Question 1100224
<br>If AB=BC, then (AB)^2 = (AC)^2; so we can solve the equation that says the squares of the distances are equal.<br>
Using the distance formula on AB, we find {{{(AB)^2 = 5^2+14^2 = 221}}}
Using the distance formula on BC, we find {{{(BC)^2 = 10^2 + (13-y)^2}}}<br>
Then
{{{221 = 100 + (13-y)^2}}}
{{{121 = (13-y)^2}}}
{{{11 = 13-y}}}
{{{y = 2}}}<br>
Point C is (-7,2)