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.

Since the vertices have integer coordinates we can use Pick's theorem.
In the border of the triangle there are the three vertices, and also $(4,4)$ on $AB$, the points $(6,2)$, $(5,1)$, $(4,0)$ and $(3,-1)$ on $BC$ and no other points in $AC$. These make $8$ points in the border.
On the other hand the interior of the triangle is given by
$$\begin{cases}
x+3y<16\\
x-y<4\\
7x+y>12
\end{cases}$$
The left-most point of the triangle is $A(1,5)$. For $x=2$ we have six interior points $(2,-1),\ldots,(2,4)$. For $x=3$ we have from $(3,0)$ to $(3,4)$. For $x=4$ we have from $(4,1)$ to $(4,3)$. For $x=5$, only $(5,2)$ and $(5,3)$, and for $x=6$, $(6,3)$. That makes $17$ interior points.
If you relax the mathematical preciseness a bit, you can easily count them in a good graph:

By Pick's theorem, the area of the triangle is $20$. The length of $BC$ is $5\sqrt 2$, so
$$5\sqrt 2\cdot AH=40$$
$$10\cdot AH=40\sqrt 2$$
$$AH=4\sqrt 2$$
