Chapter 10 Straight Lines (Concepts)

Cartesian Plane

Coordinate geometry, also known as analytical geometry, provides a powerful bridge between algebra and geometry. This field was pioneered by the French mathematician René Descartes. Its fundamental tool is the Cartesian plane (or coordinate plane), a system that uses a pair of numbers, called coordinates, to uniquely determine the position of a point on a flat, two-dimensional surface.

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The Coordinate System

The Cartesian plane is formed by two perpendicular number lines that intersect at their zero points. These lines act as a reference grid for locating any point.

The Coordinate Axes

1. The X-axis: This is the horizontal number line. Positive numbers are to the right of the intersection point, and negative numbers are to the left. It is denoted by the line XX'.

2. The Y-axis: This is the vertical number line. Positive numbers are above the intersection point, and negative numbers are below. It is denoted by the line YY'.

3. The Origin: The point where the x-axis and y-axis intersect is called the origin. It is the starting point for all measurements and is denoted by the letter O. Its coordinates are $(0, 0)$.

Coordinates of a Point

Any point P on the plane is represented by an ordered pair of numbers $(x, y)$. The order is important, for instance, the point $(2, 5)$ is different from the point $(5, 2)$.

• The first coordinate, $x$, is the abscissa or the x-coordinate. It is the perpendicular distance of the point from the y-axis. It tells us how far to move right (positive) or left (negative) from the origin.
• The second coordinate, $y$, is the ordinate or the y-coordinate. It is the perpendicular distance of the point from the x-axis. It tells us how far to move up (positive) or down (negative) from the origin.

A point with abscissa $x$ and ordinate $y$ is written as P$(x, y)$. For example, to locate the point A(3, -4), we start at the origin, move 3 units to the right along the x-axis, and then move 4 units down parallel to the y-axis.

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Quadrants

The two coordinate axes divide the entire plane into four infinite regions. These regions are called quadrants. They are numbered using Roman numerals (I, II, III, IV) in a counter-clockwise direction, starting from the upper-right region.

The signs of the x and y coordinates are unique for each quadrant.

Quadrant	X-coordinate (Abscissa)	Y-coordinate (Ordinate)	Coordinates Form
First Quadrant (I)	Positive ($x > 0$)	Positive ($y > 0$)	(+, +)
Second Quadrant (II)	Negative ($x < 0$)	Positive ($y > 0$)	(−, +)
Third Quadrant (III)	Negative ($x < 0$)	Negative ($y < 0$)	(−, −)
Fourth Quadrant (IV)	Positive ($x > 0$)	Negative ($y < 0$)	(+, −)

Points on the Axes

Points that lie exactly on the coordinate axes are special because they do not belong to any quadrant.

• Any point on the x-axis has a y-coordinate of 0. Its form is $(x, 0)$.
• Any point on the y-axis has an x-coordinate of 0. Its form is $(0, y)$.
• The origin $(0, 0)$ is the unique point that lies on both the x-axis and the y-axis.

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Distance Formula and Section Formula

The Cartesian coordinate system allows us to express geometric ideas using algebra.

We can create formulas based on coordinates to find the distance between points or to locate a point that divides a line segment.

These formulas are fundamental tools in the study of coordinate geometry.

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Distance Formula

The Distance Formula is used to calculate the length of a line segment connecting two points in a plane.

This formula is a direct application of the Pythagorean theorem.

Derivation of the Distance Formula

Given: Two points, $P(x_1, y_1)$ and $Q(x_2, y_2)$, in the Cartesian plane.

To Find: The distance between P and Q.

Construction Required:

1. Draw a line through P parallel to the x-axis.

2. Draw a line through Q parallel to the y-axis.

3. Let these two lines intersect at a point R. The coordinates of R will be $(x_2, y_1)$.

This construction forms a right-angled triangle, $\triangle PQR$, with the right angle at R.

Proof:

Now, we find the lengths of the sides PR and QR.

The length of the horizontal side PR is the difference between the x-coordinates:

$PR = |x_2 - x_1|$

The length of the vertical side QR is the difference between the y-coordinates:

$QR = |y_2 - y_1|$

According to the Pythagorean theorem, in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Substituting the lengths of PR and QR:

To find the length PQ, we take the square root of both sides. Since distance cannot be negative, we only consider the positive root.

This is the Distance Formula.

Note that the result is the same if we use $(x_1 - x_2)^2$ and $(y_1 - y_2)^2$, as squaring removes any negative sign.

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Section Formula

The Section Formula helps us find the coordinates of a point that divides a line segment in a given ratio.

The point can divide the segment either internally (within the segment) or externally (outside the segment).

Internal Division

Consider a line segment joining points $A(x_1, y_1)$ and $B(x_2, y_2)$.

Let $P(x, y)$ be a point that lies on the segment AB.

If P divides AB internally in the ratio $m:n$, it means that $\frac{AP}{PB} = \frac{m}{n}$.

The coordinates of point P are given by the formula:

Midpoint Formula (A Special Case)

The midpoint is a point that divides a line segment into two equal parts.

This means the ratio of division is $1:1$. So, $m = 1$ and $n = 1$.

By substituting $m=1$ and $n=1$ into the internal section formula, we get the midpoint formula.

$P(x, y) = \left( \frac{1 \cdot x_2 + 1 \cdot x_1}{1+1}, \frac{1 \cdot y_2 + 1 \cdot y_1}{1+1} \right)$

The coordinates of the midpoint of the segment joining $(x_1, y_1)$ and $(x_2, y_2)$ are:

External Division

Consider a line segment joining points $A(x_1, y_1)$ and $B(x_2, y_2)$.

Let $P(x, y)$ be a point that lies on the line extended from AB, but not between A and B.

If P divides AB externally in the ratio $m:n$, it means $\frac{AP}{BP} = \frac{m}{n}$, where $m \ne n$.

The coordinates of point P are given by the formula:

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Centroid, Incentre and Area of a Triangle

Coordinate geometry provides formulas to find special points inside a triangle, such as the centroid and incentre.

It also allows us to calculate the area of a triangle using only the coordinates of its vertices.

These formulas are derived from fundamental concepts like the distance and section formulas.

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Centroid of a Triangle

The centroid is a specific point inside a triangle, often considered its "center of mass".

It is the point where the three medians of the triangle intersect.

A median is a line segment that connects a vertex to the midpoint of the opposite side.

An important property is that the centroid divides each median in the ratio $2:1$.

Derivation of the Centroid Formula

Given: A triangle with vertices $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$.

To Find: The coordinates of the centroid, G.

Solution:

First, consider the median AD, where D is the midpoint of the side BC.

We find the coordinates of D using the midpoint formula:

$D = \left(\frac{x_2+x_3}{2}, \frac{y_2+y_3}{2}\right)$

The centroid G lies on the median AD.

It divides AD in the ratio $AG : GD = 2 : 1$.

Now, we use the internal section formula to find the coordinates of G.

Here, the points are $A(x_1, y_1)$ and $D\left(\frac{x_2+x_3}{2}, \frac{y_2+y_3}{2}\right)$, and the ratio is $m:n = 2:1$.

