From the point (-1,-1), two rays are sent making angles of 45^ with the line x+y=0.

Question

From the point $(-1,-1)$, two rays are sent making angles of $45^\circ$ with the line $x+y=0$. These rays get reflected from the mirror $x+2y=1$. If the equations of the reflected rays are $ax+by=9$ and $cx+dy=7$, with $a,b,c,d\in\mathbb Z$, then the value of $ad+bc$ is \_\_\_\_

Answer: 7

Accepted Answer

Line $x+y=0$ has slope $-1$. A ray of slope $m$ makes $45^\circ$ with it when
$$\tan45^\circ = \left|\frac{m+1}{1-m}\right| = 1$$
which gives $m=0$ or the ray is vertical ($x=$ constant). So the two incident rays through $(-1,-1)$ are $L_1: y=-1$ and $L_2: x=-1$.

Reflect $L_1: y=-1$ in the mirror $x+2y=1$: Reflecting the point $(-1,-1)$ (which lies on $L_1$) across $x+2y-1=0$:
$$\frac{x+1}{1}=\frac{y+1}{2}=\frac{-2(-1-2-1)}{5} = \frac{8}{5}\implies \left(\frac35,\frac{11}{5}\right)$$
This is the image point; also since $L_1$ meets the mirror at a fixed point, the reflected line through the image point and the mirror-intersection works out to
$4x+3y=9$
Comparing with $ax+by=9$: $a=4,\ b=3$.

Reflect $L_2: x=-1$ in the mirror similarly: carrying out the analogous reflection gives the reflected line
$$-3x+4y=7 \implies 3x-4y=-7$$
Comparing with $cx+dy=7$ (written as $-3x+4y=7$): $c=-3,\ d=4$.

$$ad+bc = 4\times4 + 3\times(-3) = 16-9 = 7$$

Solutions · 1