Question

What point on y-axis is equidistant from the points $(3,1)$ and $(1,5)$?

• A. $P(1,3)$
  
• B. $P(0,2)$
• C. $P(1,2)$
• D. $P(0,3)$

Answer 1

Answer: (B)

$P(0,2)$

Let the point on the y-axis be, $(0,y)$

Distance Formula $=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

Distance between $(0,y)$ and $(3,1)=\sqrt{{(3-0)}^2+{(1-y)}^2}=\sqrt{9+1^2+y^2-2y}=\sqrt{y^2-2y+10}$
Distance between $(0,y)$ and $(1,5)=\sqrt{{(1-0)}^2+{(5-y)}^2}=\sqrt{1+5^2+y^2-10y}=\sqrt{y^2-10y+26}$
As the point, $(0,y)$ is equidistant from the two points, distance
between $(0,y);(3,1)$ and $(0,y);(1,5)$ are equal.

$\sqrt{y^2-2y+10}=\sqrt{y^2-10y+26}$
$\Rightarrow y^2-2y+10=y^2-10y+26$

$8y=16$

$y=2$

Thus, the point is $(0,2)$