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

- Slope-intercept form
- Intercept form
- Normal form

Let us learn all the straight lines formulas along with the general equation of a line and different forms to find the equation of a straight line in detail here.

## 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 â‰ 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 â‰ 0 then, the standard equation of first degree Ax + By + C = 0 can be rewritten in slope-intercept form as:

y = (âˆ’Â A/B) xÂ âˆ’ (C/B)

Thus, m=Â â€“A/BÂ andÂ cÂ =Â â€“C/BÂ .

## 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 â‰ 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 Î± is given by:

**x cos Î±+y sin Î± = 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 cosÂ Î±Â =Â B sinÂ Î±Â = â€“ p**

From this we can say that cosÂ Î± = -p/AÂ and sin Î± = -p/B.

Also it can be inferred that,

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

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

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

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

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

\(p = \pm\left (\frac{AB}{\sqrt {A^{2} + B^{2}}} \right )\) , \(\cos \alpha = \pm \left (\frac{B}{\sqrt {A^{2} + B^{2}}} \right )\)Â , \(\sin \alpha = \pm \left (\frac{A}{\sqrt {A^{2} + B^{2}}} \right )\) - If two pointsÂ (x
_{1}, y_{1})Â and(x_{2}, y_{2})are said to lie on the same side of the line Ax + By + C = 0, then the expressionsÂ Ax_{1}+ By_{1}+ CÂ andÂ Ax_{2}+ By_{2}+ CÂ 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}Â )Â |
m=(y_{2}-y_{1})/(x_{2}-x_{1}), x_{1}â‰ 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}Â ) |
y-y_{1}= [(y_{2}-y_{1})/(x_{2}-x_{1})]Ã—(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 â€“ 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 Î±+y sin Î± = 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 â€“ 6y +3 = 0. Find the slope and both the intercepts.**

**Solution**:

The given equation 2x â€“ 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)Â The equation of a line is given by, 13x – y + 12 = 0. Find the slope and both the intercepts.**

**Solution:**Â 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.

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