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.
... using the equation y = mx + c
The equation is y = 2/3x - 1
... 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

(i) Gradient of AB, m = (y₂ - y₁)/x₂ - x₂ = (2 - 3)/(1 - -5) = -1/6

... m = -1/6

(ii) Since the gradient of AB = -1/6, then the equation of AB is:

(y - y₁)/(x - x₁) = m. Substituting

(y - 3)/(x - -5) = -1/6

(y - 3)/(x + 5) = -1/6 CLearing

... 6(y - 3) = -1 (x + 5) ...6y - 18 = -x -5
...6y + x = 18 - 5 ...6y + x = 13

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 problems 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. This information is then used inthe appropriate formula

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 = (y₂ - y₁)/(x₂ - x₁)

= (-1-3)/(6 - 2)

= -4/4 = -1

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

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

... M = ( (6 + 2)/2 , (-1 + 3)/2 ) = (8/2 , 2/2) = (4 , 1)

... The perpendicular bisector of AB has gradient 1 and passes through the point (4 , 1).

The equation of the perpendicular bisector is found using the formula:

(y - y₁)/(x - x₁) = m ... (y - 1)/(x - 4) = 1

... y - 1 = x - 4 ... y - x = -3

The equation of the perpendicular bisector is y - x = -3.
At the point where this line cuts th ey axis, x = 0

... Substituting, y - 0 = -3 ... y = -3
... The co-ordinates of the point is (o, -3) ... the line cuts the y-axis at (0, -3)

Remembering 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 coordiantes of X

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

Solutions

(a)

(i) the length of AB
AB² = (7 - 3)² + (1 - 5)² = 4² + (-4)² = 32
... the length of AB = square root of 32

(ii) the gradient of AB
Gradient of AB or m = (y₂ - y₁)/(x₂ - x₁) = (1 - 5)/(7 - 3) = -4/4 = -1

(iii) the coordinates of X
... the midpoint of AB or M = ( (x₂ + x₁)/2 , (y₂ + y₁)/2 )

... M = ( (7 + 3)/2 , (1 + 5)/2 ) = (5 , 3)

(b) Since the gradient of AB is -1
... the gradient of the perpendicular to AB = 1. (N.B. 1 x -1 = -1)
Since the bisector of AB passes through (5 , 3) and has gradient 1
... the equation is found by using the formula: (y - y₁)/(x - x₁) = m

By substituting: (y - 3)/(x - 5) = 1

... y - 3 = x - 5
... y = x - 2

Now for your homework.

The coordiantes of the points L and N are (5 , 6) and (8 , -2) respectively.

1. (i) State the coordinates of the midpoint, M, of the line LN.

(ii) Calculate the length of the line LN.

(iii) Calculate the gradient of the line LN

(iv) Determine the equation of the straight line which is perpendicular to LN and which passes through the point, M.

2. Determine three different features of the following equations 2y + 3x = 5

I expect, as usual, that you will find more exercises of this nature in your text books and past papers.

Please practice as many of them as you can.

Professor Gordon Shirley (left), pro vice-chancellor and principal of the University of the West Indies (UWI), Mona campus, presents the award for the Most Outstanding Researcher to Dr Minerva Thame, from the Department of Obstetrics, Gynaecology and Child Health in the Faculty of Medical Sciences, at the UWI awards ceremony, in recognition of outstanding researcher work at the Mona Visitors' Lodge, recently. - Peta-Gaye Clachar/Staff Photographer

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