Long Questions (Question 4)

Posted on April 21, 2020 by user

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)

$\begin{array}{l} P = (\frac{2 (6) + 3 h}{2 + 3}, \frac{2 (12) + 3 k}{2 + 3}) \\ (0, 6) = (\frac{12 + 3 h}{5}, \frac{24 + 3 k}{5}) \\ \frac{12 + 3 h}{5} = 0 \\ 3 h = - 12 \\ h = - 4 \\ \frac{24 + 3 k}{5} = 6 \\ 3 k = 30 - 24 \\ k = 2 \\ P = (- 4, 2) \end{array}$

(a)(ii)

$\begin{array}{l} m_{P S} = \frac{2 - (- 6)}{- 4 - 2} \\ = - \frac{8}{6} \\ = - \frac{4}{3} \\ Equation of P S : \\ y - y_{1} = - \frac{4}{3} (x - 2) \\ y - (- 6) = - \frac{4}{3} x + \frac{8}{3} \\ 3 y + 18 = - 4 x + 8 \\ 3 y = - 4 x - 10 \end{array}$

(a)(iii)

$\begin{array}{l} Area of △ P R S \\ = \frac{1}{2} | \begin{array}{l} - 4 2 6 \\ 2 - 612 \end{array} \begin{array}{l} - 4 \\ 2 \end{array} | \\ = \frac{1}{2} | (24 + 24 + 12) - (4 - 36 - 48) | \\ = \frac{1}{2} | 60 - (- 80) | \\ = 70 {unit}^{2} \end{array}$

(b)

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