A ray of light travels along the line $2 x - 3 y + 5 = 0$ and strikes a plane mirror lying along the line $x + y = 2$ . The equation of the straight line containing the reflected ray is

2 x - 3 y + 3 = 0

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3 x - 2 y + 3 = 0

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21 x - 7 y + 1 = 0

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21 x + 7 y - 1 = 0

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Solution

The correct option is D $3 x - 2 y + 3 = 0$
We use the formula,

$L_{3} = L_{1} - \frac{2 (a_{1} a_{2} + b_{1} b_{2})}{a_{2}^{2} + b_{2}^{2}} L_{2} = 0$

where,

$L_{1} = a_{1} x + b_{1} y + c_{1} = 0$

$L_{2} = a_{2} x + b_{2} y + c_{2} = 0$

Given,

$L_{1} = 2 x - 3 y + 5 = 0$

$L_{2} = x + y - 2 = 0$

$∴ L_{3} = (2 x - 3 y + 5) - \frac{2 [2 \times 1 + (- 3) \times 1]}{1^{2} + 1^{2}} (x + y - 2) = 0$

$= (2 x - 3 y + 5) - (- 1) (x + y - 2) = 0$

$\to 2 x - 3 y + 5 + x + y - 2 = 0$

$∴ 3 x - 2 y + 3 = 0$ is the required equation.

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Similar questions

Equation of the straight line passing through the point of intersection of the lines $3 x + 4 y = 7, x - y + 2 = 0$ and having slope $3$ is

Equation of the straight line whose slope is 3 and passing through the point of intersection of the lines 3x + 4y = 7 and
x − y + 2 = 0 is ____.

x - 2 y - 3 = 0

3 x - 2 y - 5 = 0

L = 0

P = 0

B

A (2, 1, 6)

L = 0

P = 0

A^{'}

L = 0

\frac{x - 2}{3} = \frac{y - 1}{4} = \frac{z - 6}{5}

P = 0

x + y - 2 z = 3

L_{1} = 0

x - 2 y + 5 = 0

3 x - 2 y + 7 = 0,

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