Question

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

Answer 1

Let (0, b) be the point on the y-axis whose distance from line $\frac{ x}{3} +\frac{ y}{4} = 1$ is 4 units.
The given line can be written as

$4x + 3y \- 12 = 0 … (1)$

On comparing equation (1) to the general equation of line $Ax + By + C = 0,$ we obtain $A = 4, B = 3, and\space C = -12.$
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

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

Therefore, if (0, b) is the point on the y-axis whose distance from line $\frac{ x}{3} +\frac{ y}{4} = 1$ is 4 units, then:

$4=\frac{\left|4(0)+3(0)-12\right|}{\sqrt{4^2+3^2}}$

$⇒ 4=\frac{\left|3b-12\right|}{5}$

$⇒ 20=|3b-12|$

$⇒ 20=±(3b-12)$
$⇒ 20=(3b-12) \space or 20=-(3b-12)$
$⇒ 3b=20+12 \space or\space 3b=-20+12$

$⇒ b=\frac{32}{3}$ or $b=\frac{-8}{3}$

Thus, the required points are$(0, \frac{32}{3})$ and $(0, \frac{-8}{3})$.