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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.

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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 )

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