Question 2:
Solution:
(a)
(b)
QR:RT=1:3Lets coordinates of T=(x, y)((1)(x)+(3)(4)1+3,(1)(y)+(3)(5)1+3)=(5, 4)x+124=5x+12=20x=8y+154=4y+15=16y=1T=(8, 1)
(c)
Gradient of PQ=5−24−3=3Equation of PQ,y−2=3(x−3)y−2=3x−9y=3x−7−−−−(1)Gradient of ST=6−17−8=−5Equation of ST,y−1=−5(x−8)y−1=−5x+40y=−5x+41−−−−(2)
Substitute (1) into (2),
In the diagram, PRS and QRT are straight lines. Given R is the midpoint of PS and
QR : RT = 1 : 3, Find
QR : RT = 1 : 3, Find
(a) the coordinates of R,
(b) the coordinates of T,
(c) the coordinates of the point of intersection between lines PQ and ST produced.Solution:
(a)
Given R is the midpoint of PS.
R=(3+72,2+62)R=(5, 4)
(b)
QR:RT=1:3Lets coordinates of T=(x, y)((1)(x)+(3)(4)1+3,(1)(y)+(3)(5)1+3)=(5, 4)x+124=5x+12=20x=8y+154=4y+15=16y=1T=(8, 1)
(c)
Gradient of PQ=5−24−3=3Equation of PQ,y−2=3(x−3)y−2=3x−9y=3x−7−−−−(1)Gradient of ST=6−17−8=−5Equation of ST,y−1=−5(x−8)y−1=−5x+40y=−5x+41−−−−(2)
Substitute (1) into (2),
3x – 7 = –5x + 41
8x = 48
x = 6
From (1),
y = 3(6) – 7 = 11
The coordinates of the point of intersection between lines PQ and ST = (6, 11).