y-coordinate is indeed $2 + 5\cos 30^\circ$, as suggested by the hint.

The y-coordinate can be calculated as:

$2 + 5 \times \frac{\sqrt{3}}{2} = 2 + \frac{5\sqrt{3}}{2} = \frac{4 + 5\sqrt{3}}{2}$

The coordinates of the foot of the perpendicular are $\left(0, \frac{4 + 5\sqrt{3}}{2}\right)$.

Final Answer:

The coordinates of the foot of the perpendiculars drawn from P₁ and P₂ on the bisector of the angle between the given lines are $\left(0, 2 + \frac{5\sqrt{3}}{2}\right)$.

Given:

Equation of the line is $\frac{x}{a} + \frac{y}{b} = 1$.

$p$ is the length of the perpendicular from the origin $(0, 0)$ to the line.

$a^2, p^2, b^2$ are in Arithmetic Progression (A.P.).

To Show:

$a^4 + b^4 = 0$

Solution:

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

We can rewrite this equation in the general form $Ax + By + C = 0$ by finding a common denominator and rearranging terms:

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

$bx + ay = ab$}

$bx + ay - ab = 0$}

Comparing this with $Ax + By + C = 0$, we have $A = b$, $B = a$, and $C = -ab$.

The length of the perpendicular ($p$) from the origin $(x_0, y_0) = (0, 0)$ to the line $Ax + By + C = 0$ is given by the formula:

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

Substituting the values, we get:

$p = \frac{|-ab|}{\sqrt{a^2 + b^2}}$

Since $p$ is a length, it is non-negative. Thus, $p = \frac{|ab|}{\sqrt{a^2 + b^2}}$.

Squaring both sides, we get:

We are given that $a^2, p^2, b^2$ are in A.P.

By the property of an Arithmetic Progression, the middle term is the average of the other two terms, or twice the middle term is equal to the sum of the other two terms.

Now, substitute the expression for $p^2$ from equation (ii) into equation (iii):

$2 \left(\frac{a^2 b^2}{a^2 + b^2}\right) = a^2 + b^2$

Multiply both sides by $(a^2 + b^2)$ (assuming $a^2 + b^2 \neq 0$; if $a^2+b^2=0$, then $a=0, b=0$, which makes the original line equation undefined):

$2a^2 b^2 = (a^2 + b^2)(a^2 + b^2)$

$2a^2 b^2 = (a^2 + b^2)^2$}

Expand the right side:

$2a^2 b^2 = (a^2)^2 + 2(a^2)(b^2) + (b^2)^2$}

$2a^2 b^2 = a^4 + 2a^2 b^2 + b^4$}

Subtract $2a^2 b^2$ from both sides of the equation:

$2a^2 b^2 - 2a^2 b^2 = a^4 + 2a^2 b^2 + b^4 - 2a^2 b^2$}

$0 = a^4 + b^4$}

Thus, we have shown that $a^4 + b^4 = 0$.

Conclusion:

Based on the given information that $p$ is the perpendicular distance from the origin to the line $\frac{x}{a} + \frac{y}{b} = 1$ and $a^2, p^2, b^2$ are in A.P., we have derived the relation $a^4 + b^4 = 0$.

Final Answer:

It is shown that $a^4 + b^4 = 0$.}

Given:

Y-intercept ($c$) = $-3$.

Tangent of the angle the line makes with the x-axis = $\frac{3}{5}$. This is the slope of the line ($m$).

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

To Find:

The equation of the line.

Solution:

The equation of a line in slope-intercept form is given by $y = mx + c$, where $m$ is the slope and $c$ is the y-intercept.

Substitute the given values of $m$ and $c$ into the equation:

$y = \left(\frac{3}{5}\right)x + (-3)$

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

To remove the fraction, multiply the entire equation by 5:

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

$5y = 3x - 15$

Rearrange the terms to match the options (general form $Ax + By + C = 0$):

$3x - 5y - 15 = 0$

Alternatively, move all terms to the left side:

$5y - 3x + 15 = 0$

Compare this equation with the given options.

Final Answer:

The equation of the line is $5y - 3x + 15 = 0$.

Correct Option: (A)

Given:

A line cuts off intercepts of equal lengths on the axes.

To Find:

The slope of the line.

Solution:

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

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

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

The problem states that the intercepts have equal lengths. The length of the x-intercept is $|a|$ and the length of the y-intercept is $|b|$.

So, we have $|a| = |b|$.

This condition implies two possibilities:

Case 1: $a = b$

In this case, the equation of the line becomes:

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

Assuming $a \neq 0$ (otherwise the intercepts are zero, which implies the line passes through the origin and the concept of non-zero intercepts of equal length is not met in the standard intercept form), we can multiply by $a$:

$x + y = a$}

Rearranging into the slope-intercept form $y = mx + c$:

$y = -x + a$}

The slope in this case is $m = -1$.

Case 2: $a = -b$}

In this case, the equation of the line becomes:

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

Assuming $b \neq 0$, we can multiply by $b$:

$-x + y = b$}

Rearranging into the slope-intercept form $y = mx + c$:

$y = x + b$}

The slope in this case is $m = 1$.

The slope of the line can be either $-1$ or $1$. Looking at the given options, only $-1$ is present.

Final Answer:

The slope of a line which cuts off intercepts of equal lengths on the axes is either $-1$ or $1$. Among the given options, the correct answer is $-1$.

Correct Option: (A)

Given:

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

The line is perpendicular to the line $y = x$.

To Find:

The equation of the line.

Solution:

The equation of the given line is $y = x$.

This equation is in the slope-intercept form $y = mx + c$, where $m$ is the slope.

The slope of the given line is $m_{given} = 1$.

If two non-vertical lines are perpendicular, the product of their slopes is $-1$.

Let $m_{required}$ be the slope of the required line.

$m_{required} \times m_{given} = -1$

$m_{required} \times 1 = -1$

$m_{required} = -1$

The required line passes through the point $(x_1, y_1) = (3, 2)$ and has a slope $m = -1$.

Using the point-slope form of the equation of a line, $y - y_1 = m(x - x_1)$:

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

Simplify the equation:

$y - 2 = -x + 3$

Rearrange the terms to match the options. Move all terms to one side or match the form $x + y = C$:

$x + y = 3 + 2$

$x + y = 5$

Final Answer:

The equation of the straight line is $x + y = 5$.

Correct Option: (B)

Given:

The line passes through the point $(1, 2)$.

The line is perpendicular to the line $x + y + 1 = 0$.

To Find:

The equation of the line.

Solution:

The equation of the given line is $x + y + 1 = 0$.

This equation is in the general form $Ax + By + C = 0$, where $A=1$, $B=1$, and $C=1$.

The slope of the given line is $m_{given} = -\frac{A}{B} = -\frac{1}{1} = -1$.

Let $m_{required}$ be the slope of the required line.

Since the required line is perpendicular to the given line, the product of their slopes is $-1$ (for non-vertical lines, which these are).

$m_{required} \times m_{given} = -1$

$m_{required} \times (-1) = -1$

$m_{required} = 1$

The required line passes through the point $(x_1, y_1) = (1, 2)$ and has a slope $m = 1$.

Using the point-slope form of the equation of a line, $y - y_1 = m(x - x_1)$:

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

Simplify the equation:

$y - 2 = x - 1$

Rearrange the terms to match the options:

$y - x - 2 + 1 = 0$

$y - x - 1 = 0$

Final Answer:

The equation of the line is $y - x - 1 = 0$.

Correct Option: (B)

Given:

Line 1 has x-intercept $a$ and y-intercept $-b$.

Line 2 has x-intercept $b$ and y-intercept $-a$.

To Find:

The tangent of the angle between the two lines.

Solution:

The equation of a line with x-intercept $x_0$ and y-intercept $y_0$ is given by $\frac{x}{x_0} + \frac{y}{y_0} = 1$.

For the first line, with x-intercept $a$ and y-intercept $-b$, the equation is:

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

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

To find the slope ($m_1$) of this line, we rearrange the equation into slope-intercept form $y = mx + c$:

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

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

$y = \frac{b}{a}x - b$}

The slope of the first line is $m_1 = \frac{b}{a}$. (Assuming $a \neq 0$)

For the second line, with x-intercept $b$ and y-intercept $-a$, the equation is:

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

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

To find the slope ($m_2$) of this line, we rearrange the equation into slope-intercept form $y = mx + c$:

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

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

$y = \frac{a}{b}x - a$}

The slope of the second line is $m_2 = \frac{a}{b}$. (Assuming $b \neq 0$)

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

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

We calculate the expression $\frac{m_1 - m_2}{1 + m_1 m_2}$:

$m_1 - m_2 = \frac{b}{a} - \frac{a}{b} = \frac{b^2 - a^2}{ab}$

$1 + m_1 m_2 = 1 + \left(\frac{b}{a}\right)\left(\frac{a}{b}\right) = 1 + 1 = 2$}

So, $\frac{m_1 - m_2}{1 + m_1 m_2} = \frac{\frac{b^2 - a^2}{ab}}{2} = \frac{b^2 - a^2}{2ab}$.

The tangent of the angle can be $\frac{b^2 - a^2}{2ab}$ or its negative $\frac{a^2 - b^2}{2ab}$, depending on which line's angle is subtracted from the other, and whether the acute or obtuse angle is considered. Since the options do not include absolute value and contain $\frac{b^2-a^2}{2ab}$, this is the expected form.

Final Answer:

The tangent of the angle between the lines is $\frac{b^2 - a^2}{2ab}$.

Correct Option: (C)

Given:

Equation of the line is $\frac{x}{a} + \frac{y}{b} = 1$.

The line passes through the points $(2, -3)$ and $(4, -5)$.

To Find:

The values of $a$ and $b$, i.e., the coordinates $(a, b)$.

