Find the equation of the line through point (3,1) and is perpendicular to the line x + 5y + 5 = 0

5 x - 2 y = 14

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5 x - y = 14

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2 x - 5 y = 14

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2 x + 5 y = 14

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none of these

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Solution

The correct option is B $5 x - y = 14$
The slope of line $x + 5 y + 5 = 0 i . e y = - \frac{1}{5} x - 1$ is $m = - \frac{1}{5}$
The slope of the line perpendicular to line $x + 5 y + 5 = 0$ is $m^{'} = - \frac{1}{m} = 5$
The equation of line having slope m and passing through the point $(a, b)$ is given by $y - b = m (x - a)$
Therefore, the equation of line having slope $m^{'} = 5$ and passing through the point $(3, 1)$ is given by $y - 1 = 5 (x - 3)$
$\Rightarrow y - 1 = 5 x - 15$
$\Rightarrow 5 x - y = 14$
Option B is correct.

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