Coordinate Geometry Long Questions (Question 1 & 2)


Question 1:
The diagram shows a straight line PQ which meets a straight line RS at the point Q. The point P lies on the y-axis.
(a) Write down the equation of RS in the intercept form.
(b) Given that 2RQ = QS, find the coordinates of Q.
(c) Given that PQ is perpendicular to RS, find the y-intercept of PQ.


Solution:
(a) 
Equation of RS
x 12 + y 6 =1 x 12 y 6 =1

(b)
Given 2RQ=QS RQ QS = 1 2 Lets coordinates of Q=(x, y) ( ( 0 )( 2 )+( 12 )( 1 ) 1+2 , ( 6 )( 2 )+( 0 )( 1 ) 1+2 )=( x, y ) x= 12 3 =4 y= 12 3 =4 Q=(4,4)

(c) 
Gradient of RS,  m RS =( 6 12 )= 1 2 m PQ = 1 m RS = 1 1 2 =2
Point Q = (4, –4), m = –2
Using y = mx + c
–4 = –2 (4) + c
c = 4
y–intercept of PQ = 4


Question 2:
The diagram shows a trapezium PQRS. Given the equation of PQ is 2y x – 5 = 0, find
(a) The value of w,
(b) the equation of PS and hence find the coordinates of P.
(c) The locus of M such that triangle QMS is always perpendicular at M.

Solution:
(a)
Equation of PQ 2yx5=0 2y=x+5 y= 1 2 x+ 5 2 m PQ = 1 2 In a trapizium,  m PQ = m SR 1 2 = 0(3) w4 w4=6 w=10

(b)
m PQ = 1 2 m PS = 1 m PQ = 1 1 2 =2

Point S = (4, –3), m = –2
yy1 = m (xx1)
y – (–3) = –2 (x – 4)
y + 3 = –2x + 8
y = –2x + 5
Equation of PS is y = –2x + 5

PS is y = –2x + 5-----(1)
PQ is 2y = x + 5-----(2)
Substitute (1) into (2)
2 (–2x + 5) = x + 5
–4x + 10 = x + 5
–5x = –5
x = 1
From (1), y = –2(1) + 5
y = 3
Coordinates of point P = (1, 3).

(c)
Let M=( x,y ) Given that QMS is perpendicular at M Thus QMS= 90 ( m QM )( m MS )=1 ( y5 x5 )( y( 3 ) x4 )=1 ( y5 )( y+3 )=1( x5 )( x4 ) y 2 +3y5y15=1( x 2 4x5x+20 ) y 2 2y15= x 2 +9x20 x 2 + y 2 9x2y+5=0

Hence, the equation of locus of the moving point M is
x2 + y2 – 9x – 2y + 5 = 0.