## What Is the Equation of a Straight Line?

**An equation of straight line is the linear equation that expresses the relationship between the coordinate points on a straight line.**

A straight line extends infinitely on both sides. It has an infinite number of points. The equation of a straight line can be expressed in a variety of ways, such as point-slope form, slope-intercept form, general form, standard form, and so on and indicates the line’s slope, x-intercept, and y-intercept.

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## Equation of a Straight Line: Definition

The equation of a straight line is the linear equation that is satisfied by all the points on the line. It is the relation between the x-coordinate and y-coordinate of any point on the line.

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## General Equation of a Line

The general equation of line is given by

$Ax + By + C = 0$

where A and B are not zero simultaneously.

A, B, and C are constants (real numbers).

When we graph this equation, we always get a straight line.

**Note: **A line represented by an equation in its general form $ax + by + c$, if $b \neq 0$ has slope $\frac{-a}{b}$ and y-intercept $\frac{-c}{b}$ .

**Example:** Consider the equation $x + 2y + 6 = 0$.

Here, $a = 1,\; b = 2,\; c = 6$

Slope of the line $= \;-\;\frac{a}{b} = \;-\;\frac{1}{2}$

y-intercept of the line $= \;-\;\frac{c}{b} = \;-\;\frac{6}{2} = \;-\;3$

## Different Forms of Equation of a Straight Line

The equation of a straight line can be written in different forms, such as point-slope form, slope-intercept form, standard form, etc.

These forms help us understand how to find the equation of a straight line with the given information. Let’s discuss each form in detail.

**Standard Form of Equation of Straight Line**

The standard form of equation of a line is given by

$Ax + By = C$

where

A, B, and C are constants (integers),

and the values of A and B cannot be zero at the same time.

**Example:** Consider an equation $2x + 3y = 12$.

If we set $x = 0$,we get $3y = 12$ and we can find that $y = 4$, which means the y-intercept is (0,4).

In a similar way, we can set $y = 0$ to get $2x = 12$ and find that the x-intercept is (6,0).

Now we can graph the line:

**Point-Slope Form Equation of Straight Line**

The point-slope form is used to determine the equation of a straight line passing through a point having coordinates $(x_{1},\; y_{1})$ and having slope m.

The point-slope form is given by

$y \;-\; y_{1} = m (x \;-\;x_{1})$

where $(x,\; y)$ is an arbitrary point on the line.

**Slope-Intercept Form Equation of Straight Line**

Suppose that a line has slope m and y-intercept b.

Thus, the line passes through the point (0, b).

Using the point-slope form, we can write

$y \;-\; b = m(x \;-\; 0)$

$y = mx + b$

The slope-intercept form of a line is $y = mx + b$.

**Two Point Form Equation of Straight Line**

Let’s understand how to find the equation of a straight line from two points. Let the line passes through the points be $A(x_{1},\;y_{1})$ and $B(x_{2},\;y_{2})$. Let $P (x,\;y)$ be any point on the straight line.

Now, the slope of the line $AB = \frac{y_{1}\;-\; y_{2}}{x_{1} \;-\; x_{2}}$

The slope of the line $AP = \frac{y_{1}\;-\; y_{2}}{x_{1} \;-\; x_{2}}$

But the three points A, B and P are collinear.

Therefore, slope of the line AP $=$ slope of the line AB

$\frac{y \;-\; y_{1}}{x \;-\; x_{1}} = \frac{y_{1}\;-\; y_{2}}{x_{1}\;-\; x_{2}}$

$(y\;-\; y_{1}) = \frac{y_{2}\;-\; y_{1}}{x_{2} \;-\; x_{1}} (x \;-\; x_{1})$

Thus, the equation of a line passing through two points $(x_{1},\;y_{1})$ and $(x_{2},\;y_{2})$ is

$y\;-\;y_{1} = \frac{y_{2} \;-\; y_{1}}{x_{2}\;-\; x_{1}} (x \;-\; x_{1})$

The above equation is satisfied by the coordinates of any point lying on the line AB and hence, represents the equation of the straight line AB.

