Lines are drawn parallel to the line 4 x - 3 y + 2 = 0 , at a distance 3 5 units from the origin. Then which one of the following points lies on any of these lines?

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The equation of a line parallel to the line $a x + b y + c = 0$ is $a x + b y + k = 0 .$

Thus, the equation of a line parallel to $4 x - 3 y + 2 = 0$ is $4 x - 3 y + c = 0 .$

The distance of a line $a x + b y + c = 0$ from origin is $\frac{|c|}{\sqrt{a^{2} + b^{2} + c^{2}}} .$

Given the line $4 x - 3 y + c = 0$ is at a distance of $\frac{3}{5}$ units from origin

$\Rightarrow \frac{|c|}{\sqrt{16 + 9}} = \frac{3}{5}$

$\Rightarrow \frac{|c|}{5} = \frac{3}{5}$

$\Rightarrow c = \pm 3 .$

Hence, the equation of the lines are $4 x - 3 y + 3 = 0$ and $4 x - 3 y - 3 = 0 .$

Clearly point $(- \frac{1}{4}, \frac{2}{3})$ satisfy the given equation $4 x - 3 y + 3 = 0 .$

Mathematics - Straight line Question with Solution | TestHub

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