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Coordinate
geometry
Clement Radcliffe, Contributor
I
expect by now you are comfortable with finding the length, gradient and mid-point
of the line joining two given points. These will be further illustrated by giving
you the solution to the homework. Homework
Given
the points A(-8, 2) and B(3 , -2), find the following with respect to the line
AB. i)
the gradient, m
ii) the mid-point, M
iii)
the length of the line AB
iv)
the Gradient of XY, which is parallel to AB (v) the gradient of AC which is perpendicular
to AB Solution:
(iii)
The length of AB²
= (x2 - x1)²
+ (y2 - y1)²
Substituting 
(v)
Since AC is perpendicular to AB, let AC = m1 and gradient of AB = m. ie.
m x m1 = -1. 
We
will continue coordinate geometry by considering: Equation
of straight lines Reminders:
- All
straight lines have the equation: y = mx + c where m is the gradient and c is
the intercept.
- When
the axes cut at the origin (0 , 0), the equation of the x axis is y = 0 and for
the y axis it is x = 0.
- y
= 2x + 3 is the equation of a line if for each point (x , y) on the line, the
y coordinate is equal to twice the x coordinate of the same point plus 3. The
points (2 , 7) and (-1 , 1) are, therefore, on the line.
This
fact about an equation is not usually emphasised but must be clearly noted. The
point (x , y) is on the line y = mx + c if it satisfies the equation. You
may show that (1 , - 2) is a point on the line y = 3x - 5 by substituting x =
1 and y = - 2 into the equation. (Substitution shows that -2 = -2). The
value of c, the intercept of a line, is found by substituting x = 0 into its equation.
Do you know why? If not, please investigate. Methods
of finding the equation The
following are the three methods which are commonly used to find the equation of
a straight line. i)
Evaluating the equation given the gradient m and the intercept c. Example:
Find
the equation given that m = 3/4 and c = 1. The
equation is y = mx + c. Substituting y = 3/4 x + 1 or 4y = 3x + 4. Answer
is 4y = 3x + 4. This
method can be extended to a given line on a graph. In this case, both the gradient
and the intercept can be found from the graph and the equation determined. ii)
A feature of the second method is: Given the coordinates of two points,
(x1
, y1) and (x2 , y2),
the equation is given.
Using
the points A(4 , - 1), B(1 ,1) in the above, then 
ie.
3y + 3 = -2x + 8 ie.
3y + 2x = 5. Answer
is 3y + 2x = 5. iii)
The formula given in (ii) may be expressed as 
This
formula is used, given the coordinates of a point on the line and the gradient
of the line. Example:
Find
the equation of the line if the gradient m = 2/3 and the point (1 , 2) is on the
line. 
ie.
3y - 6 = 2x - 2 ie.
3y - 2x = 4 Answer
is 3y - 2x = 4 Now
for your 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. No.
7(b)CXC, June 1994 General Proficiency 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. (No. 4(a) CXC, June 1995 General) I
must emphasise again that the problems based on this topic are fairly routine.
It will
do you well to practise them so as not to miss out on the opportunity to score
full marks for the question if it is presented in the June exam this year. Clement
Radcliffe is the principal of Glenmuir High School in May Pen. | 