The x-coordinate of G is:

$x = \frac{m x_D + n x_A}{m+n} = \frac{2 \left(\frac{x_2+x_3}{2}\right) + 1(x_1)}{2+1}$

$x = \frac{x_2+x_3+x_1}{3}$

The y-coordinate of G is:

$y = \frac{m y_D + n y_A}{m+n} = \frac{2 \left(\frac{y_2+y_3}{2}\right) + 1(y_1)}{2+1}$

$y = \frac{y_2+y_3+y_1}{3}$

So, the coordinates of the centroid G are the average of the coordinates of the vertices.

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Incentre of a Triangle

The incentre of a triangle is the point of concurrency of the internal bisectors of the angles of the triangle. It is equidistant from all three sides of the triangle. This equidistant point is the center of the incircle, which is the largest circle that can be inscribed within the triangle, touching all three sides.

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Formula for the Incentre

Let the vertices of a triangle $\triangle ABC$ be $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$.

Let $a, b, c$ be the lengths of the sides opposite to vertices A, B, and C, respectively. These lengths can be calculated using the distance formula:

$a = BC = \sqrt{(x_2-x_3)^2 + (y_2-y_3)^2}$

$b = AC = \sqrt{(x_1-x_3)^2 + (y_1-y_3)^2}$

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

The coordinates of the incentre, denoted by $I$, are given by the formula:

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Derivation of the Incentre Formula

The derivation uses the Angle Bisector Theorem and the Section Formula.

1. Angle Bisector Theorem: This theorem states that an angle bisector of a triangle divides the opposite side into two segments that are proportional to the other two sides of the triangle.

Let AD be the angle bisector of $\angle A$, where D is a point on BC. According to the Angle Bisector Theorem:

$\frac{BD}{DC} = \frac{\text{Side adjacent to BD}}{\text{Side adjacent to DC}} = \frac{AB}{AC} = \frac{c}{b}$

This means that the point D divides the side BC in the ratio $c:b$.

2. Find Coordinates of D: Using the section formula, the coordinates of D which divides $B(x_2, y_2)$ and $C(x_3, y_3)$ in the ratio $c:b$ are:

$D = \left( \frac{c x_3 + b x_2}{c+b}, \frac{c y_3 + b y_2}{c+b} \right)$

3. Incentre I divides AD: The incentre I lies on the angle bisector AD. Now, consider $\triangle ABD$. The line segment BI is the angle bisector of $\angle B$. Applying the Angle Bisector Theorem to $\triangle ABD$:

$\frac{AI}{ID} = \frac{AB}{BD}$

We know $AB = c$. We need to find the length of BD. Since D divides BC in the ratio $c:b$, we have:

$BD = \left( \frac{c}{c+b} \right) \times \text{length of BC} = \frac{c}{c+b} \times a = \frac{ac}{b+c}$

Now, substitute the value of BD back into the ratio:

$\frac{AI}{ID} = \frac{c}{\frac{ac}{b+c}} = \frac{c(b+c)}{ac} = \frac{b+c}{a}$

So, the incentre I divides the line segment AD in the ratio $(b+c):a$.

4. Find Coordinates of I: Now, we use the section formula again to find the coordinates of I, which divides the line segment joining $A(x_1, y_1)$ and $D\left( \frac{bx_2 + cx_3}{b+c}, \frac{by_2 + cy_3}{b+c} \right)$ in the ratio $(b+c):a$.

x-coordinate of I:

$x = \frac{(b+c) \times (\text{x-coord of D}) + a \times (\text{x-coord of A})}{(b+c)+a}$

$x = \frac{(b+c) \left( \frac{bx_2 + cx_3}{b+c} \right) + a x_1}{a+b+c} = \frac{bx_2 + cx_3 + ax_1}{a+b+c}$

y-coordinate of I:

$y = \frac{(b+c) \times (\text{y-coord of D}) + a \times (\text{y-coord of A})}{(b+c)+a}$

$y = \frac{(b+c) \left( \frac{by_2 + cy_3}{b+c} \right) + a y_1}{a+b+c} = \frac{by_2 + cy_3 + ay_1}{a+b+c}$

Combining and rearranging the terms gives the coordinates of the incentre I:

$I = \left( \frac{ax_1 + bx_2 + cx_3}{a+b+c}, \frac{ay_1 + by_2 + cy_3}{a+b+c} \right)$

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Area of a Triangle

The area of a triangle can be calculated directly from the coordinates of its three vertices without needing to know the side lengths or angles.

Let the vertices of the triangle be $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$.

The formula for the area of $\triangle ABC$ is:

The absolute value is taken because area must always be a positive quantity.

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Derivation of the Area Formula

We can derive this formula by calculating the area of the region under the vertices using trapeziums formed by dropping perpendiculars from the vertices to the x-axis.

Let's drop perpendiculars from A, B, and C to the x-axis, meeting the axis at P, Q, and R respectively. The coordinates of these points are $P(x_1, 0)$, $Q(x_2, 0)$, and $R(x_3, 0)$.

Assuming $x_1 < x_2 < x_3$ for this derivation, we can see three trapeziums are formed: ABQP, BCRQ, and ACRP.

The area of $\triangle ABC$ can be found as:

Area($\triangle ABC$) = Area(Trapezium ABQP) + Area(Trapezium BCRQ) - Area(Trapezium ACRP)

The area of a trapezium is given by $\frac{1}{2} \times (\text{sum of parallel sides}) \times (\text{height})$.

Area(Trapezium ABQP) = $\frac{1}{2} (AP + BQ) \times PQ = \frac{1}{2} (y_1 + y_2) (x_2 - x_1)$

Area(Trapezium BCRQ) = $\frac{1}{2} (BQ + CR) \times QR = \frac{1}{2} (y_2 + y_3) (x_3 - x_2)$

Area(Trapezium ACRP) = $\frac{1}{2} (AP + CR) \times PR = \frac{1}{2} (y_1 + y_3) (x_3 - x_1)$

Now, substitute these into the main expression:

Area($\triangle ABC$) = $\frac{1}{2} [(y_1 + y_2) (x_2 - x_1) + (y_2 + y_3) (x_3 - x_2) - (y_1 + y_3) (x_3 - x_1)]$

Expanding the terms:

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

$= \frac{1}{2} [x_2 y_1 - x_1 y_1 - x_1 y_2 + x_3 y_2 - x_2 y_3 - x_3 y_1 + x_1 y_1 + x_1 y_3]$

$= \frac{1}{2} [x_2 y_1 - x_1 y_2 + x_3 y_2 - x_2 y_3 - x_3 y_1 + x_1 y_3]$

Now, group the terms by $x_1, x_2, x_3$:

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

Taking a negative sign out from all terms inside the bracket to match the standard formula:

$= \frac{1}{2} [ -x_1(y_2 - y_3) - x_2(y_3 - y_1) - x_3(y_1 - y_2)]$

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

Since area must be positive, we take the absolute value of this expression, which gives the final formula:

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

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Condition for Collinearity of Three Points

Three or more points are said to be collinear if they lie on the same straight line.

If three points A, B, and C are collinear, they cannot form a triangle. Therefore, the area of the triangle formed by these three points must be zero.

This gives us a direct method to check for collinearity.

The points $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$ are collinear if and only if Area($\triangle ABC$) = 0.

