The perpendicular form of the line $3 x + 4 y - 5 = 0$ is

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$x \cos α + y \sin α = 1 where \cos α = \frac{3}{5}, \sin α = \frac{4}{5}$

$x \cos α - y \sin α = 1 where \cos α = \frac{3}{5}, \sin α = \frac{4}{5}$

$x \cos α + y \sin α = 1 where \cos α = \frac{4}{5}, \sin α = \frac{3}{5}$

$x \cos α - y \sin α = 1 where \cos α = \frac{4}{5}, \sin α = \frac{3}{5}$

answer is A.

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Detailed Solution

$\begin{array}{l} 3 x + 4 y - 5 = 0 \\ 3 x + 4 y = 5 \end{array}$

Both sides divide with $\sqrt{3^{2} + 4^{2}} = 5$

$(\frac{3}{5}) x + (\frac{4}{5}) y = 1$

$\Rightarrow x \cos α + y \sin α = 1$ where $\cos α = \frac{3}{5} and \sin α = \frac{4}{5}$

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