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

Let P (0, y) be a point on the y-axis. Then as per the question, we have

$A P=B P$

$\Rightarrow \sqrt{(0-6)^{2}+(y-5)^{2}}=\sqrt{(0+4)^{2}+(y-3)^{2}}$

$\Rightarrow \sqrt{(6)^{2}+(y-5)^{2}}=\sqrt{(4)^{2}+(y-3)^{2}}$

$\Rightarrow(6)^{2}+(y-5)^{2}=(4)^{2}+(y-3)^{2}$                 (Squaring both sides)

$\Rightarrow 36+y^{2}-10 y+25=16+y^{2}-6 y+9$

$\Rightarrow 4 y=36$

$\Rightarrow y=9$

Hence, the point on the y-axis is (0, 9).