This leads to the condition for collinearity:

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Locus of a Point and an Equation

In coordinate geometry, a locus (plural: loci) is the set of all points, and only those points, that satisfy a given geometric condition. The word "locus" is Latin for "place" or "location". In simpler terms, it is the path or curve traced by a point as it moves according to a specific rule.

The geometric condition defines the shape of the path. The algebraic representation of this path is called the equation of the locus. This equation is a relation between the x and y coordinates that is true for every point on the locus and false for every point that is not on the locus.

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Understanding the Concept

Imagine the tip of the second hand of a clock. Its "geometric condition" is that it must always be at a fixed distance (the length of the hand) from the center of the clock. As the hand moves, the path traced by its tip is a circle. Therefore, the locus is a circle.

The equation of this circle would be the algebraic formula that describes the relationship between the x and y coordinates of any point on this circular path.

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Definition of Locus

The locus of a point is the collection of all points in a plane that satisfy a given geometric condition.

In simpler terms, it is the path or shape formed by a point as it moves according to a certain rule.

Examples of Common Loci

Many common geometric shapes can be defined as the locus of a moving point.

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1. Circle

Geometric Condition: The locus of a point that is always at a constant distance (radius, $r$) from a fixed point (center, $C(h,k)$).

Equation: $(x-h)^2 + (y-k)^2 = r^2$

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2. Perpendicular Bisector

Geometric Condition: The locus of a point that is equidistant from two fixed points ($A$ and $B$).

Equation: A linear equation of the form $ax + by + c = 0$.

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3. Parabola

Geometric Condition: The locus of a point whose distance from a fixed point (the focus) is equal to its perpendicular distance from a fixed straight line (the directrix).

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4. Ellipse

Geometric Condition: The locus of a point for which the sum of its distances from two fixed points (the foci) is a constant.

Equation of a Locus

The equation of a locus is an algebraic equation.

It describes the path of a moving point that follows a specific geometric rule.

This equation creates a crucial link between the geometric idea of a locus and its algebraic representation in the coordinate plane.

This section details the systematic process for finding this equation.

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Procedure to Find the Equation of a Locus

To find the equation of a locus, we translate the geometric problem into an algebraic one by following these steps:

1. Assume a General Point: Let $P(x, y)$ be any point on the locus. The coordinates $x$ and $y$ are the variables that will form our final equation.
2. State the Geometric Condition: Clearly write down the given rule that the moving point $P(x, y)$ must follow. This rule usually involves distances, slopes, or other relationships with fixed points or lines.
3. Formulate an Algebraic Equation: Convert the geometric condition into an algebraic equation. Use the appropriate formulas from coordinate geometry to do this, such as:
   • The Distance Formula: $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$
   • The Section Formula or Midpoint Formula.
   • Slope formulas and their properties (parallelism, perpendicularity).
4. Simplify the Equation: Use algebraic methods to simplify the equation from the previous step. This often involves squaring both sides to remove square roots, expanding terms, and combining like terms to get the equation in its simplest form.

The final simplified equation in terms of $x$ and $y$ is the required equation of the locus.

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Shifting of Origin

In coordinate geometry, we usually measure the position of a point from a fixed origin (0, 0).

Sometimes, it is helpful to change this reference point by shifting the origin to a new location, without rotating the axes.

This process is called Shifting of Origin or Translation of Axes.

Shifting the origin can often simplify the equation of a curve.

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Transformation of Coordinates

Suppose we have a coordinate system with origin O and axes x and y.

Let's shift the origin to a new point $O'$ whose coordinates in the old system are $(h, k)$.

The new axes, X and Y, are parallel to the old x and y axes, respectively.

Let P be any point in the plane.

Let its coordinates be $(x, y)$ with respect to the old origin O.

Let its coordinates be $(X, Y)$ with respect to the new origin O'.

From the diagram, we can see the relationship between the old and new coordinates.

The total horizontal distance from the old y-axis to P is $x$. This distance is the sum of the horizontal shift ($h$) and the new x-coordinate ($X$).

Similarly, the total vertical distance from the old x-axis to P is $y$. This distance is the sum of the vertical shift ($k$) and the new y-coordinate ($Y$).

This gives us the following transformation formulas:

These formulas convert new coordinates $(X, Y)$ back to the old coordinates $(x, y)$.

We can also rearrange these formulas to find the new coordinates from the old ones:

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Transformation of an Equation

If we have an equation of a curve in the original system, say $f(x, y) = 0$, we can find its new equation.

To do this, we substitute the expressions for $x$ and $y$ from formulas (i) and (ii) into the original equation.

We replace $x$ with $(X+h)$ and $y$ with $(Y+k)$.

The new equation for the curve will be $f(X+h, Y+k) = 0$.

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Slope of a Straight Line

The slope, also known as the gradient, is a fundamental property of a straight line in coordinate geometry. It provides a precise numerical measure that describes both the direction and the steepness of the line. In essence, the slope quantifies the rate of change in the vertical direction (the 'rise') with respect to the change in the horizontal direction (the 'run').

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Angle of Inclination

To formally define the slope, we first introduce the concept of the angle of inclination. The angle of inclination of a non-vertical line is the angle, denoted by $\theta$, that the line makes with the positive direction of the x-axis, measured in the counter-clockwise direction.

The range of the angle of inclination is always taken as $0^\circ \le \theta < 180^\circ$ (or $0 \le \theta < \pi$ in radians).

• If a line is parallel to the x-axis, its angle of inclination is $0^\circ$.
• If a line is parallel to the y-axis (vertical), its angle of inclination is $90^\circ$.
• For any other line, the angle will be between these values. We do not consider angles greater than or equal to $180^\circ$ because the orientation of the line repeats every $180^\circ$. For example, a line with an inclination of $210^\circ$ is identical to a line with an inclination of $30^\circ$.

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Definition and Interpretation of Slope

The slope of a non-vertical line, universally denoted by the letter $m$, is defined as the tangent of its angle of inclination $\theta$.

The tangent function is used because it naturally relates the angle of a right triangle to the ratio of the opposite side (vertical rise) to the adjacent side (horizontal run), which is the intuitive meaning of slope.

Interpreting the Value of the Slope

• Positive Slope ($m > 0$): When the angle of inclination $\theta$ is acute ($0^\circ < \theta < 90^\circ$), $\tan \theta$ is positive. The line rises as viewed from left to right. A larger positive slope indicates a steeper line.
• Negative Slope ($m < 0$): When the angle of inclination $\theta$ is obtuse ($90^\circ < \theta < 180^\circ$), $\tan \theta$ is negative. The line falls as viewed from left to right. A more negative slope (e.g., -4 is steeper than -1) indicates a steeper decline.
• Zero Slope ($m = 0$): When the line is horizontal, its angle of inclination is $\theta = 0^\circ$. Since $\tan 0^\circ = 0$, the slope is zero. There is zero 'rise' for any 'run'.
• Undefined Slope: When the line is vertical, its angle of inclination is $\theta = 90^\circ$. Since $\tan 90^\circ$ is undefined, the slope of a vertical line is also undefined. This is because the 'run' (horizontal change) is zero, which would lead to division by zero.

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Derivation of Slope Formula for Two Points

While the definition $m = \tan \theta$ is fundamental, it is often more practical to calculate the slope using the coordinates of two points on the line. We can derive this formula directly from the definition and show that it holds true for both acute and obtuse angles of inclination.

