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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


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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