**Intercept Form Equation of Straight Line**

If a line cuts the x-axis and y-axis at (a, 0) and (0, b), respectively, we say that the x-intercept is a and the y-intercept is b.

The equation of a straight line having x-intercept a and y-intercept b is given by

$(\frac{x}{a}) + (\frac{y}{b}) = 1$

**Note:** If a line is represented by an equation in its general form $ax + by + c$, and $c \neq 0$, then we have x-intercept $= \;-\;\frac{c}{a}$ and y-intercept $\;-\;\frac{c}{b}$ .

## Graph of Different Equations of a Straight Line

The graph of an equation of a straight line in one variable y is a horizontal line parallel to the x-axis, and the graph of a linear equation in one variable x creates a vertical line parallel to the y-axis. A linear equation with two variables, x and y, generates a straight line on the graph.

## Equation of a Straight Line Formulas

General form | $Ax + By + C = 0$ |

Standard form | $Ax + By = C$ |

Equation of a horizontal line | $y = a (a \in R)$ |

Equation of a vertical line | $x = b (b \in R)$ |

Two point form | $y\;-\;y_{1} = \frac{y_{2} \;-\; y_{1}}{x_{2}\;-\; x_{1}} (x \;-\; x_{1})$ |

Slope-intercept form | $y = mx + c$ |

Intercept form | $(\frac{x}{a}) + (\frac{y}{b}) = 1$ |

## Facts about Equation of a Straight Line

- The equation of x-axis is $y = 0$.

The equation of y-axis is $x = 0$. - The general form of a horizontal line is $y = b$.

The general form of a vertical line is $x = a$. - If two straight lines are parallel to each other, then they have the same slope.
- If two lines are perpendicular, the product of slopes is $\;-\;1$.
- In the general form of the equation of a straight line $ax + by + c, (b \neq 0)$, rearranging the terms, we get

by $= \;-\; ax \;-\; c$

$y = (\;-\;\frac{a}{b})x \;-\; (\frac{c}{b})$ …(1)

Slope-intercept form: $y = mx + c$ …(2)

By comparing (1) and (2), we get

Slope $= \;-\;\frac{a}{b}$ and y-intercept $= \;-\;\frac{c}{b}$ .

## Conclusion

In this article, we learned about the equation of a straight line and its different forms. Let us solve some examples on the equation of a straight line to understand it better.

## Solved Examples on Equation of a Straight Line

**1. Find the slope and y-intercept of the line with equation **$3x – 5y + 6 = 0$**.**

**Solution:**The given equation of the line is $3x \;-\; 5y + 6 = 0$.

We need to convert this equation in the slope intercept form of the equation of a line.

$3x \;-\; 5y + 6 = 0$.

$3x + 6 = 5y$

$5y = 3x + 6$

$y = \frac{3x}{5} + \frac{6}{5}$

Comparing this equation with the slope-intercept form of the equation of line $y = mx + c$, we have

slope $m = \frac{3}{5}$

y-intercept $c = \frac{6}{5}$

**2. Draw the graph of the linear equation **$3x + 4y = 6$**. At which points does the graph cut the x-axis and y-axis?**

**Solution: **

Given equation is $3x + 4y = 6$.

Now, we need at least 2 points on the graph to draw this line.

**(i) x-intercept**

Substituting $y = 0$ in the equation, $3x + 4y = 6$, we get

$3x + (4 \times 0) = 6$

$3x = 6$

$x = 2$

Hence, the point at which the graph cuts x-axis $= (2,\; 0)$.

**(ii) y-intercept**

Substituting $x = 0$ in the equation, $3x + 4y = 6$, we get

$(3 \times 0) + 4y = 6$

$4y = 6$

$y = \frac{6}{4}$

$y = \frac{3}{2}$

$y = 1.5$

Hence, the point at which the graph cuts y-axis $= (0,\; 1.5)$.

Plotting the points (0, 1.5) and (2, 0) on the graph.