Let a non-vertical line $L$ pass through two distinct points $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$.

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Case 1: When the Angle of Inclination $\theta$ is Acute ($0^\circ < \theta < 90^\circ$)

In this case, the line rises from left to right. Let's assume we have chosen our points such that $x_2 > x_1$, which implies that $y_2 > y_1$.

We construct a right-angled triangle by drawing a horizontal line through $P_1$ and a vertical line through $P_2$, intersecting at point $Q(x_2, y_1)$.

In the right-angled triangle $P_1QP_2$:

• The length of the adjacent side (run) is $P_1Q = x_2 - x_1$. This value is positive.
• The length of the opposite side (rise) is $QP_2 = y_2 - y_1$. This value is also positive.

The angle $\angle QP_1P_2$ is equal to the angle of inclination $\theta$, because the line segment $P_1Q$ is parallel to the x-axis, making them corresponding angles.

From the definition of the tangent function in a right-angled triangle:

$\tan \theta = \frac{\text{Opposite Side}}{\text{Adjacent Side}} = \frac{QP_2}{P_1Q}$

Since the slope $m = \tan \theta$, we can substitute the lengths:

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

Thus, the formula holds for lines with a positive slope.

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Case 2: When the Angle of Inclination $\theta$ is Obtuse ($90^\circ < \theta < 180^\circ$)

In this case, the line falls from left to right. Let's assume we have chosen our points such that $x_2 > x_1$, which implies that $y_2 < y_1$.

We again construct a right-angled triangle by drawing a horizontal line through $P_2$ and a vertical line through $P_1$, intersecting at point $Q(x_1, y_2)$.

In the right-angled triangle $P_1QP_2$:

• The length of the adjacent side (run) is $QP_2 = x_2 - x_1$. This value is positive.
• The length of the opposite side (rise) is $P_1Q = y_1 - y_2$. This value is also positive, as $y_1 > y_2$.

Let $\alpha$ be the interior angle $\angle P_1P_2Q$. From the geometry of the figure, the angle of inclination $\theta$ and the angle $\alpha$ are supplementary, as they form a linear pair with the angle corresponding to $\alpha$. Therefore:

$\theta = 180^\circ - \alpha$

First, we find $\tan \alpha$ from the right-angled triangle:

$\tan \alpha = \frac{\text{Opposite Side}}{\text{Adjacent Side}} = \frac{P_1Q}{QP_2} = \frac{y_1 - y_2}{x_2 - x_1}$

Now, we find the slope $m$ using its definition, $m = \tan \theta$:

$m = \tan \theta = \tan(180^\circ - \alpha)$

Using the trigonometric identity $\tan(180^\circ - \alpha) = -\tan \alpha$, we get:

$m = -\tan \alpha = - \left( \frac{y_1 - y_2}{x_2 - x_1} \right)$

By distributing the negative sign into the numerator, we get:

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

This is the exact same formula as in Case 1. This proves that the formula is universally applicable for any non-vertical line, regardless of its angle of inclination.

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Conditions for Parallelism and Perpendicularity

The slope provides a powerful algebraic method to determine the geometric relationship between two lines.

1. Parallel Lines

Condition: Two distinct non-vertical lines, $L_1$ and $L_2$, are parallel if and only if their slopes, $m_1$ and $m_2$, are equal.

Proof: Let the angles of inclination of lines $L_1$ and $L_2$ be $\theta_1$ and $\theta_2$, respectively.

If the lines $L_1$ and $L_2$ are parallel, then their angles of inclination must be equal because they are corresponding angles formed by the transversal (the x-axis).

Thus, $L_1 \parallel L_2 \implies \theta_1 = \theta_2$.

Taking the tangent of both sides:

$\tan \theta_1 = \tan \theta_2 \implies m_1 = m_2$.

Conversely, if $m_1 = m_2$, then $\tan \theta_1 = \tan \theta_2$. Since $\theta_1$ and $\theta_2$ are both in the range $[0^\circ, 180^\circ)$, this implies $\theta_1 = \theta_2$. Equal angles of inclination mean the lines are parallel.

2. Perpendicular Lines

Condition: Two non-vertical lines, $L_1$ and $L_2$, are perpendicular if and only if the product of their slopes, $m_1$ and $m_2$, is equal to -1.

Proof: Let the angles of inclination of lines $L_1$ and $L_2$ be $\theta_1$ and $\theta_2$. Let their slopes be $m_1 = \tan \theta_1$ and $m_2 = \tan \theta_2$.

Assume the lines are perpendicular. As shown in the diagram, they form a triangle with the x-axis. The angle between the lines is $90^\circ$.

By the exterior angle theorem of a triangle, the exterior angle $\theta_2$ is equal to the sum of the two opposite interior angles.

$\theta_2 = \theta_1 + 90^\circ$

Now, let's relate their slopes:

$m_2 = \tan \theta_2 = \tan(\theta_1 + 90^\circ)$

Using the trigonometric identity $\tan(A + 90^\circ) = -\cot(A)$, we get:

$m_2 = -\cot \theta_1$

Since $\cot \theta_1 = \frac{1}{\tan \theta_1}$, we can write:

$m_2 = -\frac{1}{\tan \theta_1}$

Substituting $m_1 = \tan \theta_1$ into the equation:

$m_2 = -\frac{1}{m_1}$

Multiplying both sides by $m_1$ gives the condition for perpendicular lines:

$m_1 m_2 = -1$

The converse can also be proven by reversing these steps. This relationship means that the slope of one line is the negative reciprocal of the slope of the other.

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Angle Between Two Lines

When two non-parallel straight lines intersect, they form two pairs of vertically opposite angles. One pair consists of acute angles, and the other consists of obtuse angles (unless the lines are perpendicular, in which case all four angles are $90^\circ$).

By using the slopes of the two intersecting lines, we can derive a formula to calculate the measure of the acute angle between them.

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Derivation of the Formula

Let $L_1$ and $L_2$ be two intersecting, non-vertical, and non-perpendicular lines.

• Let their respective angles of inclination with the positive x-axis be $\theta_1$ and $\theta_2$.
• Let their respective slopes be $m_1 = \tan \theta_1$ and $m_2 = \tan \theta_2$.

Let $\phi$ and $\psi$ be the angles between the lines $L_1$ and $L_2$, where $\phi$ is the acute angle and $\psi = 180^\circ - \phi$ is the obtuse angle.

From the diagram, by applying the exterior angle theorem to the triangle formed by the lines and the x-axis, we can establish a relationship between the angles. Assuming $\theta_2 > \theta_1$, the exterior angle $\theta_2$ is equal to the sum of the two opposite interior angles, $\theta_1$ and $\phi$.

$\theta_2 = \theta_1 + \phi$

This can be rearranged to find $\phi$:

$\phi = \theta_2 - \theta_1$

To relate this angle to the slopes of the lines, we take the tangent of both sides of the equation.

