Equation of the line passing through point of intersection of lines $3 x + 11 y - 7 = 0$ and $5 x + 7 y + 1 = 0$ and is perpendicular to the line $7 x - 5 y + 5 = 0$ is

7 x - 5 y + 5 = 0

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11 x - 3 y + 1 = 0

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5 x + 7 y + 1 = 0

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Solution

The correct option is C $5 x + 7 y + 1 = 0$
$L_{1} : 5 x + 7 y + 1 = 0$
$L_{2} : 3 x + 11 y - 7 = 0$
$L_{3} : 7 x + 5 y + 5$
We find to $L_{4}$
Equation of line through the intersection of lines $L_{1}$ and $L_{2}$
$(5 x + 7 y + 1) + k (3 x + 11 y - 7)$ -------- (1)
$(3 + 5 k) x + (11 + 7 k) y - 7 + k = 0$
Slope $= - \frac{3 + 5 k}{11 + 7 k}$
Also perpendicular to
the line $L_{3}$
$y = \frac{7}{5} x + 1$
Whose slope is $\frac{7}{5}$
Therefore slope of $L_{4} = - \frac{5}{7}$ ( $m_{3} = - \frac{1}{m_{4}}$ )
this implies :
$- (\frac{3 + 5 k}{11 + 7 k}) = - \frac{5}{7}$
$k = 0$
Put in first
Hence the required line
$5 x + 7 y + 1 = 0$

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