Long Question 10

Question 10 (10 marks):
Diagram shows a triangle ABC. The straight line AE intersects with the straight line BC at point D. Point V lies on the straight line AE.

$\begin{array}{l}\text{It is given that }\overrightarrow{BD}=\frac{1}{3}\overrightarrow{BC},\overrightarrow{AC}=6\underset{˜}{x}\text{ and }\overrightarrow{AB}=9\underset{˜}{y}.\\ \left(a\right)\text{ Express in terms of }\underset{˜}{x}\text{ and / or }\underset{˜}{y}:\\ \left(i\right)\overrightarrow{BC},\\ \left(ii\right)\overrightarrow{AD}.\\ \left(b\right)\text{ It is given that }\overrightarrow{AV}=m\overrightarrow{AD}\text{ and }\overrightarrow{BV}=n\left(\underset{˜}{x}-9\underset{˜}{y}\right),\text{ where }m\text{ and }n\text{ are constants}\text{.}\\ \text{ Find the value of }m\text{ and of }n.\\ \left(c\right)\text{ Given }\overrightarrow{AE}=h\underset{˜}{x}+9\underset{˜}{y},\text{ where }h\text{ is a constant, find the value of }h.\end{array}$

Solution: 
(a)(i)
$\begin{array}{l}\overrightarrow{BC}=\overrightarrow{BA}+\overrightarrow{AC}\\ =-9\underset{˜}{y}+6\underset{˜}{x}\\ =6\underset{˜}{x}-9\underset{˜}{y}\end{array}$

(a)(ii)
$\begin{array}{l}\overrightarrow{AD}=\overrightarrow{AB}+\overrightarrow{BD}\\ =9\underset{˜}{y}+\frac{1}{3}\overrightarrow{BC}\\ =9\underset{˜}{y}+\frac{1}{3}\left(6\underset{˜}{x}-9\underset{˜}{y}\right)\\ =9\underset{˜}{y}+2\underset{˜}{x}-3\underset{˜}{y}\\ =2\underset{˜}{x}+6\underset{˜}{y}\end{array}$


(b)
$\begin{array}{l}\text{Given }\overrightarrow{AV}=m\overrightarrow{AD}\\ =m\left(2\underset{˜}{x}+6\underset{˜}{y}\right)\\ =2m\underset{˜}{x}+6m\underset{˜}{y}\\ \\ \overrightarrow{AV}=\overrightarrow{AB}+\overrightarrow{BV}\\ \text{ = }9\underset{˜}{y}+n\left(\underset{˜}{x}-9\underset{˜}{y}\right)\\ =9\underset{˜}{y}+n\underset{˜}{x}-9n\underset{˜}{y}\\ =n\underset{˜}{x}+\left(9-9n\right)\underset{˜}{y}\\ \\ \text{By equating the coefficients of }\underset{˜}{x}\text{ and }\underset{˜}{y}\text{, }\\ 2m\underset{˜}{x}+6m\underset{˜}{y}=n\underset{˜}{x}+\left(9-9n\right)\underset{˜}{y}\\ 2m=n\\ n=2m\mathrm{.............}\left(1\right)\\ 6m=9-9n\mathrm{.............}\left(2\right)\\ \text{Substitute (1) into (2),}\\ 6m=9-9\left(2m\right)\\ 6m=9-18m\\ 24m=9\\ m=\frac{9}{24}=\frac{3}{8}\\ \\ \text{From }\left(1\right):\\ n=2\left(\frac{3}{8}\right)=\frac{3}{4}\end{array}$


(c)
$\begin{array}{l}A,D\text{ and }E\text{ are collinear}.\\ \overrightarrow{AD}=k\left(\overrightarrow{AE}\right)\\ \overrightarrow{AD}=k\left(h\underset{˜}{x}+9\underset{˜}{y}\right)\\ 2\underset{˜}{x}+6\underset{˜}{y}=kh\underset{˜}{x}+9k\underset{˜}{y}\\ \\ \text{Equating the coefficients of }\underset{˜}{y}:\\ 9k=6\\ k=\frac{6}{9}\\ k=\frac{2}{3}\\ \\ \text{Equating the coefficients of }\underset{˜}{x}:\\ kh=2\\ \left(\frac{2}{3}\right)h=2\\ h=2\times \frac{3}{2}\\ h=3\end{array}$