$\tan \phi = \tan(\theta_2 - \theta_1)$

Using the trigonometric identity for the tangent of the difference of two angles, $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$, we get:

$\tan \phi = \frac{\tan \theta_2 - \tan \theta_1}{1 + \tan \theta_1 \tan \theta_2}$

Now, we substitute the slopes $m_1 = \tan \theta_1$ and $m_2 = \tan \theta_2$ into the equation:

$\tan \phi = \frac{m_2 - m_1}{1 + m_1 m_2}$

Considering the other angle: The other angle between the lines is $\psi = 180^\circ - \phi$. If we calculate its tangent, we get:

$\tan \psi = \tan(180^\circ - \phi) = -\tan \phi = - \left( \frac{m_2 - m_1}{1 + m_1 m_2} \right)$

This shows that the expression $\frac{m_2 - m_1}{1 + m_1 m_2}$ can be positive or negative. If it's positive, it gives $\tan \phi$ (the acute angle). If it's negative, it gives $\tan \psi$ (the obtuse angle).

To create a general formula that always gives the acute angle, we take the absolute value of the expression, since the tangent of an acute angle (in the first quadrant) is always positive.

Thus, the formula for the tangent of the acute angle $\phi$ between two lines with slopes $m_1$ and $m_2$ is:

Once the acute angle $\phi$ is found, the obtuse angle can be found using $\psi = 180^\circ - \phi$.

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Analysis of Special Cases

The formula is consistent with our prior knowledge of parallel and perpendicular lines.

Case 1: Parallel Lines

Two lines are parallel if the angle between them is $0^\circ$.

Let's check this with the formula. If lines are parallel, their slopes are equal, i.e., $m_1 = m_2$.

Substituting this into the formula:

$\tan \phi = \left| \frac{m_1 - m_1}{1 + m_1 m_1} \right| = \left| \frac{0}{1 + m_1^2} \right| = 0$

If $\tan \phi = 0$, then the angle $\phi = 0^\circ$. This confirms that the formula correctly identifies parallel lines as having a zero-degree angle between them.

Case 2: Perpendicular Lines

Two lines are perpendicular if the angle between them is $90^\circ$.

The tangent of $90^\circ$ is undefined. For the expression $\left| \frac{m_2 - m_1}{1 + m_1 m_2} \right|$ to be undefined, its denominator must be zero.

$1 + m_1 m_2 = 0$

This gives the condition $m_1 m_2 = -1$.

This confirms that the formula is consistent with the condition for perpendicular lines, where the product of their slopes is -1.

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Condition for Collinearity Using Slopes

The fundamental idea is that a single straight line has a constant slope. Therefore, if three points A, B, and C are collinear, the slope of the line segment joining A and B must be the same as the slope of the line segment joining B and C (and also the same as the slope of the segment joining A and C).

Let the three points be $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$.

The slope of the line segment AB is:

$m_{AB} = \frac{y_2 - y_1}{x_2 - x_1}$

The slope of the line segment BC is:

$m_{BC} = \frac{y_3 - y_2}{x_3 - x_2}$

For the points A, B, and C to be collinear, these slopes must be equal.

Thus, the condition for collinearity is:

Note: This condition holds provided the line is not vertical. If the line is vertical, then $x_1 = x_2 = x_3$, and the slopes would be undefined. In such cases, checking the x-coordinates is sufficient to prove collinearity.

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Alternative Condition: Using Area of a Triangle

Another way to think about collinearity is from a geometric perspective. If three points lie on the same line, they cannot form a triangle. Therefore, the area of the triangle formed by three collinear points must be zero.

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:

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

For the points to be collinear, the area must be zero. This gives the condition:

This equation is algebraically equivalent to the slope condition (after cross-multiplication and rearrangement).

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Various Forms of Equations of a Line

A straight line is a fundamental geometric object. In coordinate geometry, we represent it using an algebraic equation.

There are several standard forms for the equation of a line. The best form to use depends on the information given about the line, such as its slope, the points it passes through, or its intercepts.

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1. Horizontal and Vertical Lines

These are the simplest cases, defined by their constant orientation with respect to the coordinate axes.

Horizontal Line

A horizontal line is defined as a line parallel to the x-axis. A key characteristic of such a line is that every single point on it has the exact same y-coordinate. If this constant y-coordinate is $k$, then the condition for any point $(x, y)$ to be on the line is simply that its y-value must be $k$.

This gives the equation:

The slope of a horizontal line is 0, since the change in y (the "rise") is always zero. The equation of the x-axis itself is a special case where $k=0$, giving $y=0$.

Vertical Line

A vertical line is defined as a line parallel to the y-axis. Similarly, every point on a vertical line has the exact same x-coordinate. If this constant x-coordinate is $h$, then the condition for any point $(x, y)$ to be on the line is that its x-value must be $h$.

This gives the equation:

The slope of a vertical line is undefined, as the change in x (the "run") is zero, which would lead to division by zero. The equation of the y-axis itself is a special case where $h=0$, giving $x=0$.

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2. Point-Slope Form

This is arguably the most fundamental and versatile form for the equation of a line. It is used when we know the slope of a line and the coordinates of at least one point it passes through.

Derivation:

Let the given slope of the line be $m$, and let the line pass through a fixed point $P_1(x_1, y_1)$. Now, let $P(x, y)$ be any other arbitrary point on the same line.

By the definition of slope, the slope calculated between any two points on a line must be constant and equal to $m$. Therefore, the slope between points $P_1$ and $P$ is:

$\text{slope} = \frac{\text{change in y}}{\text{change in x}} = \frac{y - y_1}{x - x_1}$

Since we know this slope must be $m$, we can set up the equation:

$m = \frac{y - y_1}{x - x_1}$

Multiplying both sides by $(x - x_1)$, we get the standard point-slope form:

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3. Two-Point Form

This form is a direct application of the point-slope form and is used when we know the coordinates of two distinct points that the line passes through.

Derivation:

Let the two given points be $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$.

Step 1: Find the slope. Using the two given points, we first calculate the slope of the line:

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

Step 2: Use the point-slope form. Now that we have the slope $m$, we can use it along with either of the two given points. Let's choose point $P_1(x_1, y_1)$. The point-slope equation is:

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

Step 3: Substitute the slope. Substitute the expression for $m$ from Step 1 into the equation from Step 2. This gives the two-point form:

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4. Slope-Intercept Form

This popular form is useful because it explicitly states the line's slope and its y-intercept, making it easy to visualize and graph the line.

Derivation:

This form is a special case of the point-slope form. Let the slope be $m$ and the y-intercept be $c$. The y-intercept is the point where the line crosses the y-axis. The coordinates of this point are always $(0, c)$.

We now have a slope $m$ and a point $(x_1, y_1) = (0, c)$. We substitute these into the point-slope form:

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

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

Simplifying the equation gives the slope-intercept form:

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5. Intercept Form

This form is used when the x-intercept and y-intercept of the line are known.

Derivation:

Let the x-intercept be $a$ and the y-intercept be $b$. This means the line passes through two specific points: $(a, 0)$ and $(0, b)$. We can use the two-point form with $(x_1, y_1) = (a, 0)$ and $(x_2, y_2) = (0, b)$.

First, calculate the slope: $m = \frac{b - 0}{0 - a} = -\frac{b}{a}$.

Now use the point-slope form with the point $(a, 0)$:

$y - 0 = \left(-\frac{b}{a}\right)(x - a)$

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

To get the standard intercept form, rearrange the equation. Move the x-term to the left side:

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

Now, divide the entire equation by $b$ (assuming $b \neq 0$):

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

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6. Normal Form

This form defines a line in terms of its distance from the origin and the orientation of that distance. It is also known as the Perpendicular Form.

