Long Question 8

Question 8:
Diagram below shows quadrilateral OPQR. The straight line PR intersects the straight line OQ at point S.


$\begin{array}{l}\text{It is given that }\overrightarrow{OP}=7\underset{˜}{x},\overrightarrow{OR}=5\underset{˜}{y},PS:SR=3:1\text{ and }\overrightarrow{OR}\text{ is parallel to }\overrightarrow{PQ}.\\ \text{(a) Express in terms of }\underset{˜}{x}\text{ and }\underset{˜}{y},\\ \text{(i) }\overrightarrow{PR}\\ \text{(ii) }\overrightarrow{OS}\\ \\ \text{(b) Using }\overrightarrow{PQ}=m\overrightarrow{OR}\text{ and }\overrightarrow{SQ}=n\overrightarrow{OS},\text{ where }m\text{ and }n\text{ are constants,}\\ \text{ Find the value of }m\text{ and of }n.\\ \\ \text{(c) Given that }\left|\underset{˜}{y}\right|=4\text{ units and the area of }ORS{\text{ is 50 cm}}^2\text{, find the}\\ \text{ perpendicular distance from point }S\text{ to }OR\text{.}\end{array}$


Solution:
(a)(i)
$\begin{array}{l}\overrightarrow{PR}=\overrightarrow{PO}+\overrightarrow{OR}\\ =-7\underset{˜}{x}+5\underset{˜}{y}\end{array}$


(a)(ii)
$\begin{array}{l}\overrightarrow{OS}=\overrightarrow{OP}+\overrightarrow{PS}\\ =7\underset{˜}{x}+\frac{3}{4}\overrightarrow{PR}\\ =7\underset{˜}{x}+\frac{3}{4}\left(-7\underset{˜}{x}+5\underset{˜}{y}\right)\\ =7\underset{˜}{x}-\frac{21}{4}\underset{˜}{x}+\frac{15}{4}\underset{˜}{y}\\ =\frac{7}{4}\underset{˜}{x}+\frac{15}{4}\underset{˜}{y}\end{array}$


(b)
$\begin{array}{l}\overrightarrow{PS}=\overrightarrow{PQ}-\overrightarrow{SQ}\\ \frac{3}{4}\overrightarrow{PR}=m\overrightarrow{OR}-n\overrightarrow{OS}\\ \frac{3}{4}\left(-7\underset{˜}{x}+5\underset{˜}{y}\right)=m\left(5\underset{˜}{y}\right)-n\left(\frac{7}{4}\underset{˜}{x}+\frac{15}{4}\underset{˜}{y}\right)\\ -\frac{21}{4}\underset{˜}{x}+\frac{15}{4}\underset{˜}{y}=5m\underset{˜}{y}-\frac{7}{4}n\underset{˜}{x}-\frac{15}{4}m\underset{˜}{y}\\ -\frac{21}{4}\underset{˜}{x}+\frac{15}{4}\underset{˜}{y}=-\frac{7}{4}n\underset{˜}{x}+\frac{5}{4}m\underset{˜}{y}\\ -21\underset{˜}{x}+15\underset{˜}{y}=-7n\underset{˜}{x}+5m\underset{˜}{y}\\ 7n=21\\ n=3\\ 5m=15\\ m=3\end{array}$


(c)
$\begin{array}{l}\text{Area of }\Delta ORS=50\\ \frac{1}{2}\times \left(5\underset{˜}{y}\right)\times t=50\\ \frac{1}{2}\times 5\left(4\right)\times t=50\\ 10t=50\\ t=5\\ \therefore \text{ Perpendicular distance from point }S\text{ to }OR=5\text{ units}\text{.}\end{array}$