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

A
$(\frac{1}{4}, - \frac{1}{3})$
B
$(- \frac{1}{4}, - \frac{2}{3})$
C
$(- \frac{1}{4}, \frac{2}{3})$
D
$(\frac{1}{4}, \frac{1}{3})$

Solution:

The given line is $4 x - 3 y + 2 = 0$ The straight line parallel to the given line is

$4 x - 3 y + c = 0$ Distance of this line from origin is 3/5.

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

So, the parallel lines are $4 x - 3 y + 3 = 0$ or $4 x - 3 y - 3 = 0$

Now, Point $(- \frac{1}{4}, \frac{2}{3})$ given in option (c) satisfies the equation, $4 x - 3 y + 3 = 0 .$ So, $(- \frac{1}{4}, \frac{2}{3})$ lies on the line $4 x - 3 y + 3 = 0$

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

Solution:

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