Review of Coordinate Geometry
Clement Radcliffe, Contributor

We will continue the review of COORDINATE GEOMETRY with problems in the Cartesian Plane. If I may, I will begin with the homework given last week.

Homework

1. A straight line HK cuts the y axis at H(0, -1). The gradient of HK is 2/3.
Show that the equation of the line HK is 2x - 3y = 3.

Solution

The line therefore has gradient 2/3 and intercept -1.
ie. using the equation y = mx + c
The equation is y=2/3X-1
ie. 3y = 2x -3 or 2x -3y = 3.

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

(i) Determine the gradient of AB.
(ii) Write the equation of the line AB.

Solution

Given the points A(-5, 3) and B(1, 2), then;

It is clear from the above that you need to study the various methods and the appropriate formula. Having done this, you are only required to select and present the formula and perform the required substitution effectively.

As we continue to review problens in the Cartesian Plane, I wish to remind you that, in some instances, you are given information which enables you to find either a point on the line, or the gradient of the line.

Example

Given the points A (2, 3) and B (6, -1) determine the equation of the perpendicular bisector of AB and state the co-ordinates of the point at which the perpendicular bisector meets the y-axis.

Given the points A (2, 3) and B (6, -1), the gradient of AB =

Let the gradient of the line perpendicular to AB be m.
ie. mx - 1 = -1 (Product of the gradients of perpendicular lines is -1)
ie. m = 1

The equation of the perpendicular bisector is y - x = -3

At the poing where this line cuts the y axis, x = 0.

Remember the numerous poor attempts made on this topic in past examinations, I am recommending that you do conscientious review of this topic.

Let us now attempt the following together.

The coordinates of A and B are (3, 5) and (7, 1) respectively. X is the mid-point of AB.

(a) Calculate
(i) the length of AB
(ii) the gradient of AB
(iii) the coordinates of X

(b) Determine the equation of the prependicular bisector of AB and state the coordinates of the point of which the prependicular bisector meets the y axis.

Solution

Now for your homework

E is the point (-2, 5) and F is the point (2, -3).

Find by calculation.

(i) the coordinates of G, the midpoint of EF.

(ii) the gradient of EF.

(iii) Determine the equation of the perpendicular bisector of EF.

I expect, as usual, that you will find more excercises of this nature in your text books and past papers. Please practice as many of them as you can.

* Clement Radcliffe is principal of Glenmuir High School in Clarendon.