**3. Find the equation of the straight line passing through the points with coordinates (3, 4) and **$(6, \;-\;5)$**.**

**Solution: **

The equation of the straight line passing through the points (3, 4) and $(6, \;-\;5)$ is given by two-point form $\frac{y \;-\; y_{1}}{x \;-\; x_{1}} = \frac{y_{1}\;-\; y_{2}}{x_{1}\;-\; x_{2}}$.

$\frac{y \;-\; 4}{x \;-\; 3} = \frac{4 + 5}{3 \;-\; 6}$ (Using the form: $\frac{y \;-\; y_{1}}{x \;- \;x_{1}} = \frac{y_{1}\;-\; y_{2}}{x_{1}\;-\; x_{2}}$ )

$\frac{y \;-\; 4}{x \;-\; 3} = \frac{9}{\;-\;3}$

$\frac{y \;-\; 4}{x \;-\; 3} = \;-\;3$

$y \;-\; 4 = \;-\;3x + 9$

$3x + y = 13$ is the required equation.

**4. Find the equation of a line passing through the point (6, 4) with a slope of 3. **

**Solution: **

The equation of a line with slope m and one given point $(x_{1},\;y_{1})$ is $y\;-\; y_{1} = m (x\;-\; x_{1})$

The required equation of the line using this one point form is as follows

$(y\;-\;4) = 3(x\;-\;6)$

$(y\;-\;4) = 3x \;-\; 18$

$3x \;-\; 18 \;-\; y \;-\; 4 = 0$

$3x \;-\; y = 14$ is the required equation.

## Practice Problems on Equation of a Straight Line

## Equation of a Straight Line: Definition, Forms, Examples, Facts, FAQs

### The slope of a line $ax + by + c = 0$, $(b \neq 0)$ is __________.

Rearranging the terms and comparing it with the slope-intercept form, we get

$y = (\;-\;\frac{a}{b})x \;-\; (\frac{c}{b})$

Thus, slope $= m = \;-\;\frac{a}{b}$

### The slope-intercept form of a straight line is _________.

The slope-intercept form of a straight line is $y = mx + c$ , where m is the slope of the line and c is the y-intercept.

### The equation of the x-axis is _______________.

Equation of the x-axis is $y = 0$.

### The point-slope form is given by _______________.

The point-slope form is $y \;-\; y_{1} = m (x \;-\;x_{1})$, where (x, y) is an arbitrary point on the line.

### The equation of the y-axis is _______________.

The equation of the y-axis is $x = 0$. The x-coordinate of any point on the y-axis is 0.

## Frequently Asked Questions on Equation of a Straight Line

**How do you write an equation for a vertical and horizontal line?**

The equation of a horizontal line passing through (a, b) is of the form $y = b$. The equation of a vertical line passing through (a, b) is of the form $x = a$.

**How to convert standard form to slope intercept form?**

By rearrangement and comparison, we can change a line’s equation from standard form to slope intercept form. We know that the standard form of the equation of a straight line can be given as, $Ax + By + C = 0$. Rearranging the terms to find the value of “y,” we get,

$B \times y = \;-\;Ax \;-\; C$

$\Rightarrow y = (\;-\; \frac{A}{B})x + (\;-\; \frac{C}{B})$,where, $\;-\; AB$ makes the slope of the line and $(\;-\; \frac{C}{B})$ is the y-intercept.

**What is c in the slope-intercept form of the equation of a line?**

The “c” in the slope-intercept form of the equation of a line $y = mx + c$ is the y-intercept of the line. The line cuts the y-axis at the point (c, 0), and c is the distance of the point on the y-axis from the origin.

**How can we find an equation of a line given the slope and a point?**

To find an equation of a line given the slope and a point follow the following steps:

- Identify the slope and coordinates of the point.
- Substitute the values into the point-slope form, $y \;-\; y_{1} = m (x \;-\;x_{1})$.
- Write the equation in slope-intercept form.

**What is the normal form of a straight line?**

The normal form of the straight line is x cos $+$ y sin $=$ p.

Here, x and y are coordinates, p is the length of the perpendicular from origin to the straight line and is the angle between the positive x-axis and the perpendicular of the straight line from the origin.

The general form of the line $Ax + By + C = 0$ can be represented in normal form as:A cos $=$ B sin $= \;–\; p$