Coordinate geometry
Clement Radcliffe, 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 by looking at the length of a straight line.

LENGTH OF LINE

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

... AB² = (x₂ - x₁)² + (y₂ - y₁)²

Example

A straight line is drawn through the points X (- 2, 1) , Y(3 , 2) . Find the length of XY.

Length: XY² = (x₂ - x₁)² + (y₂ - y₁)² Substituting

... XY² = (3 - - 2)² + (2 - 1)²

= 5² + 1² = 26

... XY = square root of 26

We will try another example.

Example

A straight line is drawn through the points A(1 , 2) and B(-5 , 3).

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 - y₁)/x₂ - x₁

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

(ii) The mid-point of AB is M = ( (x₂ + x₁)/2, (y₂ + y₁)/2 )

... M = ( (-5 + 1)/2 , (3 + 2)/2 )

... M = (- 4/2 , 5/2) = (- 2 , 5/2)

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

... AB² = (- 5 - 1)² + (3 - 2)²

= ( - 6)² + 1² = 37

... AB = square root of 37.

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 - y₁)/(x₂ - x₁) or m = (y₁ - y₂)/(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 m₁ x m₂ = -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.

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

Example

Given the line A(3 , -2) and B(1 , 4), describe the gradient of the perpendicular bisector of AB.

Solution

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

... m = (4 - - 2)/(1 - 3) = 6/-2 = -3

... m = -3

Let the gradient of the line perpendicular to AB be m₁

... m x m₁ = -1.

... -3 x m₁ = -1.

... m₁ = -1/-3 = 1/3

The mid-point of AB, M, = ( (x₂ + x_1)/2 , (y₂ + y_1)/2) Substituting

... M = ( (1 + 3)/2 , (4 -2)/2 ) = (2, 1)

... The perpendicular bisector of AB has gradient 1/3 and passes through the point (2, 1).

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

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