Question

Find the point on the y-axis which is equidistant from $A(3,-6)$ and $B(-2,5)$

• A. $(\frac{9}{11},0)$
• B. $(0,\frac{8}{11})$
• C. $(0,\frac{-4}{3})$
• D. $(0,\frac{9}{11})$

Answer 1

The correct option is C $(0,\frac{-4}{3})$

Let the point on y-axis be $C(0,k)$
Distance between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $\sqrt{{(x_2-x_1)}^2+{{(y}_2-y_1)}^2}$
So, length of side $AC$ is $\sqrt{{(k-3)}^2+{(k+6)}^2}=\sqrt{k^2+9-6k+k^2+36+12k}=\sqrt{2k^2+45+6k}$
Length of side $BC$ is $\sqrt{{(k+2)}^2+{(k-5)}^2}=\sqrt{k^2+4+4k+k^2+25-10k}=\sqrt{2k^2+29-6k}$
Given, $AC=BC$
$=>\sqrt{2k^2+45+6k}=\sqrt{2k^2+29-6k}$
Squaring both sides, we get
$=>2k^2+45+6k=2k^2+29-6k$
$=>12k=-16$
$=>k=\frac{-4}{3}$
So, the point on y-axis is $(0,\frac{-4}{3})$