# 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 are discussed below.

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

Similarly, 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 as $x$– 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.’

#### Practise This Question

Which point is interior to ABC?