# Coordinate Geometry (Perpendicular Lines)

Updated: Jun 22, 2021

__In this note, you will learn:__

· __How to solve questions involving perpendicular lines__

__How to solve questions involving perpendicular lines__

__How to solve questions involving perpendicular lines__

In general, when we have two lines given by **l1** and **l2**, with gradients **m1** and **m2** respectively. We can show their relationship as such:

**l1 is perpendicular to l2** ->** product of their gradients m1m2 = -1**

This means that the gradients of the two lines are **complete opposites (one is positive, whereas the other is negative, and they are the reciprocals of each other)**.

However, a special exception arises if one of the lines is **vertical**, which means that the gradient is undefined and hence, we will not be able to calculate m1m2.

To show the relationship of two lines to be **perpendicular**, we denote it with the notation: **⊥**

So, for example, if lines **AB** and **CD** are perpendicular to each other, we will express it as such:

**AB ⊥ CD**

Now, let’s take a look at the application of the relationship above in the example questions shown below:

1. The vertices of triangle ABC are given by A (2, 5), B (-3, 0) and C (4, -2). Is triangle ABC a right-angled triangle? Give reasons for your answer.

Solution:

**Method 1:**

Let m1 = gradient of AB and m2 = gradient of BC.

Then m1m2 = [(5 – 0)/ (2 – (-3))] x [(0 – (-2))/ (-3 – 4)]

= (5/5) x [2/ (-7)]

= -2/7

Therefore, since m1m2 does not equal to -1, this means that AB is not perpendicular to BC and hence, triangle ABC is not a right-angled triangle.

**Method 2:**

AB = √ [(0 – 5)^2 + ((-3) – 2)^2] = √ 50 units

BC = √ [((-2) – 0)^2 + (4 – (-3))^2] = √ 53 units

AC = √ [((-2) – 5)^2 + (4 – 2)^2] = √ 53 units

Since AB^2 + BC^2 does not equal to AC^2, by the converse of Pythagoras’ Theorem, triangle ABC is not a right-angled triangle.

2. The coordinates of three points are O (0, 1), P (m, n) and Q (5, 3). Find a possible set of values of m and n if OP is perpendicular to OQ.

Solution:

Gradient of OP x gradient of OQ = -1

[(n – 1)/ (m – 0)] x [(3 – 1)/ (5 – 0)] = -1

2(n – 1)/ 5m = -1

2n – 2 = -5m

n = (-5m + 2)/ 2

Therefore, a possible set of values is m = 1 and n = -3/2.

3. Find the equation of the perpendicular bisector of the line segment joining A (2, 4) and B (12, 6).

Solution:

Gradient of AB = (y2 – y1)/ (x2 – x1)

= (6 – 4)/ (12 – 2)

= 2/ 10

= 1/ 5

Gradient of perpendicular bisector of AB = -5

Midpoint of AB = ([(x1 + x2)/ 2], [(y1 + y2)/ 2])

= ([(2 + 12)/ 2], [(4 + 6)])

= (7, 5)

Therefore, equation of perpendicular bisector of AB is **y – y1 = m (x – x1)**.

y – 5 = (-5) (x – 7)

y = -5x + 40

4. The diagram shows a quadrilateral ABCD in which the points A and D lie on the y-axis and x-axis respectively. C is the point (16, 11) and the equation of AB is y = 2x + 4. If AD and BC are perpendicular to AB, find

i) The coordinates of A and of D,

ii) The equation of BC,

iii) The coordinates of B,

iv) The length of CD.

Solution:

i) At A, x = 0, so y = 2(0) + 4 = 4

Therefore, the coordinates of A are (0, 4).

Let D be the point (k, 0)

Since AB ⊥ AD and gradient of AB = -2, then

Gradient of AD = -1/2

(4 – 0)/ (0 – k) = -1/2

k = 8

Therefore, the coordinates of D are (8, 0).

ii) Gradient of BC = Gradient of AD = -1/2

Therefore, equation of BC is **y – y1 = m (x – x1)**

y – 11 = (-1/2) (x – 16)

y = (-1/2) x + 19

iii) 2x + 4 = (-1/2) x + 19

4x + 8 = -x + 38

5x = 30

x = 6

Substitute x = 6 into y = 2x + 4:

y = 2(6) + 4 = 16

Therefore, the coordinates of B are (6, 16).

iv) Length of CD = √ [(x2 – x1)^2 + (y2 – y1)^2]

= √ [(16 – 8)^2 + (11 – 0)^2]

= √ (185)

= 13.6 units (3 s.f)

And that’s all for today, students! ** Math Lobby** hopes that after this article, you have a clear understanding on the application of the relationship between two lines which are perpendicular and how you can use it to solve questions in coordinate geometry!

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