Question

The point(s) on Y-axis which is/are at a distance of 4 units from the line $\frac{x}{3}+\frac{y}{4}=1$ will be

• A. $(0,-\frac{32}{3})$
• B. $(0,\frac{8}{3})$
• C. $(0,\frac{32}{3})$
• D. $(0,-\frac{8}{3})$

Answer 1

Answer: (C), (D)

The correct options are
C $(0,\frac{32}{3})$
D $(0,-\frac{8}{3})$

Given line is $\frac{x}{3}+\frac{y}{4}=1$
$\Rightarrow$ 4x+3y-12=0
Let any point on Y-axis
be (0, k).
$\mathrm{So},\left(\frac{4\left(0\right)+3\left(k\right)-12}{\sqrt{4^2}+\sqrt{3^2}}\right|=4\Rightarrow \frac{3k-12}{5}=\pm 4\Rightarrow 3k-12=\pm 20\Rightarrow k=\frac{32}{3}\mathrm{or}k=\frac{-8}{3}$