Question

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

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

Answer 1

The correct option is D $(-\frac{1}{4},\frac{2}{3})$

Equation of line parallel to the line
$4x-3y+2=0$ is $4x-3y+\lambda =0...\left(1\right)$
Distance of line from the origin is
$\left|\frac{4\left(0\right)-3\left(0\right)+\lambda }{\sqrt{4^2+(-3{)}^3}}\right|=\frac{3}{5}$
$\Rightarrow \lambda =\pm 3$
putting the value of $\lambda$ in equation $\left(1\right)$
$\therefore$ the required lines are $4x-3y+3=0$ and $4x-3y-3=0$
Point $(-\frac{1}{4},\frac{2}{3})$ satisfies the line $4x-3y+3=0$