Sec 3 E Maths Notes Coordinate Geometry

Sec 3 E Maths Notes Coordinate Geometry

1. 4048 3EM Coordinate Geometry (1) Math Academy®
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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)Ap- (2, 5)B
(3, 4)C ACAB=p3- ACAB=







Given two points and , the distance between these points is given by the
formula:

()
11
,xy ( )
22
,xy
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 .






ABCDEFAEF
(0,6)( , 0)s (0,2)
Bsx-D
(, 8)BsDFX
22 2
24s+=
12s=
212DF=x-D
(2 , 2)Ds (, 0)Es 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, )Ap-
2
(,)Bpp-
2
1
-B
(1,1), (1.5,2.25)BB-
2
() 1
32
pp
p
--
=-
--
5=y xy-=3 343 +=xyABC
(,0)kABk
61
(2,5), (3,5), ( ,2 )
77
ABC-
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-=
xy12=
5=y
182 +=xy
1
4
2
yx=+
1
2
105+=yx
The equation of a straight line with gradient and intercept is


m-yc ymxc=+

cmxy +=



Gradient,
must be subject
y

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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
260kx ky--=9kk18k=
(4, 2)-
ymxc=+
3yxc=+
23(4)c-= + 14c=-
314yx\=-
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
06 2
-
==-
-
ymxc=+
3
2
yxc=-+
y-
3
11
2
yx\=-+
( , 2)a 3210xy+-=a
3210xy+-=
( 1, 4)-
1a=-
35
22
yx=-+




Find grad 1st
1. Gradient
Equation of Line
2. A point
Sub point into
equation of line
Grad same, manipulate
to make the subject
y

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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, )Cp
BC
80
pBC
210yx=+