# What is Perpendicular Line Formula?

A perpendicular line is defined as a line that passes through a point on another line making an angle of 90^{o} with the original line. The slope of the perpendicular line is the negative of the reciprocal of the slope of the original line. There are two basic properties regarding the perpendicular lines that derive the formulae related to perpendicular lines. Let’s take a look at them,

### Perpendicular line formula

There are properties defined for the perpendicular lines. Below are two properties, they are the product of the slope and the equation of the perpendicular line.

- The
**product of the slope**of a perpendicular line with the slope of the original line is always**-1**.

**Proof:**

Lets the original line makes an angle of θ with the X-axis. Then, the line perpendicular to the line will make an angle of θ + 90° or θ – 90°

^{ }with the X-axis.Then, the slope of the original line is equal to tanθ.

The slope of the perpendicular line is equal to tan(θ + 90

^{o}) or tan(θ – 90^{o}).Thus, the slope of the perpendicular line is -cotθ.

The product of the slopes = tanθ × (-cotθ) = -1

Hence, theproduct of the slopes is always equal to -1.

**2.** If the equation of the original line is **ax + by + c = 0**, then the equation of its perpendicular line is **– bx + ay + d = 0**, where d is a constant.

** Proof:**

The equation of the original line is ax + by + c = 0

The slope of the original line is

-a / b.Let’s the slope of the perpendicular line is m

Since product of the slopes is -1, so, we can write,

m × (-a / b) = – 1

m = b / aSo, if the perpendicular line passes through a point (x

_{1}, y_{1}),(y – y

_{1}) / (x – x_{1}) = b / ay – y

_{1}= (b / a) × (x – x_{1})ay – ay

_{1}= bx – bx_{1}– bx + ay + (bx

_{1}– ay_{1}) = 0Let’s bx

_{1}– ay_{1}= d, where d is a constant.Thus, the equation stands as,

– bx + ay + d = 0

### Sample Problems

**Question 1: Are the lines 3x + 2y + 5 = 0 and 2x – 3y + 8 = 0 perpendicular?**

**Solution:**

The slope of the line 3x + 2y + 5 = 0 is – 3 / 2.

The slope of the line 2x – 3y + 8 = 0 is -2 / (-3) = 2 / 3

Thus, the product of the slopes are: (- 3 / 2) × (2 / 3) = -1

Since, the product of slopes are -1, the lines are perpendicular.

**Question 2: Find the line perpendicular to the line x + 2y + 5 = 0 and passing through the point (2, 5).**

**Solution:**

From property 2, we get that the equation of a line perpendicular to the line ax + by + c = 0 is – bx + ay + d = 0.

Comparing the line x + 2y + 5 = 0 with ax + by + c = 0,

- a = 1
- b = 2
- c = 5
Thus, the equation of any line perpendicular to this line is

– 2x + y + d = 0, where d is a constant.Given, this line passes through (2, 5), thus putting (2, 5) in this equation of the perpendicular line,

-2 × 2 + 5 + d = 0

- d = -1

Hence, the equation of the perpendicular line stands as -2x + y – 1 = 0

**Question 3: Find the slope of the line perpendicular to the line 3x + 9y + 7 = 0.**

**Solution:**

Given, the equation of the line is 3x + 9y + 7 = 0.

So, the slope of this line is – 3 / 9 = – 1 / 3.

Let’s, the slope of the perpendicular line be m.

From property 1, we can write,

m × (- 1 / 3) = – 1

m = 3

Thus, the slope of the line perpendicular to the given line is 3.

**Question 4: Find the angle of a line perpendicular to the line x + y + 3 = 0 with the X-axis in the range [0, 90°].**

**Solution:**

Slope of the given line = – 1 / 1 = – 1

Lets, the slope of its perpendicular line be m.

So, from property 1, we can write,

m × -1 = – 1

m = 1

So, if the angle of the line perpendicular to the given line is θ, then we can write slope as

tanθ = 1

θ = tan

^{-1}(1) = 45°

Hence, the angle made by perpendicular line with X-axis is 45°.