# Equation of a Line

Different Forms of Equation of a Straight Line

There are different forms of an equation of a line. Some of them are explained below: –

1. Equation For Horizontal Line: A horizontal line is parallel the x-axis. Hence, all the points on this line need to be equidistant from the x-axis. The ordinate defines the distance of a point from the x-axis. Since all the points of the horizontal line are equidistant from the x-axis, the ordinates for all of them will be the same, i.e. y=k for all the points in a particular horizontal line.

1. Equation For Vertical Line: A vertical line is parallel the y-axis. Hence, all the points on this line need to be equidistant from the y-axis. The abscissa defines the distance of a point from the y-axis. Since all the points of the vertical line are equidistant from the y-axis, the abscissa for all of them will be the same i.e. x=k for all the points in a particular vertical line.

1. Point – Slope Equation: The point-slope form of equation can be used for a non-horizontal, non-vertical line, when we know the values for the slope, and a point on the line. Let us consider I(x, y) as an imaginary point on this line, and P(x1, y1) as the defined point on the line. Then the point-slope equation can be represented as:

m=$\frac{(y − y_{1})}{(x − x_{1})}$

$m(x – x_{1})$ = $(y – y_{1})$

By plugging in the values of the slope and the defined point, you will get an equation in terms of x and y, which is the equation of the line.

1. Equation of Line in Two Point Form: In the two point form of equation, we take a line and consider an imaginary point I (x, y). Now, take two defined points P (x1, y1) and Q (x2, y2) , which are collinear to point I. Since these points are collinear, the slope of PI= slope of PQ. Putting this in the form of an equation:

$\frac{(y – y_{1})}{(x – x_{1})}$ = $\frac{(y_{2} – y_{1})}{(x_{2} – x_{1})}$

$(y – y_{1})$ = $(y_{2} – y_{1}) \times \frac{(x – x_{1}}{(x_{2} – x_{1}})$

1. Slope Intercept Form: Consider the following image. Line L cuts the y-axis at point P(0,a). Here, a is the distance from the origin at which the line L cuts the y-axis. This distance is referred to as the y-intercept. Here, let us take I(x, y) as some imaginary point on this line and P(0,a) as the defined point. We will plug in values in the point-slope equation:

$m(x – x_{1}) = (y – y_{1})$

m(x − 0)= (y − a)

mx – 0 = y – a

mx – y + a = 0  (or) y = mx + a

Similarly, let us take a line M that cuts the x-axis at Q(b,0). Here, b is the distance from the origin, at which the line M cuts the x-axis. This distance is called the x-intercept. Here, let us take I(x, y) as some imaginary point on this line and Q(b,0) as the defined point. We will plug in values in the point-slope equation:

m (x − x1)= (y − y1)

m (x − b)= (y − 0)

m (x – b) = y

1. Intercept Form: In this case, we are given a line with x-intercept (b,0) and y-intercept (0,a). Consider a random point I (x, y) on the line. Now, we lug in the values in the two-point form equation.

$(y – y_{1})$ =$(y_{2} – y_{1})$ x $\frac{(x – x_{1})}{(x_{2} – x_{1})}$

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

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

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

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

Solved Examples

Example 1: P(3,4) and Q(6, 4) are two collinear points. Represent it in the form of an equation.

Solution:

The ordinates are equal; hence, this is a horizontal line. The slope of a horizontal line is zero, and it is represented as y = k. Therefore, the equation is y=4.

Example 2: In a particular line, the x-intercept is 3, and the y-intercept is 4. Find the value of the line.

Solution:

Since we have both the intercepts, we can use the intercept form of equation and substitute values.

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

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

$\frac{(4x + 3)}{12}$ = 1

4x + 3y = 12

4x + 3y – 12 = 0

Hence, the equation for this line is 4x + 3y – 12 = 0.