Question

Let a ray of light passing through the point $(3, 10)$ reflects on the line $2x + y = 6$ and the reflected ray passes through the point $(7, 2)$. If the equation of the incident ray is $ax + by + 1 = 0$, then $a^2 + b^2 + 3ab$ is equal to _.

Answer 1

Given: For $B'$:

$\frac{x - 7}{2} = \frac{y - 2}{1} = -2 \left( \frac{14 + 2 - 6}{5} \right)$

$\frac{x - 7}{2} = \frac{y - 2}{1} = -4$

$x = -1, \quad y = -2 \implies B'(-1, -2)$

Incident ray $AB'$:

$M_{AB'} = 3$

$y + 2 = 3(x + 1)$

$3x - y + 1 = 0$

$a = 3, \quad b = -1$

$a^2 + b^2 + 3ab = 9 + 1 - 9 = 1$

Answer: 1