Let a ray of light passing through the point (3, 10) reflects on the line 3x + 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 is equal to __________ . [2024]

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Solution

Q.
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 $a^{2} + b^{2} + 3 a b$ is equal to __________ . [2024]

Ans.
(1)

$\frac{x - 3}{2} = \frac{y - 10}{1} = \frac{- 2 (2 \times 3 + 10 - 6)}{4 + 1} = - 4$

$\Rightarrow x = - 5, y = 6$

$∴$ A'(–5, 6) and B(7, 2)

$∴$ Equation of line A'B is

$\Rightarrow y - 6 = \frac{2 - 6}{7 + 5} (x + 5)$

$\Rightarrow y - 6 = - \frac{1}{3} (x + 5)$

$\Rightarrow 3 y - 18 = - x - 5 \Rightarrow x + 3 y = 13$

and 2x + y = 6 (Given line).

On solving, we get y = 4, x = 1

$∴ Q \equiv (1, 4)$

Equation of line AQ is

$y - 10 = \frac{10 - 4}{3 - 1} (x - 3) \Rightarrow y - 10 = 3 (x - 3)$

$\Rightarrow y - 10 = 3 x - 9 \Rightarrow 3 x - y + 1 = 0$

On comparing with given equation ax + by + 1 = 0, we get a = 3, b = –1

Hence, $a^{2} + b^{2} + 3 a b = 9 + 1 + 3 (3) (- 1) = 9 + 1 - 9 = 1$ .

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