Coordinate geometry
ClementRadcliffe, Contributor

This week, we will continue to review aspects of coordinate geometry. We will begin with the solution to the homework last week.

Homework

Given the points X(-5 , 3) and Y( 1 , 1), find the values of:

(a) Gradient, m

(b) The coordinates of the mid-point, M

Solution

(a) The gradient of XY = m = y₂-y₁/ (x₂-x₁)

Substituting ...m = 1-3/(1- -5) = -2/6 = -1/3

... m = -1/3

(b) The mid-point of XY = M = {x₂+x₁/(2), y₂+y₁/(2)}

Substituting ... M = {1-5/(2) , 1+3/(2) = -4/(2) , 4/(2)}

... M = (-2,2)

Let us now continue the review of coordinate geometry with the topic, length of a straight line.

LENGTH OF LINE

The length of AB is found by using Pythagoras' Theorem. As triangle ABC is right angled, therefore AB²=BC²+AC².

... AB² = (x2-x1)²+(y2-y1)²

(See Graph 1 below)

Example

A straight line is drawn through the points A (1,2), B(2,5). Find the length of AB.

Length: AB² = (x2-x1)² + (y2-y1)² Substituting

... AB² = (2-1)² + (5-2)²

= 1²+3² = 10

... AB = …10

We will try another example.

Example

A straight line is drawn through the points A(5,-2) and B(3,4).

Find (i) the gradient of AB

(ii) the mid-point of AB

(iii) the length of AB

Solution:

(i) The gradient of AB = m = y₂-y_1/(x₂-x₁)

... m = (4 - -2)/3-5 = 6/-2 = -3

(ii) The mid-point of AB is:

M = {(x₂+x_1)/2, (y₂+y_1)/2}

... M = {(3+5)/2, (4-2)/2}

... M = {8/2, 2/2} = 4,1

(iii) In order to find the length of AB, we use the formula AB2 = (y2-y1)2+(x2-x1)2

... AB² = (4 - -2)²+(3-5)²

= 6²+(-2)² = 36+4

... AB = suqare root of 40.

If you are to do well on the topic, you must bear the following in mind:

• Always begin by presenting the required formula.
• To calculate the gradient, you may use one of the following:

m = (y₂-y_1)/x₂-x₁ or m = (y₁-y_2)/ x₁-x₂

I am sure you can prove that both are correct.

• In evaluating the values, be careful to ensure the accuracy of the substitution and please watch the negative signs. (Directed numbers).
• Kindly note the following points with respect to the gradient of a straight line:
• Parallel lines have equal gradient.
• If perpendicular lines have gradients m₁ and m₂, then m₁xm₂ = -1.
• It is clear then that given two lines with gradients m₁ and m₂, if they are parallel and m₁ =3/2, then m₂ = 3/2. If they are perpendicular and m₁ =2, then I am sure you agree that m₂ = -1/2

Given a straight, let us now consider its interception on the y axis.

Intercept

This is the coordinate of the point where the line cuts the y axis, that is the point (o, y). This y value is denoted as c.

Graph 2 is a plot of the points A and B on the Cartesian plane which will illustrate the concepts previously reviewed.

(See Graph 2 below)

Homework

Given the points A(-8,2) and B(3, -2), find the following with respect to the line AB:

(i) Gradient, m

(ii) mid-point, M

(iii) Length of the line AB

(iv) Gradient of XY which is parallel to AB

(v) The gradient of AC which is perpendicular to AB.

A Kingston College student (right) and his St Andrew High School friend pose for the camera at a gospel concert put on by Kingston College's Inter-schools' Christian Fellowship recently. -Anthony Minott/Freelance Photographer

Clement Radcliffe is the principal of Glenmuir High School in May Pen.