JEE Main · 2023 · 11 Apr 2023 · Shift 2 · Straight Lines

The lines ${\ell }_1$ and ${\ell }_2$ intersect at point P. Solving:
$3y-2x=3$ and $x-y+1=0$ gives $P=(0,1)$.

Since ${\ell }_3$ passes through P: $\alpha (0)+\beta (1)+17=0$, so $\beta =-17$.

Take a point $Q=(-1,0)$ on ${\ell }_2$. The image of Q about the angular bisector ${\ell }_1$ lies on ${\ell }_3$.

Using the reflection formula for point $(-1,0)$ about line $2x-3y+3=0$:
$\frac{x+1}{2}=\frac{y-0}{-3}=-2\cdot \frac{2(-1)-3(0)+3}{13}=-\frac{2}{13}$

This gives $Q^{\prime }=\left(-\frac{17}{13},\frac{6}{13}\right)$.

Since $Q^{\prime }$ lies on ${\ell }_3$: $\alpha \left(-\frac{17}{13}\right)+(-17)\left(\frac{6}{13}\right)+17=0$

Solving: $-17\alpha -102+221=0$, giving $\alpha =7$.

Therefore: ${\alpha }^2+{\beta }^2-\alpha -\beta =49+289-7+17=348$.