Solution:

Since the line $\frac{x}{a} + \frac{y}{b} = 1$ passes through the point $(2, -3)$, we substitute these coordinates into the equation:

$\frac{2}{a} + \frac{-3}{b} = 1$

Since the line also passes through the point $(4, -5)$, we substitute these coordinates into the equation:

$\frac{4}{a} + \frac{-5}{b} = 1$

We now have a system of two linear equations with variables $\frac{1}{a}$ and $\frac{1}{b}$. Let $u = \frac{1}{a}$ and $v = \frac{1}{b}$. The system becomes:

$2u - 3v = 1$

$4u - 5v = 1$

To solve this system, we can use the elimination method. Multiply the first equation by 2:

$2 \times (2u - 3v) = 2 \times 1$

$4u - 6v = 2$

Now subtract the second original equation ($4u - 5v = 1$) from this new equation:

$(4u - 6v) - (4u - 5v) = 2 - 1$

$4u - 6v - 4u + 5v = 1$

$-v = 1$

$v = -1$

Substitute the value of $v = -1$ into the first equation ($2u - 3v = 1$):

$2u - 3(-1) = 1$

$2u + 3 = 1$}

$2u = 1 - 3$

$2u = -2$

$u = -1$

Now we find $a$ and $b$ using $u = \frac{1}{a}$ and $v = \frac{1}{b}$:

$u = \frac{1}{a} \implies -1 = \frac{1}{a} \implies a = -1$

$v = \frac{1}{b} \implies -1 = \frac{1}{b} \implies b = -1$

So, the coordinates $(a, b)$ are $(-1, -1)$.

Final Answer:

The coordinates $(a, b)$ are $(-1, -1)$.

Correct Option: (D)

Given:

Line 1: $2x - 3y + 5 = 0$

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

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

To Find:

The distance of the point of intersection of Line 1 and Line 2 from Line 3.

Solution:

First, find the point of intersection of Line 1 and Line 2. We have the system of equations:

Multiply equation (i) by 4 and equation (ii) by 3 to eliminate $y$:

$4 \times (2x - 3y) = 4 \times (-5) \implies 8x - 12y = -20$

$3 \times (3x + 4y) = 3 \times 0 \implies 9x + 12y = 0$

Add the two resulting equations:

$(8x - 12y) + (9x + 12y) = -20 + 0$

$17x = -20$

$x = -\frac{20}{17}$

Substitute the value of $x$ into equation (ii):

$3\left(-\frac{20}{17}\right) + 4y = 0$

$-\frac{60}{17} + 4y = 0$

$4y = \frac{60}{17}$

$y = \frac{60}{17 \times 4} = \frac{15}{17}$

The point of intersection $(x_0, y_0)$ is $\left(-\frac{20}{17}, \frac{15}{17}\right)$.

Now, we need to find the distance from this point to the line $5x - 2y = 0$.

The equation of Line 3 is $5x - 2y + 0 = 0$. Comparing this with the general form $Ax + By + C = 0$, we have $A = 5$, $B = -2$, and $C = 0$.

The distance from a point $(x_0, y_0)$ to the 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 intersection point $\left(-\frac{20}{17}, \frac{15}{17}\right)$ and the coefficients of Line 3 into the distance formula:

$d = \frac{\left|-\frac{100}{17} - \frac{30}{17}\right|}{\sqrt{25 + 4}}$

$d = \frac{\left|-\frac{130}{17}\right|}{\sqrt{29}}$

$d = \frac{\frac{130}{17}}{\sqrt{29}}$

$d = \frac{130}{17\sqrt{29}}$

Final Answer:

The distance of the point of intersection from the line $5x - 2y = 0$ is $\frac{130}{17\sqrt{29}}$.

Correct Option: (A)

Given:

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

The angle between the required line and the line $\sqrt{3} x + y = 1$ is $60^\circ$.

To Find:

The equations of the required lines.

Solution:

The equation of the given line is $\sqrt{3} x + y = 1$. We can write this in slope-intercept form $y = mx + c$:

$y = -\sqrt{3} x + 1$

The slope of the given line is $m_1 = -\sqrt{3}$.

Let $m_2$ be the slope of the required line passing through $(3, -2)$.

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 $\theta = 60^\circ$, so $\tan 60^\circ = \sqrt{3}$.

Substitute the values of $m_1$ and $\theta$ into the formula:

$\sqrt{3} = \left|\frac{m_2 - (-\sqrt{3})}{1 + (-\sqrt{3}) m_2}\right|$

This gives two cases based on the absolute value:

Case 1: $\frac{m_2 + \sqrt{3}}{1 - \sqrt{3} m_2} = \sqrt{3}$

$m_2 + \sqrt{3} = \sqrt{3} (1 - \sqrt{3} m_2)$

$m_2 + \sqrt{3} = \sqrt{3} - 3 m_2$}

$m_2 + 3 m_2 = \sqrt{3} - \sqrt{3}$

$4 m_2 = 0$

$m_2 = 0$

Case 2: $\frac{m_2 + \sqrt{3}}{1 - \sqrt{3} m_2} = -\sqrt{3}$

$m_2 + \sqrt{3} = -\sqrt{3} (1 - \sqrt{3} m_2)$

$m_2 + \sqrt{3} = -\sqrt{3} + 3 m_2$}

$\sqrt{3} + \sqrt{3} = 3 m_2 - m_2$}

$2 \sqrt{3} = 2 m_2$}

$m_2 = \sqrt{3}$

So, the slopes of the required lines are $m = 0$ and $m = \sqrt{3}$. Both lines pass through the point $(x_1, y_1) = (3, -2)$.

Using the point-slope form $y - y_1 = m(x - x_1)$:

For $m = 0$:

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

$y + 2 = 0$

For $m = \sqrt{3}$:

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

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

Rearrange into the general form:

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

The equations of the required lines are $y + 2 = 0$ and $\sqrt{3}x - y - 2 - 3\sqrt{3} = 0$.

Final Answer:

The equations of the lines are $y + 2 = 0$ and $\sqrt{3} x - y - 2 - 3 \sqrt{3} = 0$.

Comparing with the given options, option (A) matches our results.

Correct Option: (A)

Given:

The line passes through the point $(1, 0)$.

The distance from the origin $(0, 0)$ to the line is $\frac{\sqrt{3}}{2}$.

To Find:

The equations of the lines.

Solution:

Let the equation of the line passing through the point $(1, 0)$ be $y - y_1 = m(x - x_1)$, where $(x_1, y_1) = (1, 0)$ and $m$ is the slope.

$y - 0 = m(x - 1)$

$y = m(x - 1)$

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

$mx - y - m = 0$

Here, $A = m$, $B = -1$, and $C = -m$.

The distance from the origin $(0, 0)$ to the 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 origin $(0, 0)$ and the coefficients of the line into the formula:

$d = \frac{|-m|}{\sqrt{m^2 + 1}}$

We are given that the distance is $\frac{\sqrt{3}}{2}$. So,

Square both sides of equation (ii):

$\left(\frac{|m|}{\sqrt{m^2 + 1}}\right)^2 = \left(\frac{\sqrt{3}}{2}\right)^2$

$\frac{m^2}{m^2 + 1} = \frac{3}{4}$

Cross-multiply:

$4m^2 = 3(m^2 + 1)$

$4m^2 = 3m^2 + 3$

$4m^2 - 3m^2 = 3$

$m^2 = 3$

Take the square root of both sides:

$m = \pm \sqrt{3}$

We have two possible values for the slope $m$: $m_1 = \sqrt{3}$ and $m_2 = -\sqrt{3}$.

Now, we find the equation of the line for each slope, using the point $(1, 0)$ and the point-slope form $y - y_1 = m(x - x_1)$.

Case 1: $m = \sqrt{3}$

$y - 0 = \sqrt{3}(x - 1)$

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

Rearrange into the general form:

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

Case 2: $m = -\sqrt{3}$

$y - 0 = -\sqrt{3}(x - 1)$

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

Rearrange into the general form:

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

These are the two equations of the lines satisfying the given conditions.

Final Answer:

The equations of the lines are $\sqrt{3} x - y - \sqrt{3} = 0$ and $\sqrt{3} x + y - \sqrt{3} = 0$.

Comparing with the given options, option (A) matches our derived equations.

Correct Option: (A)

Given:

Equation of Line 1: $y = mx + c_1$

Equation of Line 2: $y = mx + c_2$

To Find:

The distance between the two lines.

Solution:

The equations of the given lines are:

Line 1: $y = mx + c_1$

Line 2: $y = mx + c_2$

These lines have the same slope $m$, so they are parallel lines.

We can rewrite the equations in the general form $Ax + By + C = 0$:

Line 1: $mx - y + c_1 = 0$

Line 2: $mx - y + c_2 = 0$

Comparing these with $Ax + By + C = 0$, we have $A = m$, $B = -1$.

For Line 1, $C_1 = c_1$.

For Line 2, $C_2 = c_2$.

The distance $d$ between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by the formula:

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

Substituting the values for our lines:

$d = \frac{|c_1 - c_2|}{\sqrt{m^2 + 1}}$

Final Answer:

The distance between the lines $y = mx + c_1$ and $y = mx + c_2$ is $\frac{|c_1 - c_2|}{\sqrt{m^2 + 1}}$.

Comparing with the given options, option (B) matches the result.

Correct Option: (B)

Given:

Point P is $(2, 3)$.

Equation of the line L is $y = 3x + 4$.

To Find:

The coordinates of the foot of the perpendicular from P on line L.

Solution:

Let the equation of the given line be $L_1: y = 3x + 4$. We can rewrite this as $3x - y + 4 = 0$.

The slope of $L_1$ is $m_1 = 3$.

Let the required line, which is perpendicular to $L_1$ and passes through $(2, 3)$, be $L_2$.

The slope of $L_2$, say $m_2$, is related to the slope of $L_1$ by the condition for perpendicular lines: $m_1 \times m_2 = -1$.

$3 \times m_2 = -1$

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

Now we find the equation of the line $L_2$ passing through the point $(x_1, y_1) = (2, 3)$ with slope $m_2 = -\frac{1}{3}$. Using the point-slope form $y - y_1 = m_2(x - x_1)$:

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

Multiply both sides by 3:

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

$3y - 9 = -x + 2$

Rearrange the terms to get the general form of $L_2$:

$x + 3y - 9 - 2 = 0$

The foot of the perpendicular is the point of intersection of the lines $L_1$ ($y = 3x + 4$) and $L_2$ ($x + 3y - 11 = 0$).

