Question

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

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

Answer 1

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

Required line is $4x-3y+\lambda =0$
$\left|\frac{\lambda }{5}\right|=\frac{3}{5}\Rightarrow \lambda =\pm 3$
So required equation of line is $4x-3y+3=0$ and $4x-3y-3=0$
$\left(1\right)4(-\frac{1}{4})-3\left(\frac{2}{3}\right)+3=0$