Derivation:

Let $p$ be the length of the perpendicular segment drawn from the origin $(0,0)$ to the line. Let this perpendicular segment make an angle $\alpha$ with the positive x-axis. The point where the perpendicular meets the line, let's call it $Q$, has coordinates found using trigonometry: $Q(p \cos \alpha, p \sin \alpha)$.

The slope of the perpendicular segment OQ is $m_{OQ} = \tan \alpha = \frac{\sin \alpha}{\cos \alpha}$.

The line itself is perpendicular to OQ. Therefore, its slope, $m_L$, must be the negative reciprocal of $m_{OQ}$:

$m_L = -\frac{1}{\tan \alpha} = -\frac{\cos \alpha}{\sin \alpha}$.

Now we have a slope ($m_L$) and a point ($Q$) on the line. We use the point-slope form:

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

$y - p \sin \alpha = \left(-\frac{\cos \alpha}{\sin \alpha}\right)(x - p \cos \alpha)$

Multiply both sides by $\sin \alpha$ to clear the fraction:

$y \sin \alpha - p \sin^2 \alpha = -\cos \alpha(x - p \cos \alpha)$

$y \sin \alpha - p \sin^2 \alpha = -x \cos \alpha + p \cos^2 \alpha$

Group the x and y terms on one side and the constant terms on the other:

$x \cos \alpha + y \sin \alpha = p \cos^2 \alpha + p \sin^2 \alpha$

Factor out $p$ on the right side:

$x \cos \alpha + y \sin \alpha = p (\cos^2 \alpha + \sin^2 \alpha)$

Using the Pythagorean identity $\cos^2 \alpha + \sin^2 \alpha = 1$, we arrive at the Normal Form:

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7. Symmetrical Form (or Parametric / Distance Form)

This form is particularly useful for finding the coordinates of a point on a line that is a specific distance away from a known point on that same line. It defines points on the line parametrically in terms of this distance.

Derivation:

Let a line pass through a fixed point $P_1(x_1, y_1)$ and have an angle of inclination $\theta$. Let $P(x, y)$ be any other point on the line.

Let $r$ be the directed distance between $P_1$ and $P$. The value of $r$ is positive if $P$ is in the direction of inclination from $P_1$, and negative if it is in the opposite direction.

We can form a right-angled triangle by drawing a horizontal line through $P_1$ and a vertical line through $P$. The horizontal distance is $x - x_1$ and the vertical distance is $y - y_1$. The hypotenuse is the distance $r$.

Using basic trigonometry in this right-angled triangle:

$\cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{x - x_1}{r} \implies x - x_1 = r \cos \theta$

$\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{y - y_1}{r} \implies y - y_1 = r \sin \theta$

By rearranging these two relations, we can express the line's equation in a symmetric form where each part is equal to the parameter $r$:

From this, we get the parametric coordinates of any point on the line:

$x = x_1 + r \cos \theta$

$y = y_1 + r \sin \theta$

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General Equation of a Line

We have explored several specific forms for the equation of a line. However, there is one universal form that can represent any straight line in the Cartesian plane.

This is known as the general equation of a line, which is a linear equation in two variables, $x$ and $y$.

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The General Form

Every straight line can be represented by a first-degree equation in $x$ and $y$. Conversely, any such equation represents a straight line.

The general form of the equation of a line is:

Here, $A, B,$ and $C$ are real numbers that are constant coefficients for a specific line.

A crucial condition is that $A$ and $B$ cannot both be zero simultaneously. If both A and B were zero, the equation would become $C = 0$. This would either be a contradiction (e.g., $5=0$) if C is non-zero, or an identity ($0=0$) true for all points in the plane, neither of which represents a line.

Proof that Every Line has a General Form

Any non-vertical line can be written in the point-slope form $y - y_1 = m(x - x_1)$. This can be rearranged as $mx - y + (y_1 - mx_1) = 0$. This is in the form $Ax + By + C = 0$ with $A=m, B=-1,$ and $C=y_1-mx_1$. A vertical line has the equation $x=h$, which can be written as $1x + 0y - h = 0$, also in the general form. Thus, any line can be expressed in the general form.

Proof that the General Form Represents a Line

Consider the equation $Ax + By + C = 0$, where A and B are not both zero.

• Case 1: $B \neq 0$. We can rearrange the equation to solve for y: $y = (-\frac{A}{B})x + (-\frac{C}{B})$. This is the slope-intercept form $y=mx+c$, which represents a straight line with slope $m = -A/B$ and y-intercept $c = -C/B$.
• Case 2: $B = 0$. Since A and B cannot both be zero, we must have $A \neq 0$. The equation becomes $Ax + C = 0$, which simplifies to $x = -C/A$. This is the equation of a vertical line.

In both possible cases, the equation represents a straight line.

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Converting the General Equation to Other Forms

We can rearrange the general equation $Ax + By + C = 0$ to find important properties of the line and to express it in other standard forms.

1. Slope-Intercept Form ($y = mx + c$)

To find the slope ($m$) and y-intercept ($c$), we convert the general equation into the slope-intercept form. This is possible as long as the line is not vertical (i.e., $B \neq 0$).

Starting with the general equation:

$Ax + By + C = 0$

Isolate the term with $y$.

$By = -Ax - C$

Divide the entire equation by $B$.

$y = \left(-\frac{A}{B}\right)x + \left(-\frac{C}{B}\right)$

By comparing this with $y = mx + c$, we can see that:

2. Intercept Form ($\frac{x}{a} + \frac{y}{b} = 1$)

To convert to intercept form, we find the x-intercept ($a$) and y-intercept ($b$). This is possible if $A, B,$ and $C$ are all non-zero.

Starting with the general equation:

$Ax + By + C = 0$

Move the constant term to the right side.

$Ax + By = -C$

Divide the entire equation by $-C$.

$\frac{Ax}{-C} + \frac{By}{-C} = 1$

Rearrange to match the intercept form.

$\frac{x}{-C/A} + \frac{y}{-C/B} = 1$

Comparing this with $\frac{x}{a} + \frac{y}{b} = 1$, we see that:

3. Normal Form ($x \cos \alpha + y \sin \alpha = p$)

This conversion finds the length of the perpendicular from the origin to the line ($p$) and the angle this perpendicular makes with the x-axis ($\alpha$).

Derivation:

The general equation $Ax + By + C = 0$ and the normal form $x \cos \alpha + y \sin \alpha = p$ (which can be written as $x \cos \alpha + y \sin \alpha - p = 0$) represent the same line. This means their corresponding coefficients must be proportional. Let the constant of proportionality be $k$.

$\frac{A}{\cos \alpha} = \frac{B}{\sin \alpha} = \frac{C}{-p} = k$

From this proportionality, we can write:

$A = k \cos \alpha$, $B = k \sin \alpha$, and $C = -k p$.

To find $k$, we square and add the first two relations:

$A^2 + B^2 = (k \cos \alpha)^2 + (k \sin \alpha)^2 = k^2 (\cos^2 \alpha + \sin^2 \alpha)$

Since $\cos^2 \alpha + \sin^2 \alpha = 1$, we have $A^2 + B^2 = k^2$, which implies $k = \pm\sqrt{A^2 + B^2}$.

