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8.2.2 Coordinates, PT3 Focus Practice

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Question 6:
The point M (x, 4), is the midpoint of the line joining straight line Q (-2, -3) and R (14, y).
The value of x and y are

Solution:

x= −2+14 2 x= 12 2 x=6 4= −3+y 2 8=−3+y y=11


Question 7:
In diagram below, PQR is a right-angled triangle. The sides QR and PQ are parallel to the y-axis and the x-axis respectively. The length of QR = 6 units.

Given that M is the midpoint of PR, then the coordinates of M are

Solution:
x-coordinate of R = 3
y-coordinate of R = 1 + 6 = 7
R = (3, 7)

P( 1,1 ),R( 3,7 ) Coordinates of M =( 1+3 2 , 1+7 2 ) =( 2,4 )


Question 8:
Given points P (–2, 8) and Q (10, 8), find the length of PQ.

Solution:
Length of PQ = [ 10−( −2 ) ] 2 + ( 8−8 ) 2 = ( 14 ) 2 +0 =14 units


Question 9:
In diagram below, ABC is an isosceles triangle.

Find
(a) the value of k,
(b) the length of BC.

Solution:
( a ) For an isosceles triangle,  y−coordinate of C is the midpoint of straight line AB. 2+k 2 =−3 2+k=−6   k=−8 ( b ) B=( −2,−8 ) BC= [ 10−( −2 ) ] 2 + [ −3−( −8 ) ] 2  = 12 2 + 5 2  =13 units


Question 10:
Diagram below shows a rhombus PQRS drawn on a Cartesian plane. PS is parallel to x-axis.

Given the perimeter of PQRS is 40 units, find the coordinates of point R.

Solution:
All sides of rhombus have the same length, therefore length of each side= 40 4 =10 units PQ=10 ( 9− x 1 ) 2 + ( 7−( −1 ) ) 2 = 10 2 81−18 x 1 + x 1 2 +64=100 x 1 2 −18 x 1 +45=0 ( x 1 −3 )( x 1 −15 )=0 x 1 =3,15 x 1 =3 Q=( 3,−1 ),R=( x 2 ,−1 ) QR=10 ( x 2 −3 ) 2 + [ −1−( −1 ) ] 2 = 10 2 x 2 2 −6 x 2 +9+0=100 x 2 2 −6 x 2 −91=0 ( x 2 +7 )( x 2 −13 )=0 x 2 =−7,13 x 2 =13 ∴R=( 13,−1 )