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

$\begin {array} {1 1} (A)\;\bigg(0, \large\frac{32}{3} \bigg)\: and \: \bigg( 0, -\large\frac{8}{3} \bigg) & \quad (B)\;\bigg( 0, -\large\frac{32}{3} \bigg) \; and \: \bigg( 0, \large\frac{8}{3} \bigg) \\ (C)\;\bigg( 0, -\large\frac{32}{3} \bigg) \; and \: \bigg( 0,- \large\frac{8}{3} \bigg) & \quad (D)\;\bigg( 0, \large\frac{32}{3} \bigg) \; and \: \bigg( 0, \large\frac{8}{3} \bigg) \end {array}$

Toolbox:
• Perpendicular distance of a line $Ax+By+C=0$ from a point $(x_1, y_1)$ is given by $d = \large\frac{|Ax_1+By_1+C|}{\sqrt {A^2+B^2}}$
Let (o, b) be the point on the $y$ - axis given that the distance from the line
$\large\frac{x}{3}$$+\large\frac{y}{4}$$=1$ is 4 units
This can be written as $4x+3y-12=0$--------(1)
Comparing this equation with the general equation
$Ax+By+C=0$ we get,
$A=4, B=3$ and $C = -12$
If (o, b) is the point on the $y$ - axis, then $d = \large\frac{|Ax_1+By_1+C|}{\sqrt {A^2+B^2}}$
$4 = \large\frac{|4(0)+3(b)-12|}{\sqrt{4^2+3^2}}$