Shorter method to find the orthocentre of this triangle? The given vertices are (0,0) (5,-1) (-2,3).

My approach::

I assumed the orthocentre of the triangle as (a,b).
Now,
I used the fact that the line through each vertex and orthocentre is perpendicular to the opposite side.

So,

m (AH).m (BC) = -1

m (BH).m (AC) = -1

From this, I got two equations and I solved them to get the orthocentre. 

However,
THIS METHOD IS VERY LENGTHY.
Can someone please suggests me easier or shorter method to find it?

Thanks.
I don't think there is a much better way to do that. One particular case, that is very convenient, is when the origin lies on the circumcenter, on that case because of the Euler's Line we get:

$$\text{orthocenter}=A'+B'+C'$$

But, in order to use that result you have to make a translation to bring the origin to the circumcenter and use that translation to find the new vertex $A',B',C'$.
Use formulas with determinants
$$
D=\text{Det}\begin{pmatrix} 0 &amp;0 &amp;0\\ 5 &amp;-1 &amp;1\\ -2 &amp;3 &amp;1\end{pmatrix}\\
D=13\\
xH=\frac{1}{D}.\left[\text{Det}\begin{pmatrix} 0 &amp;0.0 &amp;1\\ 5 &amp;5.(-1) &amp;1\\-2&amp;-2.3&amp;1\end{pmatrix}+\text{Det}\begin{pmatrix}0&amp;0^2&amp;1\\-1&amp;(-1)^2&amp;1\\3 &amp;3^2 &amp;1\end{pmatrix}\right]=-4\\
yH=-\frac{1}{D}.\left[\text{Det}\begin{pmatrix}0 &amp;0.0&amp;1\\-1&amp;5.(-1)&amp;1\\3&amp;-2.3&amp;1\end{pmatrix}+\text{Det}\begin{pmatrix}0&amp;0^2&amp;1\\5&amp;5^2&amp;1\\-2&amp;(-2)^2&amp;1\end{pmatrix}\right]=-7\\
H=(-4,-7)$$