Substitute the expression for $y$ from $L_1$ into equation (i):

$x + 3(3x + 4) - 11 = 0$}

$x + 9x + 12 - 11 = 0$}

$10x + 1 = 0$}

$10x = -1$

$x = -\frac{1}{10}$

Now substitute the value of $x$ back into the equation of $L_1$ to find $y$:

$y = 3\left(-\frac{1}{10}\right) + 4$}

$y = -\frac{3}{10} + 4$}

$y = -\frac{3}{10} + \frac{40}{10}$}

$y = \frac{37}{10}$

The coordinates of the foot of the perpendicular are $\left(-\frac{1}{10}, \frac{37}{10}\right)$.

Final Answer:

The coordinates of the foot of the perpendicular are $\left(-\frac{1}{10}, \frac{37}{10}\right)$.

Comparing with the given options, option (B) matches our result.

Correct Option: (B)

Given:

The midpoint of the portion of a line intercepted between the coordinate axes is $(3, 2)$.

To Find:

The equation of the line.

Solution:

Let the equation of the line be in the intercept form:

where $a$ is the x-intercept and $b$ is the y-intercept.

The line intersects the x-axis at the point $A(a, 0)$ and the y-axis at the point $B(0, b)$.

The middle point of the line segment intercepted between the axes is the midpoint of the segment AB.

Using the midpoint formula, the coordinates of the midpoint of AB are $\left(\frac{a + 0}{2}, \frac{0 + b}{2}\right) = \left(\frac{a}{2}, \frac{b}{2}\right)$.

We are given that this midpoint is $(3, 2)$.

Equating the coordinates:

From equation (ii), we solve for $a$:

$a = 3 \times 2 = 6$

From equation (iii), we solve for $b$:

$b = 2 \times 2 = 4$

Now substitute the values of $a = 6$ and $b = 4$ into the intercept form equation (i):

$\frac{x}{6} + \frac{y}{4} = 1$}

To eliminate the denominators, find the LCM of 6 and 4, which is 12.

Multiply the entire equation by 12:

$12\left(\frac{x}{6}\right) + 12\left(\frac{y}{4}\right) = 12(1)$

$\cancel{12}^2 \left(\frac{x}{\cancel{6}^1}\right) + \cancel{12}^3 \left(\frac{y}{\cancel{4}^1}\right) = 12$}

$2x + 3y = 12$

This is the equation of the line.

Final Answer:

The equation of the line is $2x + 3y = 12$.

Comparing with the given options, option (A) matches our result.

Correct Option: (A)

Given:

The line passes through the point $(1, 2)$.

The line is parallel to the line $y = 3x - 1$.}

To Find:

The equation of the line.

Solution:

The equation of the given line is $y = 3x - 1$.}

This equation is in the slope-intercept form $y = mx + c$, where $m$ is the slope.

The slope of the given line is $m_{given} = 3$.

If two non-vertical lines are parallel, they have the same slope.

Let $m_{required}$ be the slope of the required line.

$m_{required} = m_{given}$

$m_{required} = 3$

The required line passes through the point $(x_1, y_1) = (1, 2)$ and has a slope $m = 3$.

Using the point-slope form of the equation of a line, $y - y_1 = m(x - x_1)$:

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

This equation is in one of the forms given in the options.

Let's check the options:

(A) $y + 2 = x + 1 \implies y = x - 1$ (Slope is 1)

(B) $y + 2 = 3(x + 1) \implies y + 2 = 3x + 3 \implies y = 3x + 1$ (Slope is 3, but y-intercept is 1, does not pass through (1,2))

(C) $y - 2 = 3(x - 1)$ (Slope is 3, passes through (1,2))

(D) $y - 2 = x - 1 \implies y = x + 1$ (Slope is 1)

Option (C) directly matches the equation we derived using the point-slope form.

Final Answer:

The equation of the line is $y - 2 = 3(x - 1)$.

Correct Option: (C)

Given:

The lines forming a square are $x = 0$, $y = 0$, $x = 1$, and $y = 1$.

To Find:

The equations of the diagonals of the square.

Solution:

The given lines intersect to form the vertices of a square.

The line $x=0$ is the y-axis.

The line $y=0$ is the x-axis.

The line $x=1$ is a vertical line parallel to the y-axis.

The line $y=1$ is a horizontal line parallel to the x-axis.

The vertices of the square are the points of intersection of these lines:

• Intersection of $x=0$ and $y=0$ is $(0, 0)$.
• Intersection of $x=0$ and $y=1$ is $(0, 1)$.
• Intersection of $x=1$ and $y=0$ is $(1, 0)$.
• Intersection of $x=1$ and $y=1$ is $(1, 1)$.

Let the vertices be A$(0, 0)$, B$(1, 0)$, C$(1, 1)$, and D$(0, 1)$.

The diagonals of the square connect opposite vertices.

Diagonal 1 connects A$(0, 0)$ and C$(1, 1)$.

Diagonal 2 connects B$(1, 0)$ and D$(0, 1)$.

Find the equation of Diagonal 1 passing through $(0, 0)$ and $(1, 1)$ using the two-point form $\frac{y - y_1}{x - x_1} = \frac{y_2 - y_1}{x_2 - x_1}$:

$\frac{y - 0}{x - 0} = \frac{1 - 0}{1 - 0}$

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

$y = x$

Find the equation of Diagonal 2 passing through $(1, 0)$ and $(0, 1)$ using the two-point form:

$\frac{y - 0}{x - 1} = \frac{1 - 0}{0 - 1}$

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

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

$y = -(x - 1)$

$y = -x + 1$

Rearranging the terms, we get $x + y = 1$.

The equations of the diagonals are $y = x$ and $x + y = 1$.

Final Answer:

The equations of the diagonals of the square are $y = x$ and $x + y = 1$.

Comparing with the given options, option (A) matches our result.

Correct Option: (A)

Explanation:

A straight line in a 2-dimensional plane is uniquely determined by two independent parameters. These parameters can be:

• The slope ($m$) and the y-intercept ($c$) in the form $y = mx + c$. (2 parameters)
• The coordinates of two distinct points $(x_1, y_1)$ and $(x_2, y_2)$ through which the line passes. Although this involves 4 numbers, these define the line, and the line itself can be represented by an equation with 2 parameters (like slope and intercept). Geometrically, two points are needed.
• A point $(x_1, y_1)$ on the line and its slope ($m$) in the form $y - y_1 = m(x - x_1)$. Geometrically, a point and a direction define a line. (2 parameters: the point's position requires 2 values, but defining a line requires the point *and* the slope, where the slope is a single value representing direction. The position of the line as a whole is pinned down by one point and its orientation by the slope).
• The x-intercept ($a$) and the y-intercept ($b$) in the form $\frac{x}{a} + \frac{y}{b} = 1$ (assuming the line does not pass through the origin or is parallel to an axis). (2 parameters)

In all standard parameterizations of a straight line (excluding redundant representations), two independent parameters are required to uniquely specify its position and orientation in a plane.

Final Answer:

For specifying a straight line, 2 geometrical parameters should be known.

Correct Option: (B)

Given:

Initial point $P_0 = (4, 1)$.

Transformation 1: Reflection about the line $y = x$.

Transformation 2: Translation by 2 units along the positive x-axis.

To Find:

The final coordinates of the point after the two successive transformations.

Solution:

Let the initial point be $P_0 = (x_0, y_0) = (4, 1)$.

Step 1: Reflection about the line $y = x$.

When a point $(x, y)$ is reflected about the line $y = x$, its coordinates are swapped. The reflected point becomes $(y, x)$.

Applying this transformation to $P_0(4, 1)$, the coordinates become $(1, 4)$.

Let the point after the first transformation be $P_1 = (x_1, y_1)$.

$P_1 = (1, 4)$

Step 2: Translation through a distance 2 units along the positive x-axis.

A translation through a distance $d$ along the positive x-axis means that the x-coordinate is increased by $d$, and the y-coordinate remains unchanged.

Here, the translation distance is $d = 2$, and it is along the positive x-axis.

Applying this transformation to $P_1(1, 4)$, the new x-coordinate becomes $1 + 2 = 3$. The y-coordinate remains 4.

Let the point after the second transformation (the final point) be $P_2 = (x_2, y_2)$.

$x_2 = x_1 + 2 = 1 + 2 = 3$

$y_2 = y_1 = 4$

The final coordinates of the point are $(3, 4)$.

Final Answer:

The final coordinates of the point are $(3, 4)$.

Correct Option: (B)

Given:

Three lines: $L_1: 4x + 3y + 10 = 0$, $L_2: 5x - 12y + 26 = 0$, and $L_3: 7x + 24y - 50 = 0$.

To Find:

A point that is equidistant from these three lines.

Solution:

We can find the point equidistant from the lines by checking the given options. The distance of a point $(x_0, y_0)$ from a line $Ax + By + C = 0$ is given by the formula $d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$.

Let's calculate the normalized denominators for each line:

For $L_1: 4x + 3y + 10 = 0$, $\sqrt{A^2 + B^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$.

For $L_2: 5x - 12y + 26 = 0$, $\sqrt{A^2 + B^2} = \sqrt{5^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13$.

For $L_3: 7x + 24y - 50 = 0$, $\sqrt{A^2 + B^2} = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25$.

Now, let's test the options.

Option (A): Point (1, -1)

Distance from $L_1$: $d_1 = \frac{|4(1) + 3(-1) + 10|}{5} = \frac{|4 - 3 + 10|}{5} = \frac{|11|}{5} = \frac{11}{5}$.

Distance from $L_2$: $d_2 = \frac{|5(1) - 12(-1) + 26|}{13} = \frac{|5 + 12 + 26|}{13} = \frac{|43|}{13} = \frac{43}{13}$.

Since $\frac{11}{5} \neq \frac{43}{13}$, the point (1, -1) is not equidistant.

Option (B): Point (1, 1)

Distance from $L_1$: $d_1 = \frac{|4(1) + 3(1) + 10|}{5} = \frac{|4 + 3 + 10|}{5} = \frac{|17|}{5} = \frac{17}{5}$.

Distance from $L_2$: $d_2 = \frac{|5(1) - 12(1) + 26|}{13} = \frac{|5 - 12 + 26|}{13} = \frac{|19|}{13} = \frac{19}{13}$.

Since $\frac{17}{5} \neq \frac{19}{13}$, the point (1, 1) is not equidistant.

Option (C): Point (0, 0)

Distance from $L_1$: $d_1 = \frac{|4(0) + 3(0) + 10|}{5} = \frac{|10|}{5} = \frac{10}{5} = 2$.

