Question

Find the points on the $x$-axis, whose distances from the line $\frac{x}{3}+\frac{y}{4}=1$ are $4$ units.

Answer 1

The given equation of line is $\frac{x}{3}+\frac{y}{4}=1$ or $4x+3y-12=0$

Let $(a,0)$ be the point on the $x$-axis whose distance from the given line is $4$ units.

Thus, using distance formula,

$4=\frac{|4a+3\cdot 0-12|}{\sqrt{4^2+3^2}}$

$\Rightarrow 4=\frac{|4a-12|}{5}$

$\Rightarrow |4a-12|=20$

$\Rightarrow \pm (4a-12)=20$

$\Rightarrow (4a-12)=20\text{or}-(4a-12)=20$

$\Rightarrow 4a=20+12\text{or}4a=-20+12$

$\Rightarrow a=8\text{or}-2$

Thus, the required points on the $x$-axis are $(-2,0)$ and $(8,0)$