Question 11 (5 marks):
Diagram 4 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)
$\begin{array}{l}2y=3x+15\\ y=\frac{3}{2}x+\frac{15}{2}\\ \text{Thus }m=\frac{3}{2}\\ \text{At point }\left(3,0\right),y_1=m{x}_1+c\\ 0=\frac{3}{2}\left(3\right)+c\\ \frac{9}{2}+c=0\\ c=-\frac{9}{2}\\ \\ \text{Thus, the linear equation is}\\ y=\frac{3}{2}x-\frac{9}{2}\\ 2y=3x-9\end{array}$

Question 12 (6 marks):
Diagram 6 shows two parallel straight lines, JK and LM, drawn on a Cartesian plane.
The straight line KM is parallel to the x-axis.

Diagram 6

Find
(a) the equation of the straight line KM,
(b) the equation of the straight line LM,
(c) the value of k.



Solution:
(a)
The equation of the straight line KM is y = 3

(b)
$\begin{array}{l}\text{Given, equation of }JK:\\ 2y=4x+3\\ y=2x+\frac{3}{2}\\ \text{Thus, }{m}_{JK}=2\\ m_{LM}=m_{JK}=2\\ \\ y=mx+c\\ \text{At }M\left(5,3\right)\\ 3=2\left(5\right)+c\\ 3=10+c\\ c=-7\\ \\ \therefore \text{ Equation of the straight line }LM\\ \text{is }y=2x-7.\end{array}$


(c)
$\begin{array}{l}\text{Substitute }\left(k,0\right)\text{ into }y=2x-7\\ 0=2\left(k\right)-7\\ 7=2k\\ k=\frac{7}{2}\end{array}$

Question 9:


The diagram above shows that the straight line PMQ intersects with PNR at N. Given that OQ = OR and M is the midpoint of PQ. Find
(a) the coordinate of P
(b) the value of m and n.

Solution:
$\begin{array}{l}\text{(a)}\\ \text{Given }M\text{ is midpoint of }PQ\\ x\text{-coordinate of }P=0\\ \text{For }y\text{-coordinate of }P,\\ \frac{y+0}{2}=3\\ y=6\\ \text{Coordinates of }P=\left(0,6\right).\end{array}$

$\begin{array}{l}\text{(b)}\\ \frac{0+4}{2}=m\\ 2m=4\\ m=2\\ \\ \text{Gradient of }PR\\ =\frac{6-0}{0-\left(-4\right)}\\ =\frac{6}{4}\\ =\frac{3}{2}\\ \\ \text{At point }N\left(n,4\right)\text{, }\\ \text{using }{y}_1=m{x}_1+c\\ 4=\frac{3}{2}n+6\\ 8=3n+12\\ 3n=-4\\ n=-\frac{4}{3}\end{array}$

Question 10 (5 marks):
Diagram 6 shows two parallel straight lines, POQ and RMS drawn on a Cartesian plane.

Diagram 6

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 7:


The diagram above shows a parallelogram on a Cartesian plane. MP and NO are parallel to the y-axis. Given that the distance of MZ is 4 units. Find
(a) the value of p and q.
(b) the equation of the straight line MN.

Solution:
$\begin{array}{l}\text{(a)}\\ \text{Line }NO\text{ is parallel to }y\text{-axis,}\\ p=2\\ \\ MP=\sqrt{3^2+4^2}\\ =\sqrt{9+16}\\ =\sqrt{25}\\ =5\\ NO=MP=5\text{ units}\\ q=7-5=2\end{array}$


$\begin{array}{l}\text{(b)}\\ \text{Point }O=\left(2,2\right)\\ \text{Gradient }PO=\frac{2-0}{2-0}=1\\ \\ \text{Gradient }MN=g\text{radient }PO=1\\ y_1=m{x}_1+c\\ 7=1\left(2\right)+c\\ c=5\\ \\ \text{Equation of line }MN\text{ is}\\ y=x+5\end{array}$

Question 8:


The diagram above shows that two straight line intersect at point (0 , -2). Find
(a) the value of b
(b) the x-intercept of the straight line XY if the gradient of XY is equal to 2.
(c) the equation of XY.

Solution:
(a)
Value of b
= 2 units + 3 units
= 5 units
= –5

$\begin{array}{l}(b)\\ \text{Given }m=2,c=-2\\ \text{For }x\text{-intercept, }y=0\\ 0=2x+\left(-2\right)\\ 0=2x-2\\ 2x=2\\ x=1\\ \text{Therefore }x\text{ intercept of }XY=1.\end{array}$

$\begin{array}{l}\text{(c)}\\ \text{Substitute }m=2\text{ and }(0,-2)\text{ into }y=mx+c\\ y=2x+\left(-2\right)\\ y=2x-2\\ \text{Therefore equation of }XY:y=2x-2\end{array}$