Distance from $L_2$: $d_2 = \frac{|5(0) - 12(0) + 26|}{13} = \frac{|26|}{13} = \frac{26}{13} = 2$.

Distance from $L_3$: $d_3 = \frac{|7(0) + 24(0) - 50|}{25} = \frac{|-50|}{25} = \frac{50}{25} = 2$.

Since $d_1 = d_2 = d_3 = 2$, the point (0, 0) is equidistant from the three lines.

We do not need to check Option (D) as we have found a point that satisfies the condition.

Final Answer:

The point equidistant from the three lines is (0, 0).

Correct Option: (C)

Given:

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

The line is perpendicular to the line $3x + y = 3$.

To Find:

The y-intercept of the required line.

Solution:

The equation of the given line is $3x + y = 3$.

We can rewrite this in the slope-intercept form $y = mx + c$:

$y = -3x + 3$

The slope of the given line is $m_{given} = -3$.

Let $m_{required}$ be the slope of the required line. Since the required line is perpendicular to the given line, the product of their slopes is $-1$ (for non-vertical lines, which these are).

$m_{required} \times m_{given} = -1$

$m_{required} \times (-3) = -1$

$m_{required} = \frac{-1}{-3} = \frac{1}{3}$

The required line passes through the point $(x_1, y_1) = (2, 2)$ and has a slope $m = \frac{1}{3}$.

Using the point-slope form of the equation of a line, $y - y_1 = m(x - x_1)$:

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

To find the y-intercept, we set $x = 0$ in this equation:

$y - 2 = \frac{1}{3}(0 - 2)$

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

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

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

$y = \frac{6}{3} - \frac{2}{3}$

$y = \frac{6 - 2}{3}$

$y = \frac{4}{3}$

The y-intercept is the value of $y$ when $x=0$, which is $\frac{4}{3}$.

Final Answer:

The y-intercept of the line is $\frac{4}{3}$.

Comparing with the given options, option (D) matches our result.

Correct Option: (D)

Given:

Line 1 ($L_1$): $3x + 4y + 5 = 0$

Line 2 ($L_2$): $3x + 4y - 5 = 0$

Line 3 ($L_3$): $3x + 4y + 2 = 0$

To Find:

The ratio in which $L_3$ divides the distance between $L_1$ and $L_2$.

Solution:

The given lines are $L_1: 3x + 4y + 5 = 0$, $L_2: 3x + 4y - 5 = 0$, and $L_3: 3x + 4y + 2 = 0$.

All three lines have the same coefficients for $x$ and $y$ (A=3, B=4). This means they are parallel lines.

The distance between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by the formula $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$.

For the given lines, $A=3$ and $B=4$. So, $\sqrt{A^2 + B^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.

The constant terms for the three lines are $C_1 = 5$, $C_2 = -5$, and $C_3 = 2$.

Since $-5 < 2 < 5$, the line $L_3$ lies between the lines $L_1$ and $L_2$.

The ratio in which $L_3$ divides the distance between $L_1$ and $L_2$ is the ratio of the distance from $L_1$ to $L_3$ and the distance from $L_3$ to $L_2$.

Distance between $L_1$ and $L_3$ ($d_1$):

$d_1 = \frac{|C_1 - C_3|}{\sqrt{A^2 + B^2}} = \frac{|5 - 2|}{5} = \frac{|3|}{5} = \frac{3}{5}$

Distance between $L_3$ and $L_2$ ($d_2$):

$d_2 = \frac{|C_3 - C_2|}{\sqrt{A^2 + B^2}} = \frac{|2 - (-5)|}{5} = \frac{|2 + 5|}{5} = \frac{|7|}{5} = \frac{7}{5}$

The ratio in which $L_3$ divides the distance between $L_1$ and $L_2$ is $d_1 : d_2$.

Ratio $= \frac{3}{5} : \frac{7}{5} = \frac{3/5}{7/5} = \frac{3}{7}$

The ratio is 3 : 7.

Final Answer:

The ratio in which the line $3x + 4y + 2 = 0$ divides the distance between the lines $3x + 4y + 5 = 0$ and $3x + 4y - 5 = 0$ is 3 : 7.

Correct Option: (B)

Given:

An equilateral triangle.

Centroid is at the origin $(0, 0)$.

Equation of one side is $x + y - 2 = 0$.

To Find:

The coordinates of one vertex of the triangle.

Solution:

Let the equilateral triangle be ABC, and let the vertex opposite to the side $x + y - 2 = 0$ be A$(h, k)$. Let O$(0, 0)$ be the centroid of the triangle.

In an equilateral triangle, the centroid coincides with the circumcenter, incenter, and orthocenter. The median from a vertex is also the altitude to the opposite side.

Let D$(\alpha, \beta)$ be the foot of the perpendicular from A to the side BC (which is the line $x + y - 2 = 0$). D is the midpoint of BC.

The centroid O divides the median AD in the ratio $2 : 1$ (AO : OD = 2 : 1).

Using the section formula for internal division, the coordinates of the centroid O$(0, 0)$ that divides the segment AD [A$(h, k)$, D$(\alpha, \beta)$] in the ratio $2:1$ are:

$0 = \frac{1 \cdot h + 2 \cdot \alpha}{1 + 2} = \frac{h + 2\alpha}{3}$

$0 = \frac{1 \cdot k + 2 \cdot \beta}{1 + 2} = \frac{k + 2\beta}{3}$

From these equations, we get:

The point D$(\alpha, \beta)$ lies on the line $x + y - 2 = 0$. So, substitute the coordinates of D into the line equation:

The line segment AD is the altitude from A to BC, so AD is perpendicular to the line $x + y - 2 = 0$.

The slope of the line $x + y - 2 = 0$ is $m_{BC} = -\frac{\text{Coefficient of } x}{\text{Coefficient of } y} = -\frac{1}{1} = -1$.

The slope of the line AD passing through A$(h, k)$ and D$(\alpha, \beta)$ is $m_{AD} = \frac{\beta - k}{\alpha - h}$.

Since AD is perpendicular to BC, the product of their slopes is $-1$:

$\frac{\beta - k}{\alpha - h} \times (-1) = -1$

$\frac{\beta - k}{\alpha - h} = 1$}

$\beta - k = \alpha - h$}

Substitute the values of $h = -2\alpha$ and $k = -2\beta$ from equations (i) and (ii):

$\beta - (-2\beta) = \alpha - (-2\alpha)$

$\beta + 2\beta = \alpha + 2\alpha$}

$3\beta = 3\alpha$}

$\beta = \alpha$}

Now substitute $\beta = \alpha$ into equation (iii):

$\alpha + \alpha = 2$}

$2\alpha = 2$}

$\alpha = 1$}

Since $\beta = \alpha$, we have $\beta = 1$.

Now find the coordinates $(h, k)$ of the vertex A using equations (i) and (ii):

$h = -2\alpha = -2(1) = -2$

$k = -2\beta = -2(1) = -2$

The coordinates of the vertex A are $(-2, -2)$.

Final Answer:

One vertex of the equilateral triangle is $(-2, -2)$.

Comparing with the given options, option (C) matches our result.

Correct Option: (C)

Given:

a, b, and c are in Arithmetic Progression (A.P.).

The equation of a straight line is $ax + by + c = 0$.

To Find:

The fixed point through which the line $ax + by + c = 0$ always passes.

Solution:

If a, b, and c are in A.P., the condition that relates them is that the difference between consecutive terms is constant. This can be written as:

$b - a = c - b$

Rearranging this equation, we get:

$2b = a + c$

Or, equivalently:

$a - 2b + c = 0$

The equation of the straight line is given by:

$ax + by + c = 0$

We are looking for a point $(x, y)$ such that this equation is satisfied for any values of a, b, and c that are in A.P. This means that when we substitute the coordinates of the point $(x, y)$ into the line equation, the resulting expression should be equivalent to the A.P. condition $a - 2b + c = 0$.

Comparing the A.P. condition $a(1) + b(-2) + c(1) = 0$ with the line equation $a(x) + b(y) + c(1) = 0$, we can see that if we set $x=1$ and $y=-2$, the line equation becomes $a(1) + b(-2) + c = 0$, which is $a - 2b + c = 0$. This is exactly the condition for a, b, c to be in A.P.

Thus, the point $(1, -2)$ lies on the line $ax + by + c = 0$ whenever a, b, and c are in A.P.

Alternatively, let the line pass through a fixed point $(x_0, y_0)$. Then, substituting this point into the line equation, we get:

$ax_0 + by_0 + c = 0$

Since a, b, c are in A.P., we have $a - 2b + c = 0$.

We can express c from the A.P. condition: $c = 2b - a$. Substituting this into the equation of the line passing through $(x_0, y_0)$:

$ax_0 + by_0 + (2b - a) = 0$

Rearranging the terms to group a and b:

$a(x_0 - 1) + b(y_0 + 2) = 0$

This equation must hold true for any a and b (as long as c is determined by the A.P. condition, e.g., $c = 2b - a$). For this linear combination of a and b to be always zero, the coefficients of a and b must both be zero.

$x_0 - 1 = 0 \implies x_0 = 1$

$y_0 + 2 = 0 \implies y_0 = -2$

Thus, the fixed point is $(1, -2)$.

The straight lines $ax + by + c = 0$ will always pass through $(1, -2)$.

Given:

The straight line cuts off equal intercepts from the axes.

The straight line passes through the point $(1, -2)$.

To Find:

The equation of the straight line.

Solution:

Let the equation of the straight line in intercept form be:

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

where 'a' is the x-intercept and 'b' is the y-intercept.

According to the problem, the line cuts off equal intercepts from the axes. This means the x-intercept and the y-intercept are equal.

So, $a = b$.

Substituting $b=a$ into the intercept form equation, we get:

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

To simplify this equation, we can multiply both sides by 'a':

$x + y = a$

We are also given that the line passes through the point $(1, -2)$. This means that when we substitute $x=1$ and $y=-2$ into the equation of the line, the equation must be satisfied.

Substituting the coordinates $(1, -2)$ into the equation $x + y = a$:

$1 + (-2) = a$

$1 - 2 = a$

$-1 = a$

Now we have the value of 'a'. Substitute this value back into the equation $x + y = a$ to get the equation of the line.

