# Equation of a Line

The standard forms of the **equation of a line** are:

- Slope-intercept form
- Intercept form
- Normal form

## General Equation of a Line

The general equation of a line in two variables of the 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 straight-line formulas Ax + By + C = 0 in different forms:

## 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= –A/B and c = –C/B .

## Intercept Form

The intercept of a line is the point through which the line crosses the x-axis or y-axis. Suppose a line cuts the x-axis and y-axis at (a,0) and (0,b) respectively. Then, the equation of a line making intercepts equal to a and b on the x-axis and the y-axis respectively is given by:

**x/a + y/b =c**

Now in case of the general form of the equation of the straight line, i/e. Ax+Bx+C = 0, if C ≠ 0, then Ax + By + C = 0 can be written as;

**x/(-C/A) + y/(-C/A) = 1**

where a = -C/A and b = – C/B

## Normal Form

The equation of the line whose length of the perpendicular from the 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 the line.

In case of 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 α = -p/A and sin α = -p/B.

Also it can be inferred that,

cos^{2}α + sin^{2}α = (p/A)^{2} + (p/B)^{2}

1 = p^{2} (A^{2} + B^{2}/A^{2} .B^{2})

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

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

- The slope is given by -A/B, given that B ≠ 0.
- The x-intercept is given by -C/A and the y-intercept is given by -C/B.
- It can be seen from the above discussion that:

\(p = \pm\left (\frac{AB}{\sqrt {A^{2} + B^{2}}} \right )\) , \(\cos \alpha = \pm \left (\frac{B}{\sqrt {A^{2} + B^{2}}} \right )\) , \(\sin \alpha = \pm \left (\frac{A}{\sqrt {A^{2} + B^{2}}} \right )\) - If two points (x
_{1},y_{1}) and(x_{2},y_{2})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.

## Straight Line Formulas

Let us accumulate the straight line formulas we have discussed so far:

Slope (m) of a non-vertical line passing through the points (x_{1} , y_{1} ) and (x_{2}, y_{2} ) |
m=(y_{2}-y_{1})/(x_{2}-x_{1}), x_{1}≠x_{2} |

Equation of a horizontal line | y = a or y=-a |

Equation of a vertical line | x=b or x=-b |

Equation of the line passing through the points (x_{1} , y_{1} ) and (x_{2}, y_{2} ) |
y-y_{1}= [(y_{2}-y_{1})/(x_{2}-x_{1})]/(x-x_{1}) |

Equation of line with slope m and intercept c | y = mx+c |

Equation of line with slope m makes x-intercept d. | y = m (x – d). |

Intercept form of the equation of a line | x/a+y/b=1 |

The normal form of the equation of a line | x cos α+y sin α = p |

### Example of Straight Lines

To understand this concept better let us take an example:

**(i) 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 = x/3 + 1/2

Comparing it with y = mx + c, we can say that,

the slope of the line, m = 1/3

Also, the above equation can be re-framed in intercept form and be written as;

x/a + y/b = 1

x – y/(1/3) = -3/2

Thus x-intercept is given as a = 1 and y-intercept as b = -1/3.

(ii) **The equation of a line is given by, 13x – y + 12 = 0. Find the slope and both the intercepts.**

Solution: The given equation 13x – y + 12 = 0 can be represented in slope-intercept form as:

y = 13x + 12

Comparing it with y = mx + c, we can say that,

the slope of the line, m = 13

Also, the above equation can be re-framed in intercept form and be written as;

x/a + y/b = 1

x/1 – y/13 = -12/13

Thus x-intercept is given as a = 1 and y-intercept as b = -13.