Question

Point $A(3,4)$ is reflected about the line $x-y=0$. The reflected point is $B$. Point $B$ translates a distance equal to $OB$ along the line $L$, away from origin, upto point $C$. The line $y=x$ is acute angle bisector of $OA$ and $L$. Find the ratio of area of triangle $OAB$ to the area of triangle $BNC$, where $N$ is the mid point of $AB$.

Answer 1

Answer: 0002

$B(4,3)$
$\frac{x_C-4}{4/5}=\frac{y_C-3}{3/5}=5$
$x_C=8,y_C=6$
$\therefore C(8,6)$

Area of $△OAB=\frac{1}{2}\left|\begin{array}{ccc}3& 4& 1\\ 0& 0& 1\\ 4& 3& 1\end{array}\right|=\frac{7}{2}$
Area of $△BNC=\frac{1}{2}\left|\begin{array}{ccc}\frac{7}{2}& \frac{7}{2}& 1\\ 4& 3& 1\\ 8& 6& 1\end{array}\right|$

$=\frac{1}{2}\left[\frac{7}{2}(-3)-\frac{7}{2}(-4)+24-24\right]=\frac{1}{2}\left(\frac{-21}{2}+\frac{28}{2}\right)=\frac{7}{4}.$
$\therefore \frac{\text{ Area of }△OAB}{\text{ Area of }△BNC}=\frac{7/2}{7/4}=2.$