$x + y = -1$

We can write this equation in the standard form $Ax + By + C = 0$ by moving the constant term to the left side:

$x + y + 1 = 0$

Conclusion:

The equation of the line which cuts off equal intercepts from the axes and passes through the point $(1, -2)$ is $x + y + 1 = 0$.

The blank should be filled with $x + y + 1 = 0$.

Given:

Point $(x_1, y_1) = (3, 2)$ through which the required lines pass.

Angle $\theta = 45^\circ$ that the required lines make with the line $x - 2y = 3$.

The equation of the given line is $L_1: x - 2y = 3$.

To Find:

The equations of the straight lines passing through $(3, 2)$ and making an angle of $45^\circ$ with the line $x - 2y = 3$.

Solution:

First, let's find the slope of the given line $L_1: x - 2y = 3$. We can rewrite this equation in the slope-intercept form $y = mx + c$ to find its slope.

$x - 2y = 3$

$-2y = -x + 3$

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

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

The slope of the given line $L_1$ is $m_1 = \frac{1}{2}$.

Let $L_2$ be a required line passing through $(3, 2)$ and making an angle $\theta = 45^\circ$ with $L_1$. Let the slope of $L_2$ be $m_2$.

The formula for the angle $\theta$ 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 the given values $\theta = 45^\circ$ and $m_1 = \frac{1}{2}$ into the formula:

$\tan 45^\circ = \left|\frac{m_2 - \frac{1}{2}}{1 + \frac{1}{2} m_2}\right|$

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

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

This absolute value equation implies two possible cases:

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

$2m_2 - 1 = 1 \cdot (2 + m_2)$

$2m_2 - 1 = 2 + m_2$

$2m_2 - m_2 = 2 + 1$

$m_2 = 3$

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

$2m_2 - 1 = -1 \cdot (2 + m_2)$

$2m_2 - 1 = -2 - m_2$

$2m_2 + m_2 = -2 + 1$

$3m_2 = -1$

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

So, there are two possible slopes for the required lines: $m_2 = 3$ and $m_2 = -\frac{1}{3}$.

Now, we use the point-slope form of the equation of a line, $y - y_1 = m(x - x_1)$, with the point $(x_1, y_1) = (3, 2)$ to find the equations of the two lines.

For $m_2 = 3$:

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

$y - 2 = 3x - 9$

$y = 3x - 9 + 2$

$y = 3x - 7$

This can be written in the standard form $3x - y - 7 = 0$.

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

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

Multiply both sides by 3 to clear the fraction:

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

$3y - 6 = -x + 3$

Rearrange the terms to get the standard form:

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

$x + 3y - 9 = 0$

Conclusion:

The equations of the lines through the point $(3, 2)$ and making an angle of $45^\circ$ with the line $x - 2y = 3$ are $3x - y - 7 = 0$ and $x + 3y - 9 = 0$.

The blanks should be filled with $3x - y - 7 = 0$ and $x + 3y - 9 = 0$.

Given:

Point 1: $(x_1, y_1) = (3, 4)$

Point 2: $(x_2, y_2) = (2, -6)$

Equation of the line: $3x - 4y - 8 = 0$

To Find:

Whether the points $(3, 4)$ and $(2, -6)$ are situated on the same side or opposite sides of the line $3x - 4y - 8 = 0$.

Solution:

To determine the position of two points relative to a line $Ax + By + C = 0$, we substitute the coordinates of each point into the expression $Ax + By + C$. If the results have the same sign, the points lie on the same side of the line. If the results have opposite signs, the points lie on opposite sides of the line.

For the given line $3x - 4y - 8 = 0$, we have $A=3$, $B=-4$, and $C=-8$.

Substitute the coordinates of the first point $(3, 4)$ into the expression $3x - 4y - 8$:

$3(3) - 4(4) - 8$

$= 9 - 16 - 8$

$= -7 - 8$

$= -15$

The result is negative.

Substitute the coordinates of the second point $(2, -6)$ into the expression $3x - 4y - 8$:

$3(2) - 4(-6) - 8$

$= 6 + 24 - 8$

$= 30 - 8$

$= 22$

The result is positive.

Since the results of substituting the two points into the expression $3x - 4y - 8$ have opposite signs ($-15$ and $22$), the points $(3, 4)$ and $(2, -6)$ are on opposite sides of the line $3x - 4y - 8 = 0$.

Conclusion:

The points $(3, 4)$ and $(2, -6)$ are situated on the opposite sides of the line $3x - 4y - 8 = 0$.

Given:

Fixed point $P_1 = (3, -2)$.

Fixed line $L: 5x - 12y - 3 = 0$.

Condition for the locus of a point $P(x, y)$: Square of the distance from $P$ to $P_1$ is numerically equal to the distance from $P$ to $L$.

To Find:

The equation of the locus of point $P(x, y)$.

Solution:

Let the moving point be $P(x, y)$.

The distance between the point $P(x, y)$ and the point $P_1(3, -2)$ is given by the distance formula:

$d(P, P_1) = \sqrt{(x - 3)^2 + (y - (-2))^2} = \sqrt{(x - 3)^2 + (y + 2)^2}$

The square of this distance is:

$d(P, P_1)^2 = (x - 3)^2 + (y + 2)^2$

The distance from the point $P(x, y)$ to the line $5x - 12y - 3 = 0$ is given by the formula:

$d(P, L) = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$

Here, $(x_0, y_0) = (x, y)$, $A=5$, $B=-12$, and $C=-3$.

$d(P, L) = \frac{|5x - 12y - 3|}{\sqrt{5^2 + (-12)^2}}$

$d(P, L) = \frac{|5x - 12y - 3|}{\sqrt{25 + 144}}$

$d(P, L) = \frac{|5x - 12y - 3|}{\sqrt{169}}$

$d(P, L) = \frac{|5x - 12y - 3|}{13}$

According to the given condition, the square of the distance from $P$ to $P_1$ is equal to the distance from $P$ to $L$:

$d(P, P_1)^2 = d(P, L)$

$(x - 3)^2 + (y + 2)^2 = \frac{|5x - 12y - 3|}{13}$

Expand the squared terms on the left side:

$(x^2 - 6x + 9) + (y^2 + 4y + 4) = \frac{|5x - 12y - 3|}{13}$

$x^2 + y^2 - 6x + 4y + 13 = \frac{|5x - 12y - 3|}{13}$

Multiply both sides by 13:

$13(x^2 + y^2 - 6x + 4y + 13) = |5x - 12y - 3|$

$13x^2 + 13y^2 - 78x + 52y + 169 = |5x - 12y - 3|$

Due to the absolute value, we have two possible cases:

Case 1: $5x - 12y - 3 \geq 0$

$13x^2 + 13y^2 - 78x + 52y + 169 = 5x - 12y - 3$

Move all terms to the left side:

$13x^2 + 13y^2 - 78x - 5x + 52y + 12y + 169 + 3 = 0$

$13x^2 + 13y^2 - 83x + 64y + 172 = 0$

Case 2: $5x - 12y - 3 < 0$

$13x^2 + 13y^2 - 78x + 52y + 169 = -(5x - 12y - 3)$

$13x^2 + 13y^2 - 78x + 52y + 169 = -5x + 12y + 3$

Move all terms to the left side:

$13x^2 + 13y^2 - 78x + 5x + 52y - 12y + 169 - 3 = 0$

$13x^2 + 13y^2 - 73x + 40y + 166 = 0$

The locus consists of points satisfying either of these two equations.

Conclusion:

The equation of the locus is $13x^2 + 13y^2 - 83x + 64y + 172 = 0$ or $13x^2 + 13y^2 - 73x + 40y + 166 = 0$.

The blank should be filled with $13x^2 + 13y^2 - 83x + 64y + 172 = 0$ or $13x^2 + 13y^2 - 73x + 40y + 166 = 0$.

Given:

The equation of the line is $x \sin \theta + y \cos \theta = p$.

We need to find the locus of the mid-points of the segment of this line intercepted between the coordinate axes.

To Find:

The equation of the locus of the mid-points.

Solution:

The line intercepts the x-axis when $y=0$. Substituting $y=0$ into the equation of the line:

$x \sin \theta + 0 \cdot \cos \theta = p$

$x \sin \theta = p$

Assuming $\sin \theta \neq 0$ and $p \neq 0$, the x-intercept is at $x = \frac{p}{\sin \theta}$. Let this point be $A = \left(\frac{p}{\sin \theta}, 0\right)$.

The line intercepts the y-axis when $x=0$. Substituting $x=0$ into the equation of the line:

$0 \cdot \sin \theta + y \cos \theta = p$

$y \cos \theta = p$

Assuming $\cos \theta \neq 0$ and $p \neq 0$, the y-intercept is at $y = \frac{p}{\cos \theta}$. Let this point be $B = \left(0, \frac{p}{\cos \theta}\right)$.

Let $M(h, k)$ be the mid-point of the segment $AB$. Using the mid-point formula, the coordinates of $M$ are:

$h = \frac{\frac{p}{\sin \theta} + 0}{2} = \frac{p}{2 \sin \theta}$

$k = \frac{0 + \frac{p}{\cos \theta}}{2} = \frac{p}{2 \cos \theta}$

From these two equations, we can express $\sin \theta$ and $\cos \theta$ in terms of $h$, $k$, and $p$ (assuming $h \neq 0$, $k \neq 0$, $p \neq 0$):

$\sin \theta = \frac{p}{2h}$

$\cos \theta = \frac{p}{2k}$

We know the fundamental trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$. Substitute the expressions for $\sin \theta$ and $\cos \theta$ into this identity:

$\left(\frac{p}{2h}\right)^2 + \left(\frac{p}{2k}\right)^2 = 1$

$\frac{p^2}{4h^2} + \frac{p^2}{4k^2} = 1$

Divide the entire equation by $p^2$ (assuming $p \neq 0$):

$\frac{1}{4h^2} + \frac{1}{4k^2} = \frac{1}{p^2}$

Multiply the entire equation by 4:

$\frac{1}{h^2} + \frac{1}{k^2} = \frac{4}{p^2}$

To get the equation of the locus of the point $M(h, k)$, we replace $h$ with $x$ and $k$ with $y$:

$\frac{1}{x^2} + \frac{1}{y^2} = \frac{4}{p^2}$

This equation represents the locus of the mid-points under the assumption that a finite segment is intercepted between the axes (i.e., $p \neq 0$, $\sin\theta \neq 0$, and $\cos\theta \neq 0$). If $p=0$, the line passes through the origin, and the locus of the midpoint is just the origin $(0,0)$.

