Question

Find that point on y axis which as equidistant from point $(6,5)$ and $(-4,3)$

Answer 1

$\left(0.4\right)(6,5)(4,3)$

$d=\sqrt{(0-6{)}^2+(y-5{)}^2}$ …….$\left(1\right)$

$d=\sqrt{(0+4{)}^2+(y-3{)}^2}$ ……$\left(2\right)$

$\left(1\right)$ $=$ $\left(2\right)$

squaring both sides

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

$36-24+y^2+25-10y=y^2+9-64$

$28=4y$

$y=7$.