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

$\begin{array}{1 1}(A)\;(-2,8)(0,0) \\(B)\; (2,0)(8,0) \\(C)\; (-2,0)(8,0) \\(D)\;(2,8)(0,0) \end{array}$

Toolbox:
• Distance of a point $(x_1,y_1)$ from the given line is $d=\large\frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}$
The given equation of the line is
$\large\frac{x}{3}$$+\large\frac{y}{4}$$=1$
This can be written as $4x+3y-12=0$
The general form of an equation of a straight line is $Ax+By+C=0$
The perpendicular distance $(d)$ of the line from a point $(x_1, y_1)$ is
$d = \large\frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}$
Let (a, o)be the point on the $x$ - axis.
Comparing both the equations we get,
$A=4, B=3\: and \: C=-12$ and given $d = 4$ units.
Substituting the values we get,
$4 = \large\frac{|4a+3x0-12|}{\sqrt{4^2+3^2}}$