[tex]The\ distance\ between\ A(x_A;\ y_A)\ and\ B(x_B;\ y_B):\\\\d=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}\\\\------------------------------\\\\A(-4;\ y);\ B(4;\ 2);\ d=10\\\\subtitute\\\\\sqrt{(4-(-4))^2+(2-y)^2}=10\\\\\sqrt{8^2+(2-y)^2}=10\\\\\sqrt{64+(2-y)^2}=10\Rightarrow64+(2-y)^2=10^2\\\\64+(2-y)^2=100\ \ \ \ |subtract\ 64\ from\ both\ sides\\\\(2-y)^2=36\iff2-y=-\sqrt{36}\ or\ 2-y=\sqrt{36}\\\\2-y=-6\ or\ 2-y=6\ \ \ \ |subtract\ 2\ from\ both\ sides\\\\-y=-8\ or\ -y=4\ \ \ |change\ the\ signs\\\\y=8\ or\ y=-4[/tex]
[tex]Answer:\boxed{(-4;\ 8)\ and\ (-4;-4).}[/tex]