Now we can express $\cos \alpha$, $\sin \alpha$, and $p$ in terms of A, B, and C:

$\cos \alpha = \frac{A}{k} = \frac{A}{\pm\sqrt{A^2 + B^2}}$

$\sin \alpha = \frac{B}{k} = \frac{B}{\pm\sqrt{A^2 + B^2}}$

$p = \frac{-C}{k} = \frac{-C}{\pm\sqrt{A^2 + B^2}}$

Since $p$ represents a distance, it must be non-negative ($p \ge 0$). Therefore, we choose the sign of the square root $(\pm)$ to make the final expression for $p$ positive. This is achieved by taking the sign opposite to the sign of $C$. A simpler convention is to first move $C$ to the right side of the original equation, $Ax+By=-C$, and then divide by $+\sqrt{A^2+B^2}$. This ensures the right-hand side, which corresponds to $p$, is positive.

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Special Cases of the General Equation

The values of the coefficients A, B, and C determine the type of line.

• If $A = 0$ (and $B \neq 0$), the equation becomes $By + C = 0$, or $y = -\frac{C}{B}$. This is a horizontal line.
• If $B = 0$ (and $A \neq 0$), the equation becomes $Ax + C = 0$, or $x = -\frac{C}{A}$. This is a vertical line.
• If $C = 0$ (and A, B not both zero), the equation becomes $Ax + By = 0$. This line passes through the origin (0, 0), because $A(0) + B(0) = 0$.

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Intersection of Two Straight Lines

If two straight lines in a plane are not parallel, they will intersect at a single, unique point. The coordinates of this point of intersection must satisfy the equations of both lines simultaneously.

Method of Solution

To find the point of intersection, we solve the system of linear equations formed by the two lines.

Let the two lines be:

$L_1: A_1x + B_1y + C_1 = 0$

$L_2: A_2x + B_2y + C_2 = 0$

We need to find the unique pair $(x, y)$ that solves this system. Common methods include substitution, elimination, and cross-multiplication.

General Solution by Cross-Multiplication:

The solution to the above system can be written directly as:

$\frac{x}{B_1C_2 - B_2C_1} = \frac{y}{C_1A_2 - C_2A_1} = \frac{1}{A_1B_2 - A_2B_1}$

From this, we can solve for $x$ and $y$:

$x = \frac{B_1C_2 - B_2C_1}{A_1B_2 - A_2B_1}$

$y = \frac{C_1A_2 - C_2A_1}{A_1B_2 - A_2B_1}$

Condition for a Unique Intersection

A unique solution exists if and only if the denominator in the general solution is non-zero:

$A_1B_2 - A_2B_1 \neq 0$

This is equivalent to the condition $\frac{A_1}{A_2} \neq \frac{B_1}{B_2}$.

Geometrically, this is the condition that the slopes of the two lines are not equal ($m_1 \neq m_2$), which means the lines are not parallel and must intersect at exactly one point.

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Families of Lines

A family of lines, also known as a pencil of lines, is an infinite set of lines that all share a common geometric property. This shared property can be defined by one or more constraints. The concept of a family is described by an equation with one or more variable parameters. For a single family of lines, we typically use one parameter.

One-Parameter Families of Lines

An equation that is linear in $x$ and $y$ but also contains a variable parameter represents a family of lines.

This means the equation does not describe a single line, but rather an infinite collection of lines. For each specific real number assigned to the parameter, we get the equation of a single, unique line from the family.

All lines belonging to the same family share a common, defining geometric property.

1. Family of Lines with a Fixed Slope (Parallel Lines)

This is a set of all lines that are parallel to each other. Their defining common property is that they all have the same slope.

Geometrically, this means they all have the same angle of inclination and will never intersect each other.

Using Slope-Intercept Form:

The equation is given by the slope-intercept form where the slope $m$ is a fixed constant, and the y-intercept $c$ is the variable parameter: $y = mx + c$.

As the parameter $c$ takes on different real values, the y-intercept changes, which effectively slides the line vertically up or down the plane without altering its tilt.

For example, the family of lines with a slope of 2 is given by the equation $y = 2x + c$.

• If $c=1$, we get the line $y = 2x + 1$.
• If $c=-3$, we get the line $y = 2x - 3$.
• If $c=0$, we get the line $y = 2x$, which passes through the origin.

All these lines are parallel to each other.

Using General Form:

The slope of a line given by $Ax + By + C = 0$ is $m = -\frac{A}{B}$. To keep the slope constant, the ratio of A to B must remain constant.

The simplest way to achieve this is to keep A and B fixed and vary the constant term. Therefore, the family of lines parallel to $Ax + By + C = 0$ is given by $Ax + By + k = 0$, where $k$ is the parameter.

2. Family of Lines Passing Through a Fixed Point (Concurrent Lines)

This is a set of all lines that are concurrent, meaning they all pass through the same single point. This fixed point is the defining common property of the family.

Using Point-Slope Form:

The equation is given by the point-slope form where the point $(x_1, y_1)$ is fixed and the slope $m$ is the variable parameter: $y - y_1 = m(x - x_1)$.

As the parameter $m$ takes on different real values, the slope of the line changes. Geometrically, this means the line rotates around the fixed pivot point $(x_1, y_1)$.

For example, the family of lines passing through the point (2, 3) is given by $y - 3 = m(x - 2)$.

• If $m=1$, we get the line $y - 3 = 1(x - 2) \implies y = x + 1$.
• If $m=-2$, we get the line $y - 3 = -2(x - 2) \implies y = -2x + 7$.
• If $m=0$, we get the horizontal line $y - 3 = 0 \implies y = 3$.

All these lines intersect at the point (2, 3).

Limitation: This form can represent every line through $(x_1, y_1)$ except for the vertical line $x=x_1$, because a vertical line has an undefined slope, so no real value of $m$ can describe it.

3. Family of Lines Perpendicular to a Given Line

This is a set of all lines that are perpendicular to a given line. Their defining property is that their slope is the negative reciprocal of the given line's slope.

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

The slope of any line perpendicular to it must be $m_{\perp} = -\frac{1}{m} = - \frac{1}{-A/B} = \frac{B}{A}$.

To construct a line with this new slope, its general form will be $Bx - Ay + k = 0$. This is because the slope of this new family is $-\frac{B}{-A} = \frac{B}{A}$, which is the required perpendicular slope.

Therefore, the rule to find the family of lines perpendicular to $Ax + By + C = 0$ is to swap the coefficients of $x$ and $y$, change the sign of one of them, and replace the constant term with a parameter $k$.

The family is given by $Bx - Ay + k = 0$.

For example, the family of lines perpendicular to $2x + 5y - 7 = 0$ is $5x - 2y + k = 0$.

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Family of Lines Through the Intersection of Two Lines

One of the most powerful and useful applications of this concept is in finding the equation of a line that passes through the point of intersection of two other given lines, without actually calculating the coordinates of the intersection point.

