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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.
Homework
1.
A straight line HK cuts the y axis
at H(0 , -1). The gradient of HK is
?.
Show
that the equation of the line HK is
2x - 3y = 3.
Solution
The
line, therefore, has gradient ? and
intercept - 1.
\\
using the equation y = mx + c
The
equation is y = ?x - 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 - x1)
= (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
... 6y + x = 18 - 5 ... 6y + x
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 in the 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 the
y axis, x = 0.
\\
Substituting, y - 0 = -3 \\ y = -3
\\
The co-ordinates of the point is (0,
-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 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 = v32
(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:
By
substituting: (y - 3)/ (x - 5) = 1
(y
- y1)/(x - x1)
= m
Now
for your homework.
The
coordinates 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
textbooks and past papers. Please
practise as many of them as you can.
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St
George's College choir performs
a musical interlude at the ground-breaking
ceremony for the Archbishop
Lawrence A. Burke Centre, in
honour of Archbishop Emeritus
Burke, at St George's College,
Winchester Park, North Street,
Kingston.
- Rudolph Brown/Chief Photographer
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Clement
Radcliffe is the principal of Glenmuir
High School in May Pen.
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