The equation of lines passing through the point of intersection of lines $3 x - y - 20 = 0$ and $x - 2 y - 5 = 0$ Which are at a distance of $5$ units from the origin is/are:

4 x + 3 y = 25

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3 x - 4 y = 25

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4 x - 3 y = 25

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3 x + 4 y = 25

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Solution

The correct option is D $3 x + 4 y = 25$
$3 x - y = 20$ , $x - 2 y = 5$
2) $6 x - 2 y = 40$
$_+_- -------------- -$
$x = 7$ , $y = 1$
Let equation be,
$y = m x + c$ passes $(7, 1)$
$\Rightarrow 7 m - 1 + c = 0$
1 dist. $5 = ∣ ∣ ∣ \frac{c}{\sqrt{1 + m^{2}}} ∣ ∣ ∣$
$\Rightarrow 25 (1 + m^{2}) = {(- 7 m + 1)}^{2}$
$\Rightarrow 25 + 25 m^{2} = 49 m^{2} + 1 - 14 m$
$\Rightarrow {24 m}^{2} - 14 m - 24 = 0$
$m = \frac{64}{48}, \frac{- 36}{48}$
$c = 25 / 3, 25 / 4$
So, $y = \frac{4 x}{3} + \frac{25}{3}$
or, $y = \frac{- 3 x}{4} + \frac{25}{4}$

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