Question 9:

The diagram above shows that the straight line PMQ intersects with PNR at N. Given that OQ = OR and M is the midpoint of PQ. Find
(a) the coordinate of P
(b) the value of m and n.
Solution:
(a)Given M is midpoint of PQx-coordinate of P=0For y-coordinate of P,y+02=3y=6Coordinates of P=(0,6).(a)Given M is midpoint of PQx-coordinate of P=0For y-coordinate of P,y+02=3y=6Coordinates of P=(0,6).
(b)0+42=m2m=4m=2Gradient of PR=6−00−(−4)=64=32At point N(n,4), using y1=mx1+c4=32n+68=3n+123n=−4n=−43(b)0+42=m2m=4m=2Gradient of PR=6−00−(−4)=64=32At point N(n,4), using y1=mx1+c4=32n+68=3n+123n=−4n=−43

The diagram above shows that the straight line PMQ intersects with PNR at N. Given that OQ = OR and M is the midpoint of PQ. Find
(a) the coordinate of P
(b) the value of m and n.
Solution:
(a)Given M is midpoint of PQx-coordinate of P=0For y-coordinate of P,y+02=3y=6Coordinates of P=(0,6).(a)Given M is midpoint of PQx-coordinate of P=0For y-coordinate of P,y+02=3y=6Coordinates of P=(0,6).
(b)0+42=m2m=4m=2Gradient of PR=6−00−(−4)=64=32At point N(n,4), using y1=mx1+c4=32n+68=3n+123n=−4n=−43(b)0+42=m2m=4m=2Gradient of PR=6−00−(−4)=64=32At point N(n,4), using y1=mx1+c4=32n+68=3n+123n=−4n=−43
Question 10:
Diagram 10 shows two parallel straight lines, JK and LM, drawn on a Cartesian plane.
The straight line KM is parallel to the x-axis.
Diagram 10
Find
(a) the equation of the straight line KM,
(b) the equation of the straight line LM,
(c) the value of k.
Solution:
(a)
The equation of the straight line KM is y = 3
(b)
Given, equation of JK:2y=4x+3y=2x+32Thus, mJK=2mLM=mJK=2y=mx+cAt M(5, 3)3=2(5)+c3=10+cc=−7∴ Equation of the straight line LMis y=2x−7.Given, equation of JK:2y=4x+3y=2x+32Thus, mJK=2mLM=mJK=2y=mx+cAt M(5, 3)3=2(5)+c3=10+cc=−7∴ Equation of the straight line LMis y=2x−7.
(c)
Substitute (k,0) into y=2x−70=2(k)−77=2kk=72
Diagram 10 shows two parallel straight lines, JK and LM, drawn on a Cartesian plane.
The straight line KM is parallel to the x-axis.

Find
(a) the equation of the straight line KM,
(b) the equation of the straight line LM,
(c) the value of k.
Solution:
(a)
The equation of the straight line KM is y = 3
(b)
Given, equation of JK:2y=4x+3y=2x+32Thus, mJK=2mLM=mJK=2y=mx+cAt M(5, 3)3=2(5)+c3=10+cc=−7∴ Equation of the straight line LMis y=2x−7.Given, equation of JK:2y=4x+3y=2x+32Thus, mJK=2mLM=mJK=2y=mx+cAt M(5, 3)3=2(5)+c3=10+cc=−7∴ Equation of the straight line LMis y=2x−7.
(c)
Substitute (k,0) into y=2x−70=2(k)−77=2kk=72