Looking for other approaches to find the height $AH$ in triangle $ ABC $ where $A(1,5)$ , $B(7,3)$, $C(2,-2)$ 
We have a triangle with vertices $A(1,5)$ , $B(7,3)$, $C(2,-2)$. What
is the length of the height $AH$ in the triangle $ABC$ ?
$1)4\qquad\qquad2)3\sqrt2\qquad\qquad3)5\qquad\qquad4)4\sqrt2$

This is a problem from a timed exam.
Here is my approach:
The equation of the line passes through $B$ and $C$ is $y_=x-4$ so slope of $AH$ is $-1$ and its a line passes through $A(1,5)$ so the equation for this line becomes $y=-x+6$. in order to find coordinate of $H$ we should equate the formulas of those lines:
$$x-4=-x+6\quad\to\quad x=5\quad\text{and $\quad y=1\quad$ So$\quad H (5,1)$}$$
$$\text{Distance from $A$ to $H$:$\qquad$}AH=\sqrt{(5-1)^2+(1-5)^2}=4\sqrt2$$
Although I believe my answer is quick but I'm looking for other ideas to solve this problem.

Area of triangle is \begin{align}\frac12|x_1y_2+x_2y_3+x_3y_1-x_2y_1-x_3y_2-x_1y_3|&=\frac12|3-14+10-35-6+2|\\
&=20 \end{align}
Hence
$$AH \times BC=40$$
$$AH \times \sqrt{5^2+5^2}=40$$
$$5\sqrt{2} AH = 40$$
$$AH=\frac{8}{\sqrt2}=4\sqrt2$$
