1. 4048 3EM Coordinate Geometry (1) Math Academy®
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- Brian Tracy
Notes: Coordinate Geometry
[A] Distance Formula
This formula is an application of Pythagoras' theorem for right triangles:
Note that the distance is taken to be positive.
Example 1: Given that is an isosceles triangle with vertices , and
and , find the value of .
Ans:
Solution:
ABC ( , 1)A p - (2, 5)B
(3, 4)C ACAB = p
3-
ACAB =
Given two points and , the distance between these points is given by the
formula:
( )1 1,x y ( )2 2,x y
2
21
2
21 )()( yyxxPQ -+-=
Length same, apply distance formula

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Example 2: In the diagram below, is a regular hexagon. , and has
coordinates , and respectively.
(i) State the coordinates of , in terms of . [1]
(ii) Justify, showing all workings clearly, why the coordinate of will not be
an integer. [2]
Solution: (i)
(ii) Let the midpoint of be .
By Pythagoras theorem,
Since 12 is not a perfect square, the coordinate of will not be an
integer.
Or
can apply distance formula to and , whereby .
ABCDEF A E F
(0,6) ( , 0)s (0,2)
B s
x - D
( , 8)B s
DF X
2 2 2
2 4s + =
12s =
2 12DF =
x - D
(2 , 2)D s ( , 0)E s 4DE =

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Example 6: Given that the gradient of and is , find the possible
coordinates of . Ans:
Solution:
Example 7: The diagram, which is not drawn to scale, shows the three lines
, and .
(a) Find the coordinates of , and . [4]
(b) The point is the same distance from as it is from
Find the value of . [1]
Ans: (a) (b) 0.5
(3, )A p- 2
( , )B p p-
2
1
-
B ( 1,1), (1.5,2.25)B B-
2
( ) 1
3 2
p p
p
- -
= -
- -
5=y xy -= 3 343 += xy
A B C
( ,0)k A B
k
6 1
( 2,5), (3,5), ( ,2 )
7 7
A B C-
Casio mode 3, 3
y
x
O
A B
C
horizontal Positive
gradient
Negative
gradient

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[C] Manipulate the Equation of a straight line
Example 8: Determine the gradient and intercept for each of the straight lines in the
table below.
y -
Equation Gradient intercept
No need transform
No need transform 12 0
No need transform 0 5
4
cmxy += -y
xy
3
1
5 -=
xy 12=
5=y
182 += xy
1
4
2
y x= +
1
2
105 += yx
The equation of a straight line with gradient and intercept ism -y c
y mx c= +
cmxy +=
Gradient,
must be subjecty

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Example 9: The gradient of the line , is . Find the value of ,
where is a whole number. Ans:
Ws 2
[D] Form equation of a straight line
(1) Given gradient and pass through a point
Example 10: Find the equation of the straight line whose gradient is 3 and passes through
the point .
Solution:
l 2
2 6 0k x ky- - = 9 k
k 18k =
(4, 2)-
y mx c= +
3y x c= +
2 3(4) c- = +
14c = -
3 14y x = -
1. Gradient
Equation of Line
2. A point
Sub grad 1st
Sub point 2nd

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(2) Given 2 points
Example 11: Find the equation of the line passing through and .
Solution: gradient
Observe that the intercept is 11
Example 12: The point lies on the line .
(a) Find the value of . [1]
(b) Find the equation of the line parallel to , passing
through the point . [2]
Ans: (a) (b)
(0,11)A )2,6(B
11 2 3
0 6 2
-
= = -
-
y mx c= +
3
2
y x c= - +
y -
3
11
2
y x = - +
( , 2)a 3 2 1 0x y+ - =
a
3 2 1 0x y+ - =
( 1, 4)-
1a = -
3 5
2 2
y x= - +
Find grad 1st
1. Gradient
Equation of Line
2. A point
Sub point into
equation of line
Grad same, manipulate
to make the subjecty

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Example 13: The diagram shows a regular hexagon, , where
and .
Given that the length of is units, find
(a) the value of . [2]
(b) the equation of , [2]
Ans: (a) 18 (b)
ABCDEF (0,10)B
(4, )C p
BC 80
p
BC
2 10y x= +