The Straight Line Long Questions (Question 1 - 3)


Question 1:
In diagram below, ABCD is a trapezium drawn on a Cartesian plane. BC is parallel to AD and O is the origin. The equation of the straight line BC is
3y = kx + 7 and the equation of the straight line AD is   y= 1 2 x+3

Find
(a) the value of k,
(b) the x-intercept of the straight line BC.

Solution:
(a)
Equation of BC:
3y = kx + 7
y= k 3 x+ 7 3 Gradient of BC= k 3 Equation of AD: y= 1 2 x+3 Gradient of AD= 1 2

Gradient of BC = gradient of AD
k 3 = 1 2 k= 3 2

(b) Equation of BC 3y= 3 2 x+7
For x-intercept, y = 0
3(0)= 3 2 x+7 3 2 x=7 x= 14 3
Therefore x-intercept of BC 14 3

Question 2:
In diagram below, O is the origin. Straight line MN is parallel to a straight line OK.
Find
(a) the equation of the straight line MN,
(b) the x-intercept of the straight line MN.

Solution:
(a) Gradient of MN = gradient of OK
Gradient of MN
= 50 30 = 5 3

Substitute m = 5/3 and (–2, 5) into y = mx + c
5= 5 3 ( 2 )+c
15 = – 10 + 3c
3c = 25
c = 25/3

Therefore equation of MN y= 5 3 x+ 25 3

(b) 
For x-intercept, y = 0
0= 5 3 x+ 25 3 5 3 x= 25 3
5x = –25
x = –5
Therefore x-intercept of MN = –5


Question 3:
Diagram below shows a straight line JK and a straight line ST drawn on a Cartesian plane. JK is parallel to ST.
Find
(a) the equation of the straight ST,
(b) the x-intercept of the straight line ST.

Solution:
(a) JK is parallel to ST, therefore gradient of JK = gradient of ST.
= 80 04 =2

Substitute m = –2 and S(5, 6) into y = mx + c
6 = –2 (5) + c
c = 16
Therefore equation of ST: y = –2x + 16

(b) For x-intercept, y = 0
0 = –2x + 16
2x = 16
x = 8
Therefore x-intercept of ST = 8