The Equation and its Derivation

Suppose we have two distinct, intersecting straight lines given by their general equations:

$L_1 \equiv A_1x + B_1y + C_1 = 0$

$L_2 \equiv A_2x + B_2y + C_2 = 0$

Let the point of intersection of these two lines be $P(x_0, y_0)$. Since this point lies on both lines, its coordinates must satisfy both equations:

$A_1x_0 + B_1y_0 + C_1 = 0$

$A_2x_0 + B_2y_0 + C_2 = 0$

Now, consider the following equation, where $\lambda$ (lambda) is a real number called a parameter:

Expanding this, we get:

$(A_1x + B_1y + C_1) + \lambda (A_2x + B_2y + C_2) = 0$

We can prove this equation represents the family of all lines (except $L_2$) passing through the intersection of $L_1$ and $L_2$.

Proof:

1. It represents a straight line. Let's rearrange the equation by grouping terms with x, y, and the constant terms:

   $(A_1 + \lambda A_2)x + (B_1 + \lambda B_2)y + (C_1 + \lambda C_2) = 0$

   This is in the form $Ax + By + C = 0$, which is a first-degree equation in $x$ and $y$. Therefore, for any real value of $\lambda$, it represents a straight line.

2. It passes through the point of intersection $P(x_0, y_0)$. Let's substitute the coordinates of the intersection point $(x_0, y_0)$ into the family equation:

   $(A_1x_0 + B_1y_0 + C_1) + \lambda (A_2x_0 + B_2y_0 + C_2) = 0$

   We already know that the expressions in the brackets are both zero. So, the equation becomes:

   $0 + \lambda(0) = 0$, which simplifies to $0=0$.

   Since this is always true, the point of intersection $P(x_0, y_0)$ satisfies the equation for any value of $\lambda$. This proves that every line generated by this equation must pass through the intersection point of $L_1$ and $L_2$.

Note: By varying the value of $\lambda$ from $-\infty$ to $+\infty$, we can generate every line that passes through the intersection point, with one exception: the line $L_2$ itself. This is because no finite value of $\lambda$ can eliminate the $L_1$ part of the equation. However, this is rarely a limitation in solving problems.

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Problem-Solving Strategy

This concept provides a powerful shortcut. Instead of first solving for the point of intersection (which can involve fractions and be prone to error) and then using the point-slope form, we can work directly with the family equation.

The typical problem-solving steps are:

1. Write the general equation of the family of lines passing through the intersection of the two given lines: $L_1 + \lambda L_2 = 0$.
2. Use the third piece of information given in the problem (e.g., the line passes through another point, is parallel to a third line, is perpendicular to a third line, etc.) to create an equation and solve for the specific value of $\lambda$.
3. Substitute this value of $\lambda$ back into the family equation and simplify to get the final equation of the required line.

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Answer:

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Distance of a Point from a Line

In coordinate geometry, a common task is to determine the distance between a point and a straight line. It's important to understand that there are infinitely many possible distances from a given point to the infinite number of points on a line.

However, when we speak of "the distance," we are always referring to the shortest possible distance. This shortest distance is always found along the path that is perpendicular to the line, connecting the given point to a point on the line at a right angle.

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Derivation of the Perpendicular Distance Formula

Let's derive the formula to calculate the perpendicular distance from a point $P(x_1, y_1)$ to a line $L$ whose equation is given in the general form, $Ax + By + C = 0$.

Step 1: Convert the line equation to Normal Form.

The normal form of the line $Ax + By + C = 0$ is $x \cos \alpha + y \sin \alpha - p = 0$.

Here, $p$ is the length of the perpendicular from the origin to the line $L$. By comparing coefficients, we know:

$\cos \alpha = \pm\frac{A}{\sqrt{A^2+B^2}}$, $\sin \alpha = \pm\frac{B}{\sqrt{A^2+B^2}}$, and $p = \frac{|C|}{\sqrt{A^2+B^2}}$.

Step 2: Construct a parallel line through the given point.

Let's draw a line $L'$ that is parallel to $L$ and passes through the point $P(x_1, y_1)$. Since $L'$ is parallel to $L$, the perpendicular from the origin to $L'$ will also make the same angle $\alpha$ with the x-axis.

Let the perpendicular distance from the origin to this new line $L'$ be $p'$. The normal form of line $L'$ is $x \cos \alpha + y \sin \alpha - p' = 0$.

Step 3: Find the distance $p'$.

Since the point $P(x_1, y_1)$ lies on the line $L'$, its coordinates must satisfy the equation of $L'$.

$x_1 \cos \alpha + y_1 \sin \alpha - p' = 0 \implies p' = x_1 \cos \alpha + y_1 \sin \alpha$.

Step 4: Find the distance between the point and the line.

The required distance $d$ from point $P$ to line $L$ is the perpendicular distance between the parallel lines $L$ and $L'$. This distance is the absolute difference of their perpendicular distances from the origin.

$d = |p' - p| = |(x_1 \cos \alpha + y_1 \sin \alpha) - p|$

Step 5: Substitute the coefficients A, B, C.

Now, we substitute the expressions for $\cos \alpha, \sin \alpha,$ and $p$ in terms of A, B, and C. For consistency, let's use the form $Ax + By = -C$ and divide by $\sqrt{A^2+B^2}$. Then $p = \frac{-C}{\sqrt{A^2+B^2}}$ (assuming the sign of the root is chosen to make this positive).

$d = \left| x_1 \left(\frac{A}{\sqrt{A^2+B^2}}\right) + y_1 \left(\frac{B}{\sqrt{A^2+B^2}}\right) - \left(\frac{-C}{\sqrt{A^2+B^2}}\right) \right|$

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

This gives us the final formula, where the absolute value applies to the numerator:

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Distance from the Origin

A common special case is finding the distance from the origin $(0, 0)$ to a line. We can use the main formula by substituting $(x_1, y_1) = (0, 0)$.

$d_{\text{origin}} = \frac{|A(0) + B(0) + C|}{\sqrt{A^2 + B^2}}$

This simplifies to a very direct formula:

This result is consistent with the definition of $p$ in the normal form of a line.

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Answer:

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Distance Between Two Parallel Lines

Since parallel lines never intersect, the perpendicular distance between them is constant everywhere. We can derive a direct formula for this distance.

Derivation of the Formula

Let the equations of two parallel lines be:

$L_1: Ax + By + C_1 = 0$

$L_2: Ax + By + C_2 = 0$

Note: For this derivation, the coefficients of $x$ (A) and $y$ (B) must be identical in both equations.

The strategy is to find any point on one line (say, $L_1$) and then calculate its perpendicular distance to the other line ($L_2$).

Let's find a point $(x_1, y_1)$ on $L_1$. Assuming $B \neq 0$, we can set $x_1=0$. Substituting into the equation for $L_1$ gives:

$A(0) + By_1 + C_1 = 0 \implies By_1 = -C_1 \implies y_1 = -\frac{C_1}{B}$.

So, the point $P(0, -\frac{C_1}{B})$ lies on line $L_1$.

Now, we find the distance from this point $P$ to the line $L_2: Ax + By + C_2 = 0$.

Using the point-to-line distance formula: $d = \frac{|A(0) + B(-\frac{C_1}{B}) + C_2|}{\sqrt{A^2 + B^2}}$

Simplify the numerator: $d = \frac{|-C_1 + C_2|}{\sqrt{A^2 + B^2}} = \frac{|C_2 - C_1|}{\sqrt{A^2 + B^2}}$.

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

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Answer:

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