Find the points on the x-axis,

Solution:

The given equation of line is

$\frac{x}{3}+\frac{y}{4}=1$

or, $4 x+3 y-12=0$ $\ldots(1)$

On comparing equation (1) with general equation of line Ax + By + C = 0, we obtain A = 4, B = 3, and C = –12.

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

It is known that the perpendicular distance $(d)$ of a line $A x+B y+C=0$ from a point $\left(x_{1}, y_{1}\right)$ is given by $d=\frac{\left|A x_{1}+B y_{1}+C\right|}{\sqrt{A^{2}+B^{2}}}$.

Therefore,

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

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

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

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

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

$\Rightarrow 4 a=20+12$ or $4 a=-20+12$

$\Rightarrow a=8$ or $-2$

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