SPM Mathematics 2018, Paper 2 (Questions 7 & 8)

Question 7 (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 8 (6 marks):
$\begin{array}{l}\text{Given }A=\left(\begin{array}{l}4-2\\ 3-1\end{array}\right),B=m\left(\begin{array}{l}-1n\\ -34\end{array}\right)\\ \text{and }I=\left(\begin{array}{l}\text{1 0}\\ \text{0 }1\end{array}\right).\end{array}$
(a) Find the value of m and of n if AB = I.
(b) Write the following simultaneous linear equation as matrix equation:
4x – 2y = 3
3x – y = 2
Hence, by using matrix method, calculate the value of x and of y.

Solution:
(a)
$\begin{array}{l}\text{If }AB=I,\text{ then }B=A^{-1}\\ A^{-1}=\frac{1}{4\left(-1\right)-\left(-2\right)\left(3\right)}\left(\begin{array}{l}-12\\ -3\text{ 4}\end{array}\right)\\ A^{-1}=\frac{1}{2}\left(\begin{array}{l}-12\\ -3\text{ 4}\end{array}\right)\\ \\ \text{By comparison:}\\ B=A^{-1}\\ m\left(\begin{array}{l}-1n\\ -34\end{array}\right)=\frac{1}{2}\left(\begin{array}{l}-12\\ -3\text{ 4}\end{array}\right)\\ m=\frac{1}{2};n=2\end{array}$


(b)
$\begin{array}{l}4x-2y=3\\ 3x-y=2\\ \left(\begin{array}{l}4-2\\ 3-1\end{array}\right)\left(\begin{array}{l}x\\ y\end{array}\right)=\left(\begin{array}{l}3\\ 2\end{array}\right)\\ \left(\begin{array}{l}x\\ y\end{array}\right)=\frac{1}{2}\left(\begin{array}{l}-12\\ -3\text{ 4}\end{array}\right)\left(\begin{array}{l}3\\ 2\end{array}\right)\\ \left(\begin{array}{l}x\\ y\end{array}\right)=\frac{1}{2}\left(\begin{array}{l}\left(-1\right)\left(3\right)\text{+}\left(2\right)\left(2\right)\\ \left(-3\right)\left(3\right)\text{+}\left(4\right)\left(2\right)\end{array}\right)\\ \left(\begin{array}{l}x\\ y\end{array}\right)=\frac{1}{2}\left(\begin{array}{l}1\\ -1\end{array}\right)\\ \left(\begin{array}{l}x\\ y\end{array}\right)=\left(\begin{array}{l}\frac{1}{2}\\ -\frac{1}{2}\end{array}\right)\\ x=\frac{1}{2}\text{ and }y=-\frac{1}{2}\end{array}$