JEE Main 2026 — Straight Lines Question with Solution

The given line is $x+y=0$, which has a slope of $m_1=-1$.

Let the slope of an incident ray be $m$. Since it makes an angle of $45^{\circ }$ with the line $x+y=0$, we have:
$\left|\frac{m-(-1)}{1+m(-1)}\right|=\tan 45^{\circ }=1$
$\Rightarrow \frac{m+1}{1-m}=\pm 1$

Taking $+1$: $m+1=1-m\Rightarrow 2m=0\Rightarrow m=0$
Taking $-1$: $m+1=-1+m\Rightarrow 1=-1$ (which is not possible, implying $m\to \infty$).

The incident rays pass through $P(-1,-1)$. Thus, their equations are:
Ray 1: $y=-1$
Ray 2: $x=-1$

Next, we find the points of intersection of these incident rays with the mirror $x+2y=1$.
For Ray 1 ($y=-1$): $x+2(-1)=1\Rightarrow x=3$. The point of incidence is $A(3,-1)$.
For Ray 2 ($x=-1$): $-1+2y=1\Rightarrow 2y=2\Rightarrow y=1$. The point of incidence is $B(-1,1)$.

The reflected rays will appear to originate from the image of $P(-1,-1)$ in the mirror $x+2y-1=0$. Let the image be $P^{\prime }(x^{\prime },y^{\prime })$.
$\frac{x^{\prime }-(-1)}{1}=\frac{y^{\prime }-(-1)}{2}=-2\frac{1(-1)+2(-1)-1}{1^2+2^2}$
$\frac{x^{\prime }+1}{1}=\frac{y^{\prime }+1}{2}=-2\left(\frac{-4}{5}\right)=\frac{8}{5}$
$\Rightarrow x^{\prime }=\frac{8}{5}-1=\frac{3}{5}$ and $y^{\prime }=\frac{16}{5}-1=\frac{11}{5}$
So, $P^{\prime }=\left(\frac{3}{5},\frac{11}{5}\right)$.

The reflected rays pass through $P^{\prime }$ and their respective points of incidence.

Equation of Reflected Ray 1 (passing through $A$ and $P^{\prime }$):
Slope $m_1^{\prime }=\frac{\frac{11}{5}-(-1)}{\frac{3}{5}-3}=\frac{\frac{16}{5}}{-\frac{12}{5}}=-\frac{4}{3}$
$y-(-1)=-\frac{4}{3}(x-3)\Rightarrow 3y+3=-4x+12\Rightarrow 4x+3y=9$
Comparing with $ax+by=9$, we get $a=4,b=3$.

Equation of Reflected Ray 2 (passing through $B$ and $P^{\prime }$):
Slope $m_2^{\prime }=\frac{\frac{11}{5}-1}{\frac{3}{5}-(-1)}=\frac{\frac{6}{5}}{\frac{8}{5}}=\frac{3}{4}$
$y-1=\frac{3}{4}(x-(-1))\Rightarrow 4y-4=3x+3\Rightarrow 3x-4y=-7\Rightarrow -3x+4y=7$
Comparing with $cx+dy=7$, we get $c=-3,d=4$.

Finally, we calculate $ad+bc$:
$ad+bc=(4)(4)+(3)(-3)=16-9=7$

Answer: $7$