The midpoint of AB where \(A(x_A;\ y_A)\) and \(B(x_B;\ y_B)\)
\(\left(\dfrac{x_A+x_B}{2};\ \dfrac{y_A+y_B}{2}\right)\)
\(H(-1;\ 3);\ X(7;-1)\)
The midpoint of HX: \(\left(\dfrac{-1+7}{2};\ \dfrac{3+(-1)}{2}\right)=\left(\dfrac{6}{2};\ \dfrac{2}{2}\right)=(3;\ 1)\)
Answer: A(3, 1)
