# What are the points on the y-axis whose distance from the line

Question:

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

Solution:

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 C = –12.

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, 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:

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

$\Rightarrow 4=\frac{|3 b-12|}{5}$

$\Rightarrow 20=|3 b-12|$

$\Rightarrow 20=\pm(3 b-12)$

$\Rightarrow 20=(3 b-12)$ or $20=-(3 b-12)$

$\Rightarrow 3 b=20+12$ or $3 b=-20+12$

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

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