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Coordinate
geometry
ClementRadcliffe,
Contributor
I do expect that 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) 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
Solution
(i)
Gradient, m of AB = y2
- y1/x2 - x1
Substituting
... m = -2 - 2/3 - -8 = -4/11
... m = -4/11
(ii) Mid-point, M of AB
= [(x2 + x1)/2,
(y2 + y1)/2]
Substituting
... M = [(3 -8)/2, (-2 + 2)/2]
= [-5/2 , 0]
... the coordinates of the midpoint
of AB are [-5/2 , 0]
(iii) The length of AB2
= (x2 - x1)2
+ (y2 - y1)2
Substituting
... AB2 = square
root of (3 - -8)2 + (-2
- 2)2 = (11)2
+ (-4)2
... AB2 =
121 + 16 = 137
... AB = Square root
of 137 = 11.7
... the length of AB = 11.7
(iv) Since XY is parallel to AB
Then XY = -4/11
(v)
Since AC is perpendicular to AB, let
AC = m1 and gardient of
AB = m.
... m x m1 = -1
... m1 x -4/11
=
-1
...
m1 = 11/4
... The gradient of AC = 11/4
We will continue COORDINATE
GEOMETRY by considering:
Equation
of Straight lines
Reminders
- All
straight lines have hte 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
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.
Method
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 x)/4 + 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.
(y
- y1)/x - x1
= (y2 - y1)/x2
- x1
Using
the points A(4 , - 1), B(1 ,1) in
the above, then
[y
- ( -1)]/x - 4 = [1- (-1)]/1 - 4 =
2/-3
...
(y + 1)/x - 4 = -2/3
...
3y + 3 = -2x + 8
...
3y + 2x = 5.
Answer
is 3y + 2x = 5.
(iii)
The formula given in (ii) may be expressed
as (y - y1)/x - x1 = m
I
am sure you realise that m = (y2
- y1)/(x2 -
x1 )
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.
(y
- y1)/(x - x1) = m, that
is (y - 2)/(x - 1) = 2/3
...
3y - 6 = 2x - 2
...
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.
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.
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.
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