Question

Find a point on the y-axis which is equidistant from the points $(-3,4)$ and $(2,3)$.

• A. $(0,-6)$
• B. $(0,6)$
• C. $(0,3)$
• D. $(0,-3)$

Answer 1

Answer: (B)

$(0,6)$

Let the point on the x-axis be $(0,y)$
Distance between $(0,y)$ and $(-3,4)=\sqrt{{(-3-0)}^2+{(4-y)}^2}=\sqrt{9+4^2+y^2-8y}=\sqrt{y^2-8y+25}$
Distance between $(0,y)$ and $(2,3)=\sqrt{{(2-0)}^2+{(3-y)}^2}=\sqrt{4+3^2+y^2-6y}=\sqrt{y^2-6y+13}$
As the point $(0,y)$ is equidistant from the two points, both the distances
calculated are equal.
$\sqrt{y^2-8y+25}=\sqrt{y^2-6y+13}$
$=>y^2-8y+25=y^2-6y+13$
$12=2y$
$y=6$
Thus, the point is $(0,6)$