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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. Questions
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
Since
HK cuts the y axis at H(0 , -1), then its intercept c is -1. Given that the gradient
is 2/3, then using the equation y = mx + c, by substitution, 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
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. Example
Given
the points A (2 , 3) and B (6 , - 1) 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. 
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 
ie.
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: 
The
equation of the perpendicular bisector is y - x = - 3. At
the point where this line cuts the y axis, x = 0. ie.
Substituting, y - 0 = -3 ...y = -3
ie.
The coordinates of the point is (0, -3)
ie.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 coordinates 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 =  32
(b)
Since the gradient of AB is - 1 ie.
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 ie.
the equation is found by using the formula
y - y1 = m
x - x1 By
substituting:
y - 3 = 1
x
- 5 ie.
y - 3 = x - 5
ie.
y = x - 2 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, and (iii)
determine the equation of the perpendicular bisector of EF. I
expect, as usual, that you will find more exercises of this nature in your text
books and past papers. Please practise as many of them as you can. Clement
Radcliffe is the principal of Glenmuir High School in May Pen. | 