Conclusion:

The locus of the mid-points of the portion of the line $x \sin \theta + y \cos \theta = p$ intercepted between the axes is $\frac{1}{x^2} + \frac{1}{y^2} = \frac{4}{p^2}$.

The blank should be filled with $\frac{1}{x^2} + \frac{1}{y^2} = \frac{4}{p^2}$.

Statement: If the vertices of a triangle have integral coordinates, then the triangle can not be equilateral.

Answer: True

Justification:

Let the vertices of the triangle be $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$, where $x_1, y_1, x_2, y_2, x_3, y_3$ are all integers.

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

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

Since all coordinates $x_i, y_i$ are integers, the expression inside the absolute value, $x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)$, is an integer. Let this integer be $I$. Thus, the area of a triangle with integral coordinates is $\frac{1}{2}|I|$, which is always a rational number (either an integer or a half-integer).

Now, let's consider an equilateral triangle with side length $s$. The area of an equilateral triangle is given by the formula:

Area $= \frac{\sqrt{3}}{4} s^2$

If the vertices $A, B, C$ have integral coordinates, the square of the side length $s^2$ must be an integer. For example, $AB^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$. Since $x_1, y_1, x_2, y_2$ are integers, $(x_2 - x_1)$ and $(y_2 - y_1)$ are integers, and their squares are non-negative integers. Therefore, their sum $(x_2 - x_1)^2 + (y_2 - y_1)^2$ is a non-negative integer. Let $s^2 = N$, where $N$ is a positive integer (for a non-degenerate triangle).

Substituting $s^2 = N$ into the area formula for an equilateral triangle, we get:

Area $= \frac{\sqrt{3}}{4} N$

We now have two expressions for the area of the triangle:

Area $= \frac{1}{2}|I|$ (Rational number)

Area $= \frac{\sqrt{3}}{4} N$ (Involving $\sqrt{3}$)

Equating these two expressions:

$\frac{1}{2}|I| = \frac{\sqrt{3}}{4} N$

$|I| = \frac{\sqrt{3}}{2} N$

Assuming $N \neq 0$ (since it's a triangle), we can write:

$\sqrt{3} = \frac{2|I|}{N}$

The left side of the equation, $\sqrt{3}$, is an irrational number.

The right side of the equation, $\frac{2|I|}{N}$, is a rational number because $I$ is an integer, $|I|$ is a non-negative integer, $N$ is a positive integer, and the quotient of two integers (with a non-zero denominator) is a rational number.

An irrational number cannot be equal to a rational number. This is a contradiction.

The assumption that an equilateral triangle can have integral coordinates for its vertices leads to this contradiction. Therefore, the assumption must be false.

Thus, if the vertices of a triangle have integral coordinates, the triangle cannot be equilateral.

Statement: The points A (– 2, 1), B (0, 5), C (– 1, 2) are collinear.

Answer: False

Justification:

To check if three points are collinear, we can calculate the area of the triangle formed by these points. If the area is zero, the points are collinear. Otherwise, they are not.

Let the coordinates of the points be $A(x_1, y_1) = (-2, 1)$, $B(x_2, y_2) = (0, 5)$, and $C(x_3, y_3) = (-1, 2)$.

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

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

Substitute the coordinates of the points A, B, and C into this formula:

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

Area $= \frac{1}{2} |-2(3) + 0(1) + (-1)(-4)|$

Area $= \frac{1}{2} |-6 + 0 + 4|$

Area $= \frac{1}{2} |-2|$

Area $= \frac{1}{2} \times 2$

Area $= 1$

Since the area of the triangle formed by the points A, B, and C is $1$, which is not equal to zero, the points are not collinear.

Statement: Equation of the line passing through the point $(a\; \cos^3 θ,\; a\; \sin^3 θ)$ and perpendicular to the line $x\; \sec θ + y\; \text{cosec}\; θ = a$ is $x \cos θ \;–\; y \sin θ = a \sin 2θ$.

Answer: False

Justification:

Let the given line be $L_1$: $x \sec θ + y \text{cosec} θ = a$.

We can rewrite this equation as:

$\frac{x}{\cos θ} + \frac{y}{\sin θ} = a$

Assuming $\cos θ \neq 0$ and $\sin θ \neq 0$, the slope of this line is $m_1 = -\frac{\text{coefficient of } x}{\text{coefficient of } y} = -\frac{\sec θ}{\text{cosec} θ} = -\frac{1/\cos θ}{1/\sin θ} = -\frac{\sin θ}{\cos θ} = -\tan θ$.

Let the required line be $L_2$. Since $L_2$ is perpendicular to $L_1$, its slope $m_2$ must satisfy $m_1 m_2 = -1$.

$m_2 = -\frac{1}{m_1} = -\frac{1}{-\tan θ} = \frac{1}{\tan θ} = \cot θ = \frac{\cos θ}{\sin θ}$.

The required line $L_2$ passes through the point $(x_1, y_1) = (a \cos^3 θ, a \sin^3 θ)$ and has slope $m_2 = \cot θ = \frac{\cos θ}{\sin θ}$. Using the point-slope form $y - y_1 = m_2(x - x_1)$:

$y - a \sin^3 θ = \frac{\cos θ}{\sin θ} (x - a \cos^3 θ)$

Multiply both sides by $\sin θ$ (assuming $\sin θ \neq 0$):

$\sin θ (y - a \sin^3 θ) = \cos θ (x - a \cos^3 θ)$

$y \sin θ - a \sin^4 θ = x \cos θ - a \cos^4 θ$

Rearrange the terms to match the form $x \cos θ - y \sin θ = \dots$:

$x \cos θ - y \sin θ = a \cos^4 θ - a \sin^4 θ$

$x \cos θ - y \sin θ = a (\cos^4 θ - \sin^4 θ)$

Factor the expression on the right using the difference of squares formula $A^2 - B^2 = (A-B)(A+B)$, where $A = \cos^2 θ$ and $B = \sin^2 θ$:

$\cos^4 θ - \sin^4 θ = (\cos^2 θ)^2 - (\sin^2 θ)^2 = (\cos^2 θ - \sin^2 θ)(\cos^2 θ + \sin^2 θ)$

Using the trigonometric identities $\cos^2 θ - \sin^2 θ = \cos(2θ)$ and $\cos^2 θ + \sin^2 θ = 1$:

$\cos^4 θ - \sin^4 θ = \cos(2θ) \cdot 1 = \cos(2θ)$

Substitute this back into the equation of the line:

$x \cos θ - y \sin θ = a \cos(2θ)$

This is the equation of the line passing through the given point and perpendicular to the given line (assuming $\sin θ \neq 0$, $\cos θ \neq 0$, and $a \neq 0$).

The statement claims the equation is $x \cos θ \;–\; y \sin θ = a \sin 2θ$.

Comparing our derived equation ($x \cos θ - y \sin θ = a \cos(2θ)$) with the claimed equation ($x \cos θ - y \sin θ = a \sin 2θ$), the right-hand sides are generally different ($a \cos(2θ) \neq a \sin(2θ)$ for most values of $\theta$ and $a \neq 0$).

For example, if $\theta = \pi/4$ and $a=1$, the point is $(a \cos^3(\pi/4), a \sin^3(\pi/4)) = (1 \cdot (\frac{1}{\sqrt{2}})^3, 1 \cdot (\frac{1}{\sqrt{2}})^3) = (\frac{1}{2\sqrt{2}}, \frac{1}{2\sqrt{2}})$. The given line is $x \sec(\pi/4) + y \text{cosec}(\pi/4) = 1 \implies x\sqrt{2} + y\sqrt{2} = 1$. Its slope is $-1$. The perpendicular slope is $1$. The line through $(\frac{1}{2\sqrt{2}}, \frac{1}{2\sqrt{2}})$ with slope 1 is $y - \frac{1}{2\sqrt{2}} = 1(x - \frac{1}{2\sqrt{2}}) \implies y = x$. Our derived equation for $\theta=\pi/4, a=1$ is $x \cos(\pi/4) - y \sin(\pi/4) = 1 \cdot \cos(2 \cdot \pi/4) \implies x \frac{1}{\sqrt{2}} - y \frac{1}{\sqrt{2}} = \cos(\pi/2) = 0 \implies \frac{1}{\sqrt{2}}(x-y) = 0 \implies x-y=0 \implies y=x$. This matches. The claimed equation for $\theta=\pi/4, a=1$ is $x \cos(\pi/4) - y \sin(\pi/4) = 1 \cdot \sin(2 \cdot \pi/4) \implies x \frac{1}{\sqrt{2}} - y \frac{1}{\sqrt{2}} = \sin(\pi/2) = 1 \implies \frac{1}{\sqrt{2}}(x-y) = 1 \implies x-y = \sqrt{2}$. This does not match $y=x$.

The derived equation is $x \cos θ - y \sin θ = a \cos(2θ)$. The statement claims it is $x \cos θ - y \sin θ = a \sin(2θ)$. Since $\cos(2θ)$ is not identically equal to $\sin(2θ)$, the statement is false in general.

Statement: The straight line $5x + 4y = 0$ passes through the point of intersection of the straight lines $x + 2y – 10 = 0$ and $2x + y + 5 = 0$.

Answer: True

Justification:

To determine if the statement is true, we need to find the point of intersection of the lines $x + 2y – 10 = 0$ and $2x + y + 5 = 0$ and then check if this point lies on the line $5x + 4y = 0$.

We have the system of equations:

$x + 2y = 10$           ... (i)

$2x + y = -5$           ... (ii)

From equation (ii), we can express $y$ in terms of $x$:

$y = -5 - 2x$

Substitute this expression for $y$ into equation (i):

$x + 2(-5 - 2x) = 10$

$x - 10 - 4x = 10$

$-3x - 10 = 10$

$-3x = 10 + 10$

$-3x = 20$

$x = -\frac{20}{3}$

Now substitute the value of $x$ back into the expression for $y$:

$y = -5 - 2\left(-\frac{20}{3}\right)$

$y = -5 + \frac{40}{3}$

$y = \frac{-15}{3} + \frac{40}{3}$

$y = \frac{40 - 15}{3}$

$y = \frac{25}{3}$

The point of intersection of the two lines is $\left(-\frac{20}{3}, \frac{25}{3}\right)$.

