9.3.6 Straight Lines, PT3 Focus Practice

Question 11:
Diagram 11 shows two parallel straight lines, POQ and RMS drawn on a Cartesian plane.

Diagram 11

Find
(a) the equation of the straight line RMS,
(b) the x-intercept of the straight line MN.


Solution:
(a)
$\begin{array}{l}{m}_{RMS}=m_{POQ}\\ =\frac{1-0}{-2-0}\\ =-\frac{1}{2}\\ \\ \text{At point }\left(4,6\right)\\ y_1=m{x}_1+c\\ 6=-\frac{1}{2}\left(4\right)+c\\ 6=-2+c\\ c=8\\ \\ \text{The equation of straight line }RMS\text{ is}\\ y=-\frac{1}{2}x+8\end{array}$


(b)
$\begin{array}{l}\text{When }y=0,\\ -\frac{1}{2}x+8=0\\ \frac{1}{2}x=8\\ x=16\\ \\ x\text{-intercept of straight line }MN\text{ is }16.\end{array}$

Question 12:
Diagram 12 shows a parallelogram drawn on a Cartesian plane which represents the locations of Rahman’s house, a cinema, a school and a shop.

It is given that the scale is 1 unit : 1 km.
(a) Calculate the distance, in km, between Rahman’s house and the school.
(b) Find the equation of the straight line that links the school to the cinema.


Solution:
(a)
2y = 3x + 15
When y = 0
3x + 15 = 0
3x = –15
x = –5

Rahman’s house = (–5, 0)
School = (3, 0)

Distance, between Rahman’s house and the school
= 3 – (– 5)
= 8 units
= 8 km

(b)
$\begin{array}{l}2y=3x+15\\ y=\frac{3}{2}x+\frac{15}{2}\\ \text{Thus }m=\frac{3}{2}\\ \text{At point }\left(3,0\right),y_1=m{x}_1+c\\ 0=\frac{3}{2}\left(3\right)+c\\ \frac{9}{2}+c=0\\ c=-\frac{9}{2}\\ \\ \text{Thus, the linear equation is}\\ y=\frac{3}{2}x-\frac{9}{2}\\ 2y=3x-9\end{array}$