The Straight Line Long Questions (Question 1 – 3)


Question 1:
In diagram below, ABCDis a trapezium drawn on a Cartesian plane. BCis 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 B C = k 3 Equation of A D : y = 1 2 x + 3 Gradient of A D = 1 2

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

(b)
Equation of BC 3 y = 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, 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
= 5 0 3 0 = 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.
= 8 0 0 4 = 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

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