Long Questions (Question 4)

Question 4:
Solutions by scale drawing will not be accepted.
Diagram below shows a triangle PRS. Side PR intersects the y-axis at point Q.

(a) Given PQ : QR = 2 : 3, find
(i) The coordinates of P,
(ii) The equation of the straight line PS,
(iii) The area, in unit², of triangle PRS.
(b) Point M moves such that its distance from point R is always twice its distance from point S.
Find the equation of the locus M.

Solution:
(a)(i) 
$\begin{array}{l}P=\left(\frac{2\left(6\right)+3h}{2+3},\frac{2\left(12\right)+3k}{2+3}\right)\\ \left(0,6\right)=\left(\frac{12+3h}{5},\frac{24+3k}{5}\right)\\ \frac{12+3h}{5}=0\\ 3h=-12\\ h=-4\\ \\ \frac{24+3k}{5}=6\\ \text{3}k=30-24\\ k=2\\ \\ P=\left(-4,2\right)\end{array}$

(a)(ii) 
$\begin{array}{l}{m}_{PS}=\frac{2-\left(-6\right)}{-4-2}\\ =-\frac{8}{6}\\ =-\frac{4}{3}\\ \\ \text{Equation of }PS:\\ y-y_1=-\frac{4}{3}\left(x-2\right)\\ y-\left(-6\right)=-\frac{4}{3}x+\frac{8}{3}\\ 3y+18=-4x+8\\ 3y=-4x-10\end{array}$

(a)(iii) 
$\begin{array}{l}\text{Area of }△PRS\\ =\frac{1}{2}|\begin{array}{l}-\text{4 2 6 }\\ \text{2 }-6\text{ 12 }\end{array}\begin{array}{l}-\text{4}\\ 2\end{array}|\\ =\frac{1}{2}|\left(24+24+12\right)-\left(4-36-48\right)|\\ =\frac{1}{2}|60-\left(-80\right)|\\ =70{\text{ unit}}^2\end{array}$

(b) 
$\begin{array}{l}\text{Let }P=\left(x,y\right)\\ MR=2MS\\ \sqrt{{\left(x-6\right)}^2+{\left(y-12\right)}^2}=2\sqrt{{\left(x-2\right)}^2+{\left(y+6\right)}^2}\\ {\left(x-6\right)}^2+{\left(y-12\right)}^2=4\left[{\left(x-2\right)}^2+{\left(y+6\right)}^2\right]\\ x^2-12x+36+y^2-24y+144=4\left[x^2-4x+4+y^2+12y+36\right]\\ x^2-12x+y^2-24y+180=4x^2-16x+4y^2+48y+160\\ 3x^2+3y^2-4x+72y-20=0\end{array}$