Now, we check if this point lies on the line $5x + 4y = 0$ by substituting the coordinates into the equation:

$5\left(-\frac{20}{3}\right) + 4\left(\frac{25}{3}\right)$

$= -\frac{100}{3} + \frac{100}{3}$

$= 0$

Since the substitution results in $0$, which is equal to the right-hand side of the equation $5x + 4y = 0$, the point of intersection $\left(-\frac{20}{3}, \frac{25}{3}\right)$ lies on the line $5x + 4y = 0$.

Therefore, the straight line $5x + 4y = 0$ passes through the point of intersection of the straight lines $x + 2y – 10 = 0$ and $2x + y + 5 = 0$. The statement is true.

Statement: The vertex of an equilateral triangle is (2, 3) and the equation of the opposite side is $x + y = 2$. Then the other two sides are $y \;–\; 3 = (2 ± \sqrt{3})\; (x \;–\; 2)$.

Answer: True

Justification:

Let the given vertex be $A = (2, 3)$.

The equation of the side opposite to vertex $A$ is $L_1: x + y = 2$. We can write this as $x + y - 2 = 0$.

The slope of the line $L_1$ is $m_1 = -\frac{\text{coefficient of } x}{\text{coefficient of } y} = -\frac{1}{1} = -1$.

In an equilateral triangle, each interior angle is $60^\circ$. The other two sides of the triangle pass through the vertex $(2, 3)$ and make an angle of $60^\circ$ with the opposite side $L_1$.

Let the slopes of the other two sides be $m$. The angle $\theta$ between a line with slope $m$ and the line $L_1$ with slope $m_1 = -1$ is $60^\circ$. The formula for the angle between two lines is:

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

Substitute $\theta = 60^\circ$ and $m_1 = -1$:

$\tan 60^\circ = \left|\frac{m - (-1)}{1 + m (-1)}\right|$

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

This gives two possibilities:

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

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

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

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

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

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

To rationalize the denominator, multiply the numerator and denominator by $(\sqrt{3} - 1)$:

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

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

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

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

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

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

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

To rationalize the denominator, multiply the numerator and denominator by $(\sqrt{3} + 1)$:

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

The slopes of the other two sides are $2 + \sqrt{3}$ and $2 - \sqrt{3}$.

The equation of a line passing through a point $(x_1, y_1)$ with slope $m$ is given by the point-slope form: $y - y_1 = m(x - x_1)$.

Here, $(x_1, y_1) = (2, 3)$.

For the slope $m = 2 + \sqrt{3}$:

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

For the slope $m = 2 - \sqrt{3}$:

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

These two equations can be combined and written as:

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

This matches the equation given in the statement.

Thus, the statement is true.

Statement: The equation of the line joining the point (3, 5) to the point of intersection of the lines $4x + y – 1 = 0$ and $7x – 3y – 35 = 0$ is equidistant from the points (0, 0) and (8, 34).

Answer: True

Justification:

First, we find the point of intersection of the two given lines:

$L_1: 4x + y - 1 = 0 \implies 4x + y = 1$

$L_2: 7x - 3y - 35 = 0 \implies 7x - 3y = 35$

From the equation $4x + y = 1$, we get $y = 1 - 4x$. Substitute this into the second equation:

$7x - 3(1 - 4x) = 35$

$7x - 3 + 12x = 35$

$19x = 35 + 3$

$19x = 38$

$x = \frac{38}{19} = 2$

Now substitute $x=2$ back into the expression for $y$:

$y = 1 - 4(2) = 1 - 8 = -7$

The point of intersection is $(2, -7)$.

Next, we find the equation of the line passing through the point $(3, 5)$ and the point of intersection $(2, -7)$.

Let the two points be $(x_1, y_1) = (3, 5)$ and $(x_2, y_2) = (2, -7)$.

The slope of the line is $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-7 - 5}{2 - 3} = \frac{-12}{-1} = 12$.

Using the point-slope form $y - y_1 = m(x - x_1)$ with the point $(3, 5)$ and slope $m=12$:

$y - 5 = 12(x - 3)$

$y - 5 = 12x - 36$

Rearranging into the standard form $Ax + By + C = 0$:

$12x - y - 36 + 5 = 0$

$12x - y - 31 = 0$

Now we check if this line is equidistant from the points $(0, 0)$ and $(8, 34)$.

The distance from a point $(x_0, y_0)$ to the line $Ax + By + C = 0$ is given by $d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$.

For the line $12x - y - 31 = 0$, we have $A=12$, $B=-1$, $C=-31$.

The denominator is $\sqrt{A^2 + B^2} = \sqrt{12^2 + (-1)^2} = \sqrt{144 + 1} = \sqrt{145}$.

Distance from $(0, 0)$ to the line $12x - y - 31 = 0$:

$d_1 = \frac{|12(0) - 1(0) - 31|}{\sqrt{145}} = \frac{|-31|}{\sqrt{145}} = \frac{31}{\sqrt{145}}$

Distance from $(8, 34)$ to the line $12x - y - 31 = 0$:

$d_2 = \frac{|12(8) - 1(34) - 31|}{\sqrt{145}} = \frac{|96 - 34 - 31|}{\sqrt{145}} = \frac{|62 - 31|}{\sqrt{145}} = \frac{|31|}{\sqrt{145}} = \frac{31}{\sqrt{145}}$

Since $d_1 = d_2 = \frac{31}{\sqrt{145}}$, the line $12x - y - 31 = 0$ is equidistant from the points $(0, 0)$ and $(8, 34)$.

Thus, the statement is true.

The statement is True.

Justification:

Let the equation of the given line be $\frac{x}{a} + \frac{y}{b} = 1$. We can rewrite this equation in the standard form $Ax + By + C = 0$ as $bx + ay - ab = 0$.

Let $(h, k)$ be the coordinates of the foot of the perpendicular from the origin $(0, 0)$ to the line $bx + ay - ab = 0$.

The formula for the foot of the perpendicular $(h, k)$ from a point $(x_0, y_0)$ to the line $Ax + By + C = 0$ is given by:

$\frac{h - x_0}{A} = \frac{k - y_0}{B} = -\frac{Ax_0 + By_0 + C}{A^2 + B^2}$

In this case, $(x_0, y_0) = (0, 0)$, $A = b$, $B = a$, and $C = -ab$. Substituting these values into the formula:

$\frac{h - 0}{b} = \frac{k - 0}{a} = -\frac{b(0) + a(0) - ab}{b^2 + a^2}$

$\frac{h}{b} = \frac{k}{a} = \frac{ab}{a^2 + b^2}$

From this, we can express $h$ and $k$ in terms of $a$ and $b$:

$h = \frac{b(ab)}{a^2 + b^2} = \frac{ab^2}{a^2 + b^2}$

$k = \frac{a(ab)}{a^2 + b^2} = \frac{a^2b}{a^2 + b^2}$

Now, we need to find the locus of $(h, k)$. We can find the value of $h^2 + k^2$:

$h^2 + k^2 = \left(\frac{ab^2}{a^2 + b^2}\right)^2 + \left(\frac{a^2b}{a^2 + b^2}\right)^2$

$h^2 + k^2 = \frac{a^2b^4}{(a^2 + b^2)^2} + \frac{a^4b^2}{(a^2 + b^2)^2}$

$h^2 + k^2 = \frac{a^2b^2(b^2 + a^2)}{(a^2 + b^2)^2}$

$h^2 + k^2 = \frac{a^2b^2}{a^2 + b^2}$

We are given the condition $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c^2}$.

Combining the terms on the left side:

$\frac{b^2 + a^2}{a^2 b^2} = \frac{1}{c^2}$

Taking the reciprocal of both sides:

$\frac{a^2 b^2}{a^2 + b^2} = c^2$

Now, substituting this result back into the expression for $h^2 + k^2$:

$h^2 + k^2 = c^2$

Replacing $(h, k)$ with $(x, y)$ to represent the general point on the locus, we get the equation of the locus as $x^2 + y^2 = c^2$.

This matches the equation given in the statement.

Therefore, the statement is true.

The statement is False.

Justification:

For three lines $a_1x + b_1y + c_1 = 0$, $a_2x + b_2y + c_2 = 0$, and $a_3x + b_3y + c_3 = 0$ to be concurrent, the determinant of their coefficients must be zero.

For the given lines:

$ax + 2y + 1 = 0$

$bx + 3y + 1 = 0$

$cx + 4y + 1 = 0$

The condition for concurrency is:

$\begin{vmatrix} a & 2 & 1 \\ b & 3 & 1 \\ c & 4 & 1 \end{vmatrix} = 0$

Expanding the determinant along the first row, we get:

$a(3 \times 1 - 4 \times 1) - 2(b \times 1 - c \times 1) + 1(b \times 4 - c \times 3) = 0$

$a(3 - 4) - 2(b - c) + (4b - 3c) = 0$

$-a - 2b + 2c + 4b - 3c = 0$

$-a + 2b - c = 0$

This can be written as:

$a + c = 2b$

This condition means that the coefficients $a, b, c$ must be in an Arithmetic Progression (A.P.) for the lines to be concurrent.

The statement claims that the lines are concurrent if $a, b, c$ are in Geometric Progression (G.P.).

The condition for $a, b, c$ to be in G.P. is $b^2 = ac$, assuming $b \neq 0$. If $b=0$, then $ac=0$, which implies either $a=0$ or $c=0$ (or both). If $a=b=0$, the first two lines are $2y+1=0$ and $3y+1=0$, which are parallel and not concurrent unless $c=0$. If $a=b=c=0$, the equations are $1=0$, which is impossible.

The statement "The lines are concurrent if $a, b, c$ are in G.P." means that if $a, b, c$ satisfy $b^2 = ac$, then they must also satisfy $a + c = 2b$. This is generally not true.

Consider a counterexample where $a, b, c$ are in G.P. but not in A.P.

Let $a=1, b=2, c=4$.

