This document provides examples and explanations of using the distance formula and equations of lines in coordinate geometry. It defines the distance formula and shows how to calculate the distance between two points with given coordinates. It also demonstrates how to determine the gradient and y-intercept of a line given its equation, find the equation of a line given the gradient and a point or two points, and find values related to lines parallel or intersecting given lines.
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Sec 3 E Maths Notes Coordinate Geometry
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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= +