Q. The equation of one of the lines parallel to $4x - 3y = 5$ and at a unit distance from the point $( - 1,\ - 4)$ is
A) $3x + 4y - 3 = 0$
B) $3x + 4y + 3 = 0$
C) $4x - 3y + 3 = 0$
D) $4x - 3y - 3 = 0$
Explanation
(d)
Required equation can be $4x - 3y - K = 0$
$\therefore\left| \frac{4 \times - 1 - 3 \times - 4 - K}{\sqrt{4^{2} + ( - 3)^{2}}} \right| = 1$
$\Rightarrow \frac{- 4 + 12 - K}{5} = \pm 1$
$\Rightarrow 8 - K = \pm 5$
$\Rightarrow K = 3\ or\ K = 13$
$\therefore$ Equation of lines are $4x - 3y - 3 = 0$ and $4x - 3y - 13 = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.