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$4x + 3y – 12 = 0......(1)$
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 $Ax + By + C = 0$ from a point $(x_1, y_1)$ is given by

$d=\frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}$
Therefore,

$4=\frac{|4a+3\times0-12|}{\sqrt{4^2+3^2}}$

$⇒4=\frac{|4a-12|}{5}$

$⇒|4a-12|=20$
$⇒±(4a-12)=20$
$⇒(4a-12)=20$ or $-(4a-12)=20$
$⇒4a=20+12$ or $4a=-20+12$
$⇒a=8\space or -2$
Thus, the required points on the x-axis are $(-2, 0)$and $(8, 0).$