 # Different Forms Of The Equation Of Line

We know that there are infinite points in the coordinate plane. Consider an arbitrary point P(x,y) on the XY plane and a line L. How will we confirm whether the point is lying on the line L? This is where the importance of equation of a straight line comes into the picture in two-dimensional geometry.

Equation of a straight line contains terms in x and y. If the point P(x,y) satisfies the equation of the line, then the point P lies on the line L.

## Different forms of equations of a straight line

1. Equations of horizontal and vertical lines

Equation of the lines which are horizontal or parallel to the X-axis is y = a, where a is the y – coordinate of the points on the line.

Similarly, equation of a straight line which is vertical or parallel to Y-axis is x = a, where a is the x-coordinate of the points on the line.

For example, the equation of the line which is parallel to X-axis and contains the point (2,3) is y= 3.

Similarly, the equation of the line which is parallel to Y-axis and contains the point (3,4) is x = 3. 2. Point-slope form equation of line

Consider a non-vertical line L whose slope is m, A(x,y) be an arbitrary point on the line and $P(x_1,y_1)$ be the fixed point on the same line. Slope of the line by the definition is,

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

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

For example, equation of the straight line having a slope $m$ = $2$ and passes through the point $(2,3)$ is

y – 3 = 2(x – 2)

y= 2x-4+3

2x-y-1 = 0

3. Two-point form equation of line

Let P(x,y) be the general point on the line L which passes through the points $A(x_1,y_1)$ and $B(x_2,y_2)$. Since the three points are collinear,

slope of PA = slope of AB

$\frac{y~-~y_1}{x~-~x_1}$ = $\frac{y_2~-~y_1}{x_2~-~x_1}$
$y – y_{1} = (y_{2} – y_{1}). \frac{x – x_{1} }{x_{2} – x_{1}}$

4. Slope-intercept form equation of line

Consider a line whose slope is $m$ which cuts the $Y$-axis at a distance ‘a’ from the origin. Then the distance a is called the $y$– intercept of the line. The point at which the line cuts $y$-axis will be $(0,a)$. Then, equation of the line will be

y-a = m(x-0)

y = mx+a

Similarly, a straight line having slope m cuts the X-axis at a distance b from the origin will be at the point (b,0). The distance b is called x- intercept of the line.

Equation of the line will be:

y = m(x-b)

5. Intercept form

Consider a line L having x– intercept a and y– intercept b, then the line touches X– axis at (a,0) and Y– axis at (0,b). By two-point form equation,

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

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

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

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

For example, equation of the line which has $x$– intercept $3$ and $y$– intercept $4$ is,

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

$4x~ + ~3y$ = $12$

6. Normal form

Consider a perpendicular from the origin having length l to line L and it makes an angle β with the positive X-axis. Let OP be the perpendicular from the origin to the line L.
Then,

$OQ$ = $l~ cosβ$

$PQ$ = $l~ sinβ$

Coordinates of the point $P$ are; $P(l ~cos~β,l ~sin~β)$

slope of the line $OP$ is $tan~β$

Therefore,

$Slope~ of ~the ~line ~L$ = $-\frac{1}{tan~β}$ = $-\frac{cos~β}{sin~β}$

Equation of the line $L$ having slope $-\frac{cos~β}{sin~β}$ and passing through the point $(l ~cos~β,l ~sin~β)$ is,

$y~-~l~ sin~β$ = $-\frac{cos~β}{sin~β} (x~-~l ~cosβ)$

$y ~sin~β~-~l~ sin^2~ β$ = $-x ~cos~β~+~l~cos^2~β$

$x~ cos~β~ + ~y ~sin~β$ = $l(sin^2~β ~+ ~cos^2~β)$

$x ~cos~β~ + ~y ~sin~β$ = $l$

You have learnt about the different forms of equation of a straight line. To know more about straight lines and its properties, log onto www.byjus.com.’