9.3.1 Straight Lines, PT3 Focus Practice


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

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
 

 

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