Check if $a, b, c$ are in G.P.: $b^2 = 2^2 = 4$, and $ac = 1 \times 4 = 4$. Since $b^2 = ac$, $1, 2, 4$ are in G.P.

Check if $a, b, c$ satisfy the concurrency condition ($a+c = 2b$): $a+c = 1+4 = 5$, and $2b = 2 \times 2 = 4$. Since $a+c \neq 2b$, the condition for concurrency is not met.

Thus, for $a=1, b=2, c=4$, the lines are $x + 2y + 1 = 0$, $2x + 3y + 1 = 0$, and $4x + 4y + 1 = 0$. Since the condition $a+c=2b$ is not satisfied (as $5 \neq 4$), these lines are not concurrent.

Since we have found a case where $a, b, c$ are in G.P. but the lines are not concurrent, the given statement is false. The lines are concurrent if and only if $a, b, c$ are in A.P.

The statement is False.

Justification:

Let the first line join points A$(3, -4)$ and B$(-2, 6)$.

The slope of line AB is given by the formula $m = \frac{y_2 - y_1}{x_2 - x_1}$.

Using the coordinates of A and B, the slope $m_1$ is:

$m_1 = \frac{6 - (-4)}{-2 - 3} = \frac{6 + 4}{-5} = \frac{10}{-5} = -2$.

Let the second line join points C$(-3, 6)$ and D$(9, -18)$.

Using the coordinates of C and D, the slope $m_2$ is:

$m_2 = \frac{-18 - 6}{9 - (-3)} = \frac{-24}{9 + 3} = \frac{-24}{12} = -2$.

Two non-vertical lines are perpendicular if the product of their slopes is $-1$. That is, $m_1 \times m_2 = -1$.

Let's calculate the product of the slopes $m_1$ and $m_2$:

$m_1 \times m_2 = (-2) \times (-2) = 4$.

Since $m_1 \times m_2 = 4$, which is not equal to $-1$, the lines are not perpendicular.

Therefore, the given statement is false.

Matching for Question 57:

We need to find the points described in each part of Column C1 and match them with the given options in Column C2.

Solution for (a):

The points P and Q lie on the line $x + 5y = 13$. Let a point on this line be $(x_0, y_0)$. Then $x_0 = 13 - 5y_0$. So any point on the line can be represented as $(13 - 5y, y)$.

The second line is $12x – 5y + 26 = 0$.

The distance from a point $(x_0, y_0)$ to the line $Ax + By + C = 0$ is given by the formula $\frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$.

Here, $(x_0, y_0) = (13 - 5y, y)$, $A=12$, $B=-5$, $C=26$. The distance is given as 2 units.

So, we have:

Simplifying the expression:

This gives two possible cases:

Case 1: $182 - 65y = 26$

$65y = 182 - 26$

$65y = 156$

Substitute $y = \frac{12}{5}$ into $x = 13 - 5y$:

The first point is $\left(1, \frac{12}{5}\right)$.

Case 2: $182 - 65y = -26$

$65y = 182 + 26$

$65y = 208$

Substitute $y = \frac{16}{5}$ into $x = 13 - 5y$:

The second point is $\left(-3, \frac{16}{5}\right)$.

The coordinates of the points are $\left(1, \frac{12}{5}\right)$ and $\left(-3, \frac{16}{5}\right)$.

This matches option (iii) in Column C2.

Solution for (b):

The point lies on the line $x + y = 4$. Let the point be $(x_0, y_0)$. Then $y_0 = 4 - x_0$. So any point on the line can be represented as $(x, 4 - x)$.

The second line is $4x + 3y – 10 = 0$.

The distance from a point $(x_0, y_0)$ to the line $Ax + By + C = 0$ is $\frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$.

Here, $(x_0, y_0) = (x, 4 - x)$, $A=4$, $B=3$, $C=-10$. The distance is given as 1 unit.

So, we have:

Simplifying the expression:

This gives two possible cases:

Case 1: $x + 2 = 5$

Substitute $x = 3$ into $y = 4 - x$:

The first point is $(3, 1)$.

Case 2: $x + 2 = -5$

Substitute $x = -7$ into $y = 4 - x$:

The second point is $(-7, 11)$.

The coordinates of the points are $(3, 1)$ and $(-7, 11)$.

This matches option (i) in Column C2.

Solution for (c):

We are given two points A $(-2, 5)$ and B $(3, 1)$. The points P and Q are on the line segment AB such that $AP = PQ = QB$. This means the points P and Q divide the segment AB into three equal parts.

Point P is the point that divides the line segment AB in the ratio 1:2.

Using the section formula, the coordinates of a point $(x, y)$ dividing the line segment joining $(x_1, y_1)$ and $(x_2, y_2)$ in the ratio $m:n$ are:

For point P, $(x_1, y_1) = (-2, 5)$, $(x_2, y_2) = (3, 1)$, $m=1$, $n=2$.

So, point P is $\left(-\frac{1}{3}, \frac{11}{3}\right)$.

Point Q is the point that divides the line segment AB in the ratio 2:1.

For point Q, $(x_1, y_1) = (-2, 5)$, $(x_2, y_2) = (3, 1)$, $m=2$, $n=1$.

So, point Q is $\left(\frac{4}{3}, \frac{7}{3}\right)$.

The coordinates of the points P and Q are $\left(-\frac{1}{3}, \frac{11}{3}\right)$ and $\left(\frac{4}{3}, \frac{7}{3}\right)$.

This matches option (ii) in Column C2.

Final Match:

Based on the calculations, the correct matches for Question 57 are:

(a) $\to$ (iii)

(b) $\to$ (i)

(c) $\to$ (ii)

Matching for Question 58:

The given family of lines is represented by the equation:

We can rewrite this equation in the standard form $Ax + By + C = 0$ by expanding and collecting terms:

Here, $A = 2 + 6\lambda$, $B = 3 - \lambda$, and $C = 4 + 12\lambda$.

Solution for (a): parallel to y-axis

A line is parallel to the y-axis if its equation is of the form $x = k$, which means the coefficient of $y$ is zero.

From equation (2), the coefficient of $y$ is $3 - \lambda$.

Setting the coefficient of $y$ to zero:

This matches option (iv) in Column C2.

Solution for (b): perpendicular to $7x + y – 4 = 0$

The slope of the line $Ax + By + C = 0$ is $m = -\frac{A}{B}$ (if $B \neq 0$).

The slope of the line $7x + y – 4 = 0$ is $m_1 = -\frac{7}{1} = -7$.

The slope of the family of lines from equation (2) is $m_2 = -\frac{2 + 6\lambda}{3 - \lambda}$ (if $3 - \lambda \neq 0$).

Two lines are perpendicular if the product of their slopes is -1 ($m_1 m_2 = -1$).

This matches option (iii) in Column C2.

Solution for (c): passes through $(1, 2)$

If the line passes through the point $(1, 2)$, substituting $x=1$ and $y=2$ into equation (1) must satisfy the equation.

This matches option (i) in Column C2.

Solution for (d): parallel to x-axis

A line is parallel to the x-axis if its equation is of the form $y = k$, which means the coefficient of $x$ is zero.

From equation (2), the coefficient of $x$ is $2 + 6\lambda$.

Setting the coefficient of $x$ to zero:

This matches option (ii) in Column C2.

Final Match:

Based on the calculations, the correct matches for Question 58 are:

(a) $\to$ (iv)

(b) $\to$ (iii)

(c) $\to$ (i)

(d) $\to$ (ii)

Matching for Question 59:

The equation of a line passing through the intersection of the lines $L_1 = 0$ and $L_2 = 0$ is given by $L_1 + \lambda L_2 = 0$, where $\lambda$ is a constant.

The given lines are $L_1: 2x - 3y = 0$ and $L_2: 4x - 5y - 2 = 0$.

The equation of the family of lines passing through their intersection is:

Expanding and rearranging the terms, we get:

This is the general equation of a line passing through the intersection of the given lines.

Solution for (a): through the point $(2, 1)$

If the line passes through the point $(2, 1)$, we substitute $x=2$ and $y=1$ into equation (1):

Substitute $\lambda = -1$ back into equation (2):

This matches option (iii) in Column C2.

Solution for (b): perpendicular to the line $x + 2y + 1 = 0$

The slope of the line $x + 2y + 1 = 0$ is $m_1 = -\frac{\text{coefficient of } x}{\text{coefficient of } y} = -\frac{1}{2}$.

From equation (2), the slope of the required line is $m_2 = -\frac{\text{coefficient of } x}{\text{coefficient of } y} = -\frac{2 + 4\lambda}{-3 - 5\lambda} = \frac{2 + 4\lambda}{3 + 5\lambda}$.

For perpendicular lines, the product of their slopes is -1 ($m_1 m_2 = -1$).

Substitute $\lambda = -\frac{2}{3}$ back into equation (2):

Multiplying by 3:

Multiplying by -1:

This matches option (i) in Column C2 ($2x - y = 4$).

Solution for (c): parallel to the line $3x – 4y + 5 = 0$

The slope of the line $3x – 4y + 5 = 0$ is $m_1 = -\frac{3}{-4} = \frac{3}{4}$.

The slope of the required line from equation (2) is $m_2 = \frac{2 + 4\lambda}{3 + 5\lambda}$.

For parallel lines, their slopes are equal ($m_1 = m_2$).

Substitute $\lambda = 1$ back into equation (2):

Dividing by 2:

This matches option (iv) in Column C2.

Solution for (d): equally inclined to the axes

A line equally inclined to the axes has a slope of either 1 or -1.

The slope of the required line from equation (2) is $m = \frac{2 + 4\lambda}{3 + 5\lambda}$.

Case 1: Slope $m = 1$

Substituting $\lambda = -1$ gives the line $x - y - 1 = 0$, which is option (iii). However, option (iii) is already matched with (a).

Case 2: Slope $m = -1$}

Substitute $\lambda = -\frac{5}{9}$ back into equation (2):

Multiplying by $-\frac{9}{2}$:

This matches option (ii) in Column C2.

Since options (i), (iii), and (iv) are matched with (b), (a), and (c) respectively, the only remaining option for (d) is (ii), which corresponds to the case where the slope is -1.

Final Match:

Based on the calculations, the correct matches for Question 59 are:

(a) $\to$ (iii)

(b) $\to$ (i)

(c) $\to$ (iv)

(d) $\to$ (ii)