Question 1121471
Call the point on the y-axis ( 0, y )
Use length formulas
{{{ ( -4 - 0 )^2 + ( -2 - y )^2 = ( 3 - 0 )^2 + ( 1 - y )^2 }}}
{{{ 16 + 4 + 4y + y^2 = 9 + 1 - 2y + y^2 }}}
{{{ y^2 + 4y + 20 = y^2 - 2y + 10 }}}
{{{ 4y + 20 = -2y + 10 }}}
{{{ 6y = -10 }}}
{{{ y = -5/3 }}}
The point is ( 0, -5/3 )
Check the math