Matematik Tambahan SPM 2017, Kertas 2 (Soalan 8)

Soalan 8 (10 markah):
Rajah 5 menunjukkan segi tiga ABC. Garis lurus AE bersilang dengan garis lurus BC di titik D. Titik V terletak pada garis lurus AE.

 

$\begin{array}{l}\text{Diberi bahawa }\overrightarrow{BD}=\frac{1}{3}\overrightarrow{BC},\overrightarrow{AC}=6\underset{˜}{x}\text{ dan }\overrightarrow{AB}=9\underset{˜}{y}.\\ \left(a\right)\text{ Ungkapkan dalam sebutan }\underset{˜}{x}\text{ dan / atau }\underset{˜}{y}:\\ \left(i\right)\overrightarrow{BC},\\ \left(ii\right)\overrightarrow{AD}.\\ \left(b\right)\text{ Diberi bahawa }\overrightarrow{AV}=m\overrightarrow{AD}\text{ dan }\overrightarrow{BV}=n\left(\underset{˜}{x}-9\underset{˜}{y}\right),\\ \text{ dengan keadaan }m\text{ dan }n\text{ ialah pemalar}\text{.}\\ \text{ Cari nilai }m\text{ dan nilai }n.\\ \left(c\right)\text{ Diberi }\overrightarrow{AE}=h\underset{˜}{x}+9\underset{˜}{y},\text{ dengan keadaan }h\text{ ialah pemalar, }\\ \text{ cari nilai }h.\end{array}$

 

Penyelesaian: 
(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)

$\begin{array}{l}\text{Diberi }\overrightarrow{AV}=m\overrightarrow{AD}\\ =m\left(2\underset{˜}{x}+6\underset{˜}{y}\right)\\ =2m\underset{˜}{x}+6m\underset{˜}{y}\\ \\ \overrightarrow{AV}=\overrightarrow{AB}+\overrightarrow{BV}\\ \text{ = }9\underset{˜}{y}+n\left(\underset{˜}{x}-9\underset{˜}{y}\right)\\ =9\underset{˜}{y}+n\underset{˜}{x}-9n\underset{˜}{y}\\ =n\underset{˜}{x}+\left(9-9n\right)\underset{˜}{y}\\ \\ \text{Dengan menyamakan pekali bagi }\underset{˜}{x}\text{ dan }\underset{˜}{y}\text{, }\\ 2m\underset{˜}{x}+6m\underset{˜}{y}=n\underset{˜}{x}+\left(9-9n\right)\underset{˜}{y}\\ 2m=n\\ n=2m\mathrm{\dots \dots \dots \dots .}\left(1\right)\\ 6m=9-9n\mathrm{\dots \dots \dots \dots .}\left(2\right)\\ \text{Gantikan (1) ke dalam (2),}\\ 6m=9-9\left(2m\right)\\ 6m=9-18m\\ 24m=9\\ m=\frac{9}{24}=\frac{3}{8}\\ \\ \text{Daripada }\left(1\right):\\ n=2\left(\frac{3}{8}\right)=\frac{3}{4}\end{array}$

 

(c)

$\begin{array}{l}A,D\text{ dan }E\text{ adalah segaris}.\\ \overrightarrow{AD}=k\left(\overrightarrow{AE}\right)\\ \overrightarrow{AD}=k\left(h\underset{˜}{x}+9\underset{˜}{y}\right)\\ 2\underset{˜}{x}+6\underset{˜}{y}=kh\underset{˜}{x}+9k\underset{˜}{y}\\ \\ \text{Dengan menyamakan pekali bagi }\underset{˜}{y}:\\ 9k=6\\ k=\frac{6}{9}\\ k=\frac{2}{3}\\ \\ \text{Dengan menyamakan pekali bagi }\underset{˜}{x}:\\ kh=2\\ \left(\frac{2}{3}\right)h=2\\ h=2\times \frac{3}{2}\\ h=3\end{array}$