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.
 A: 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
$$AH \times BC=40$$
$$AH \times \sqrt{5^2+5^2}=40$$
$$5\sqrt{2} AH = 40$$
$$AH=\frac{8}{\sqrt2}=4\sqrt2$$
A: 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
$$5\sqrt 2\cdot AH=40$$
$$10\cdot AH=40\sqrt 2$$
$$AH=4\sqrt 2$$
A: Other than applying law of cosine or using distance formula from a point to a line that I mentioned in comment, another approach that I would consider is to recognize that $\triangle ABC$ is an isosceles triangle with,
$AC = BC = 5\sqrt2, \ AB = 2 \sqrt{10}$. If $G$ is midpoint of $AB$ then we know that $CG^2 = AC^2 - AG^2 = 50 - 10 = 40 \implies CG = 2 \sqrt{10}$
So equating area of $\triangle ABC, AH \times BC = CG \times AB$
$AH = \dfrac{2 \sqrt{10} \times 2 \sqrt{10}}{5 \sqrt2} = 4\sqrt2$
A: The area of the triangle is
$$42-\frac72-\frac{25}2-6=20$$

I don't want to post three answers, but I have another one:

