The equation to the straight line passing through the point of intersection of the lines $5 x - 6 y - 1 = 0$ and $3 x + 2 y + 5 = 0$ and perpendicular to the line $3 x - 5 y + 11 = 0$ is?

5 x + 3 y + 8 = 0

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5 x + 3 y - 8 = 0

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5 x - 3 y + 8 = 0

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5 x + 9 y + 8 = 0

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Solution

The correct option is B $5 x + 3 y + 8 = 0$
$5 x - 6 y - 1 = 0 => x = \frac{6 y + 1}{5} 3 x + 2 y + 5 = 0 => 3 (\frac{6 y + 1}{5}) + 2 y + 5 = 0 => 18 y + 3 + 10 y + 25 = 0 => y = - 1 x = \frac{6 y + 1}{5} = \frac{6 \times (- 1) + 1}{5} = - 1$
Point of intersection $(- 1, - 1)$
Required line perpendicular to the line $3 x - 5 y + 11 = 0$ is
$5 x + 3 y + c = 0$ where $c$ is a constant
$5 (- 1) + 3 (- 1) + c = 0 => - 5 - 3 + c = 0 => c = 8$
Required line is $5 x + 3 y + 8 = 0$

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Similar questions

5 x - 6 y - 1 = 0, 3 x + 2 y + 5 = 0

3 x - 5 y + 11 = 0

The equation of the line passing through $(1, 5)$ and perpendicular to the line $3 x - 5 y + 7 = 0$ is

3 x + 5 y = 19

(3, 2)

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