JEE Main 2026 — 5 April, Evening Shift — Question 45

$\tan 45^{\circ}=\left|\frac{\mathrm{m}+1}{1-\mathrm{m}}\right|$ $\Rightarrow \pm 1=\frac{\mathrm{m}+1}{1-\mathrm{m}}$ $\Rightarrow \mathrm{m}=0$ & one line is perpendicular to x axis. $\mathrm{L}_{1}: \mathrm{x}=-1, \mathrm{~L}_{2}: \mathrm{y}=-1$ Now taking mirror image of $x+1=0$ in $x+2 y=1$ $\mathrm{A}^{\prime}: \frac{\mathrm{x}+1}{1}=\frac{\mathrm{y}+1}{2}=\frac{-2(-1-2-1)}{1^{2}+2^{2}}$ $\mathrm{x}=\frac{3}{5}, \mathrm{y}=\frac{11}{5} \mathrm{~A}^{\prime}\left(\frac{3}{5}, \frac{11}{5}\right)$ line $=3 x-4 y+7=0$ Now taking mirror image of $\mathrm{y}+1=0$ in $\mathrm{x}+2 \mathrm{y}=1$ $\mathrm{A}^{\prime \prime}: \frac{x+1}{1}=\frac{y+1}{2}=\frac{-2(-1-2-1)}{1^{2}+2^{2}}$ $\mathrm{x}=\frac{3}{5}, \mathrm{y}=\frac{11}{5}$ A" $\left(\frac{3}{5}, \frac{11}{5}\right)$ line : $(y+1)=\frac{\left(\frac{11}{5}+1\right)}{\left(\frac{3}{5}-3\right)}(x-3)$ $4 \mathrm{x}+3 \mathrm{y}=9$ comparing $4 \mathrm{x}+3 \mathrm{y}=9$ with $\mathrm{ax}+$ by $=9 \&$ $\mathrm{a}=4, \mathrm{~b}=3$ by line $-3 x+4 y=7$ comparing with $\mathrm{cx}+\mathrm{dy}=7$ $\therefore \mathrm{c}=-3, \mathrm{~d}=4$ So, $\mathrm{ad}+\mathrm{bc}$ $=4 \times 4+3 \times-3$ $=16-9=7$