A light ray is incident to a line mirror ' L ' along the line 3 x-4 y+5=0 and is reflected along the line 5 x-12 y+19=0. The equation of line 'L' may be

Question

A light ray is incident to a line mirror ' L ' along the line 3 x-4 y+5=0 and is reflected along the line 5 x-12 y+19=0. The equation of line 'L' may be (A) 7 x-4 y+1=0 (B) 7 x+4 y-15=0 (C) 4 x-7 y+10=0 (D) 4 x+7 y-18=0

Tardigrade Admin · Accepted Answer

(3 x-4 y+5/5)= ± (5 x-12 y+19/13) ⇒ 7 x+4 y-15=0 and 4 x-7 y+10=0

Q. A light ray is incident to a line mirror ' $L$ ' along the line $3 x - 4 y + 5 = 0$ and is reflected along the line $5 x - 12 y + 19 = 0$ . The equation of line 'L' may be

$7 x - 4 y + 1 = 0$

$7 x + 4 y - 15 = 0$

$4 x - 7 y + 10 = 0$

$4 x + 7 y - 18 = 0$

$\frac{3 x - 4 y + 5}{5} = \pm \frac{5 x - 12 y + 19}{13}$
$\Rightarrow 7 x + 4 y - 15 = 0$ and $4 x - 7 y + 10 = 0$

Q. A light ray is incident to a line mirror ' L ' along the line 3x−4y+5=0 and is reflected along the line 5x−12y+19=0. The equation of line 'L' may be

7x−4y+1=0

7x+4y−15=0

4x−7y+10=0

4x+7y−18=0