|
Coordinate
geometry
Clement Radcliffe, Contributor
This
week we will continue to review aspects of coordinate geometry. But, we will begin
with the solution to last week's homework. Homework:
Given
the points X(-5, 3) and Y(1,1), find the values of: (9)
Gradient, m (b) the coordinates of the mid-point, m Solution:
| (a)
The gradient of XY = m = | y2
- Y1 | Substituting | | | X2
- X1 | |
| ie.
m = | 1
- 3 | =
-2 | =
-1/3 | | | 1-
-5 | 6 | |
ie.
m = -1/3 | (b)
The mid-point of XY = M = | x2
- x1 | x2
- x1 | | | 2
| 2
|
| ie.
m = | 1
- 5 | ,
1 + 3 | =
-4, | 4 | | | 2 | 2 | 2 | 2 |
ie.
M = (-2, 2) Let
us now continue to review the length of a straight line. LENGTH
OF LINE The
length of AB is found by using Phythagoras' theorem. As triangle ABC is right
angled, therefore AB²
= BC²
+ AC².
: AB²
= (X2 - X2)²
+ (y2-Y2)² 
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 m2 =
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 line, 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. 
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. Example:
A
straight line drawn through the points A(1, 2), B(2, 5), Find the length of AB.
Length:
AB² =
(X2 - X2)²
= (Y2 - Y1)²
Substituting
ie.
AB² =
(2 -1)²
= (5 - 2)² =
1(2) + 3 (2) = 10 ie.
AB = square root of 10 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 | |
| ie.
m = | 3
- 2 | =
1 = | 1 | =
1 | | | -5
- 1 | | -6
| 6 |
| (ii)
The mid-point of AB is M = | (x2
+ x1 | ,
x2 + x1) | | | 2
| 2
|
| ie.
M = | (-5
+ 1 | ,
3 + 2) | | | 2 | 2 |
| ie.
M = | (-4 | ,
5) | =
,(-2, | 5) | | | 2 | 2 | | 2 |
(iii)
in order to find the length of AB, we use the formula AB²
= (Y2 - Y1)²
=
(x2 - X1)² ie.
AB² =
(3 -2)²
+ (-5 - 1)²
+
1² + (-6)²
= 37
ie.
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
- Y2 | or
m = | y1
- Y2 | | | x2
- x1 | | 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). Clement
Radcliffe is the principal of Glenmuir High School in May Pen. | 
