# How To Find The Equation Of A Line

## General Equation of a Line:

The general equation in two variables of first degree is represented as

Ax + By +C = 0, A, B ≠ 0 where A, B and C are constants which belong to real numbers.

When we represent the equation geometrically, we always get a straight line.

Below is a representation of Ax + By + C = 0 in different forms:

(i) Slope-intercept form:

We know that the equation of a straight line in slope intercept form is given as:

y = mx + c

Where m indicates the slope of the line and c is the y-intercept
When B ≠ 0 then, the standard equation of first degree Ax + By + C = 0 can be rewritten in slope intercept form as:

y = − AB x − CB

Thus m= –AB and c = –CB .

(ii) Intercept form: The equation of a line making intercepts equal to a and b on the x-axis and the y-axis respectively is given by:

xa + yb =1

Now if, C ≠ 0 then the general equation of line Ax + By + C = 0 can be obtained by replacing the x-intercept by –CA and y-intercept by –CB

x−CA + y−CB = 1If C = 0 then Ax + By = 0 represent the equation of a line passing through the origin.

iii) Normal Form: The equation of the line whose length of the perpendicular from origin is p and the angle made by the perpendicular with the positive x-axis is given by α is given by:

x cos α+y sin α = p

This is known as the normal form of line.
The general form of the line Ax + By + C = 0 can be represented in normal form as:

A cos α = B sin α = – p

From this we can say that $\cos \alpha = -\frac{p}{A}$ and $\sin \alpha = -\frac{p}{B}$.
Also it can be inferred that,

$\cos^{2} \alpha + \sin^{2} \alpha = \left (\frac{p}{A} \right )^{2} + \left (\frac{p}{B} \right )^{2}$

$\Rightarrow 1 = p^{2} \left (\frac{A^{2} + B^{2}}{A^{2}B^{2}} \right )$

$\Rightarrow p = \left (\frac{AB}{\sqrt {A^{2} + B^{2}}} \right )$

From the general equation of a straight line Ax + By + C = 0 , we can conclude the following:

i) The slope is given by -A/B, given that B ≠ 0.

ii) The x-intercept is given by $- \frac{C}{A}$ and the y-intercept is given by $- \frac{C}{B}$.

iii) It can be seen from the above discussion that:
$p = \pm\left (\frac{AB}{\sqrt {A^{2} + B^{2}}} \right )$ , $\cos \alpha = \pm \left (\frac{B}{\sqrt {A^{2} + B^{2}}} \right )$  , $\sin \alpha = \pm \left (\frac{A}{\sqrt {A^{2} + B^{2}}} \right )$

iv) If r is the length of perpendicular drawn from the point (x1, y1)to the line Ax + By + C = 0, then

r = |Ax1+By1+C|A2+B2√

v) If two points (x1,y1) and(x2,y2)are said to lie on the same side of the line Ax + By + C = 0, then the expressions $Ax_{1} + By_{1} + C$ and $Ax_{2} + By_{2} + C$ will have the same sign or else these points would lie on the opposite sides of the line.

To understand this concept better let us take an example:

Example: The equation of a line is given by, 2x – 6y +3 = 0. Find the slope and both the intercepts.
Solution: The given equation 2x – 6y + 3 = 0 can be represented in slope intercept form as:

y = 13x + 12
Comparing it with y = mx + c, we can say that,
m = 13
Also the above equation can be re-framed in intercept form and be written as;
x−32 + y- 12 = 1
Thus x-intercept is given as a = −32 and y-intercept as b = 12.
This is known as the normal form of line.
The general form of the line Ax + By + C = 0 can be represented in normal form as:

A cos α = B sin α = – p

From this we can say that $\cos \alpha = -\frac{p}{A}$ and $\sin \alpha = -\frac{p}{B}$.
Also it can be inferred that,

$\cos^{2} \alpha + \sin^{2} \alpha = \left (\frac{p}{A} \right )^{2} + \left (\frac{p}{B} \right )^{2}$

$\Rightarrow 1 = p^{2} \left (\frac{A^{2} + B^{2}}{A^{2}B^{2}} \right )$

$\Rightarrow p = \left (\frac{AB}{\sqrt {A^{2} + B^{2}}} \right )$

From the general equation of a straight line Ax + By + C = 0 , we can conclude the following:
i) The slope is given by -A/B, given that B ≠ 0.

ii) The x-intercept is given by $- \frac{C}{A}$ and the y-intercept is given by $- \frac{C}{B}$.

iii) It can be seen from the above discussion that:
$p = \pm\left (\frac{AB}{\sqrt {A^{2} + B^{2}}} \right )$ , $\cos \alpha = \pm \left (\frac{B}{\sqrt {A^{2} + B^{2}}} \right )$  , $\sin \alpha = \pm \left (\frac{A}{\sqrt {A^{2} + B^{2}}} \right )$

iv) If r is the length of perpendicular drawn from the point (x1, y1)to the line Ax + By + C = 0, then

r = |Ax1+By1+C|A2+B2√

v) If two points (x1,y1) and(x2,y2)are said to lie on the same side of the line Ax + By + C = 0, then the expressions $Ax_{1} + By_{1} + C$ and $Ax_{2} + By_{2} + C$ will have the same sign or else these points would lie on the opposite sides of the line.

Example: The equation of a line is given by, 2x – 6y +3 = 0. Find the slope and both the intercepts.

Solution: The given equation 2x – 6y + 3 = 0 can be represented in slope intercept form as:

y = 13x + 12
Comparing it with y = mx + c, we can say that,
m = 13
Also the above equation can be re-framed in intercept form and be written as;
x−32 + y- 12 = 1
Thus x-intercept is given as a = −32 and y-intercept as b = 12.