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  2 R Q = Q S R Q Q S = 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  R S ,   m R S = ( 6 12 ) = 1 2 m P Q = 1 m R S = 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 PQis 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  P Q 2 y x 5 = 0 2 y = x + 5 y = 1 2 x + 5 2 m P Q = 1 2 In a trapizium,  m P Q = m S R 1 2 = 0 ( 3 ) w 4 w 4 = 6 w = 10

(b)
m P Q = 1 2 m P S = 1 m P Q = 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  Q M S  is perpendicular at  M Thus  Q M S = 90 ( m Q M ) ( m M S ) = 1 ( y 5 x 5 ) ( y ( 3 ) x 4 ) = 1 ( y 5 ) ( y + 3 ) = 1 ( x 5 ) ( x 4 ) y 2 + 3 y 5 y 15 = 1 ( x 2 4 x 5 x + 20 ) y 2 2 y 15 = x 2 + 9 x 20 x 2 + y 2 9 x 2 y + 5 = 0

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

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