Table of Contents

## 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 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 = (− A/B) x − (C/B)

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 = 1**

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

**x/(-C/A) + y/(-C/B) = 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})

## 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 go through the below examples:

**(1) 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,

Slope of the line, m = 1/3

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

x/a + y/b = 1

2x – 6y = -3

x/(-3/2) – y/(-1/2) = 1

Thus, x-intercept is given as a = -3/2 and y-intercept as b = 1/2.

**(2) 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,

Slope of the line, m = 13

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

x/a + y/b = 1

13x – y = -12

x/(-12/13) + y/12 = 0

Thus, x-intercept is given as a = -12/13 and y-intercept as b = 12.