The orthocenter (intersection of the altitudes) of the triangle $ ABC $ is point:
$$ \left(0,~2\right) $$explanation
The orthocenter of a triangle is given by:
$$ H = \left( ~\frac{\begin{vmatrix} A_y & B_x C_x + A_y^2 & 1 \\ B_y & C_x A_x + B_y^2 & 1 \\ C_y & A_x B_x + C_y^2 & 1 \\ \end{vmatrix}}{ 2 \cdot \begin{vmatrix} A_x & A_y & 1 \\ B_x & B_y & 1 \\ C_x & C_y & 1 \\ \end{vmatrix}}~,~ \frac{\begin{vmatrix} A_x^2 + B_y C_y & A_x & 1 \\ B_x^2 + C_y A_y & B_x & 1 \\ C_x^2 + A_y B_y & C_x & 1 \\ \end{vmatrix}} {2 \cdot \begin{vmatrix} A_x & A_y & 1 \\ B_x & B_y & 1 \\ C_x & C_y & 1 \\ \end{vmatrix}}~ \right) $$where $ A_x $ and $ A_y $ are $ x $ and $ y $ coordinates of the point $ A $ , $ B_x $ and $ B_y $ are $ x $ and $ y $ coordinates of the point $ B $ etc..
In this example we have : $ A_x = -2 $ , $ A_y = 5 $ , $ B_x = 5 $ , $ B_y = 1 $ , $ C_x = -4 $ and $ C_y = -5 $ .
When we insert these values into the formula, we obtain the given result.