<s>[INST] 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.
 [/INST] 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 $\triangle ABC, AH \times BC = CG \times AB$
$AH = \dfrac{2 \sqrt{10} \times 2 \sqrt{10}}{5 \sqrt2} = 4\sqrt2$
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