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 $O B$ 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 $A B$.

Answer 1

$B(4, 3)$
$\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 $\triangle OAB =\frac{1}{2}\begin{vmatrix}3 & 4 & 1 \\ 0 & 0 & 1 \\ 4 & 3 & 1\end{vmatrix}=\frac{7}{2}$
Area of $\triangle B N C=\frac{1}{2}\begin{vmatrix}\frac{7}{2} & \frac{7}{2} & 1 \\ 4 & 3 & 1 \\ 8 & 6 & 1\end{vmatrix}$

$=\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} . $
$\therefore \frac{\text { Area of } \triangle OAB }{\text { Area of } \triangle BNC }=\frac{7 / 2}{7 / 4}=2 .$