A ma leżeć na osi OY, więc
A(0,y)
[tex]|AB|=\sqrt{(-6-0)^2+(1-y)^2}=\sqrt{36+1-2y+y^2}=\sqrt{y^2-2y+37}\\\\|BC|=\sqrt{(2+6)^2+(1-1)^2}=\sqrt{8^2+0^2}=\sqrt{8^2}=8\\\\|CA|=\sqrt{(0-2)^2+(y-1)^2}=\sqrt{4+y^2-2y+1}=\sqrt{y^2-2y+5}[/tex]
Z twierdzenia Pitagorasa
[tex]|AB|^2+|CA|^2=|BC|^2\\\\(\sqrt{y^2-2y+37})^2+(\sqrt{y^2-2y+5})^2=8^2\\\\y^2-2y+37+y^2-2y+5=64\\\\2y^2-4y^2+37+5-64=0\\\\2y^2-4y^2-22=0\ \ \ |:2\\\\y^2-2y-11=0\\\\\Delta=(-2)^2-4\cdot1\cdot(-11)=4+44=48\\\\\sqrt{\Delta}=\sqrt{48}=4\sqrt3\\\\y_1=\frac{2-4\sqrt3}{2}=1-2\sqrt3\\\\y_2=\frac{2+4\sqrt3}{2}=1+2\sqrt3\\\\[/tex]
Współrzędne punktu A(0,1-2√3) lub A(0,1+2√3)
