Question

What are the points on y-axis which are at a distance of 4 units from 4x+3y=12

Answer 1

Given:

The equation of line is

$4x+3y=12$

To find:

The points on y-axis which are at a distance of 4 units from the given line.

Solution:

The point lie on the y-axis. So, their x-coordinate must be zero.

Let the points are in the form of (0,k).

The distance between a point $(x_0,y_0)$ and a line $ax+by+c=0$ is

$d=\dfrac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}$

The given equation can be written as

$4x+3y-12=0$

The distance between (0,k) and $4x+3y-12=0$ is 4 units.

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

$4=\dfrac{|0+3k-12|}{\sqrt{16+9}}$

$4=\dfrac{|3k-12|}{\sqrt{25}}$

$4=\dfrac{|3k-12|}{5}$

Multiply both sides by 5.

$20=|3k-12|$

$\pm 20=3k-12$

$12\pm 20=3k$

$\dfrac{12\pm 20}{3}=k$

Now,

$k=\dfrac{12-20}{3}\text{ and }k=\dfrac{12+20}{3}$

$k=\dfrac{-8}{3}\text{ and }k=\dfrac{32}{3}$

Therefore, the two points are $\left(0,\dfrac{-8}{3}\right)\text{ and }\left(0,\dfrac{32}{3}\right)$.