Q : 4 What are the points on the -axis whose distance from the line is units.

Name: What are the points on the y-axis whose distance from the line x/3+y/4=1 is 4 units.
Uploaded: Thu, 2019-06-13 12:15
Description: What are the points on the y-axis whose distance from the line x/3+y/4=1 is 4 units.

What are the points on the y-axis whose distance from the line x/3+y/4=1 is 4 units.

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

Answers (1)

Given the equation of the line is
$\small \frac{x}{3}+\frac{y}{4}=1$
we can rewrite it as
$4x+3y=12$
Let's take point on y-axis is $(0,y)$
It is given that the distance of the point $(0,y)$ from line $4x+3y=12$ is 4 units
Now,
$d= \left | \frac{Ax_1+By_1+C}{\sqrt{A^2+B^2}} \right |$
In this problem $A = 4 , B=3 , C =-12 ,d=4\ \ and \ \ (x_1,y_1) = (0,y)$
$4 = \left | \frac{4\times 0+3\times y-12}{\sqrt{4^2+3^2}} \right |=\left | \frac{3y-12}{\sqrt{16+9}} \right |=\left | \frac{3y-12}{5} \right |$

Case (i)

$4 = \frac{3y-12}{5}$
$20=3y-12$
$y = \frac{32}{3}$
Therefore, the point is $\left ( 0,\frac{32}{3} \right )$ -(i)

Case (ii)

$4=-\left ( \frac{3y-12}{5} \right )$
$20=-3y+12$
$y = -\frac{8}{3}$
Therefore, the point is $\left ( 0,-\frac{8}{3} \right )$ -(ii)

Therefore, points on the $\small y$ -axis whose distance from the line $\small \frac{x}{3}+\frac{y}{4}=1$ is $\small 4$ units are $\left ( 0,\frac{32}{3} \right )$ and $\left ( 0,-\frac{8}{3} \right )$