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.
 A: 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 $A',B',C'$.
A: The orthocenter of triangle $ABC$ is the circumcenter of $A'B'C'$, where 
$$
A'=B+C-A,\quad B'=A+C-B,\quad C'=A+B-C,
$$
are the reflections of each vertex across the midpoint of the opposite side.
You can then insert the coordinates of $A'$, $B'$, $C'$ into the generic circle equation $x^2+y^2-2ax-2by+c=0$, to get a linear system in three unknowns $a$, $b$, $c$: solve it to obtain circumcenter $(a,b)$.
EXAMPLE.
Starting with vertices $A=(0,0)$, $B=(5,-1)$, $C=(-2,3)$ we get: $A'=(3,2)$, $B'=(-7,4)$, $C'=(7,-4)$ and the system
$$
\cases{
13-6a-4b+c=0\\
65+14a-8b+c=0\\
65-14a+8b+c=0
}.
$$
This can be readily solved to yield $a=-4$ and $b=-7$.
A: To add to @Arnaldo's method, we can find the circumcentre by assuming the circumcircle to be:
$$x^2 + y^2 + 2gx + 2fy + c = 0$$
Since all three points lie on this circle we get three equations. But, each equation has a $c$ in it, so subtracting an equation from each of the other two, we get two equations, solving which we get $g$ and $f$. Then, circumcentre is given by:
$$C \equiv (-g,-f) $$
After translating the origin, just find orthocentre using Arnaldo's formula and then translate it back to the original coordinate system.
To be honest, this method is only useful if you want both, the orthocentre and the circumcentre. Otherwise your own method is the best bet!
I would like to comment that both orthocentre and circumcentre require solving two simultaneous equations, with orthocentre being easier to find that way. So it would actually be better to first find orthocentre and then use the Euler line's property.
A: Use formulas with determinants
$$
D=\text{Det}\begin{pmatrix} 0 &0 &0\\ 5 &-1 &1\\ -2 &3 &1\end{pmatrix}\\
D=13\\
xH=\frac{1}{D}.\left[\text{Det}\begin{pmatrix} 0 &0.0 &1\\ 5 &5.(-1) &1\\-2&-2.3&1\end{pmatrix}+\text{Det}\begin{pmatrix}0&0^2&1\\-1&(-1)^2&1\\3 &3^2 &1\end{pmatrix}\right]=-4\\
yH=-\frac{1}{D}.\left[\text{Det}\begin{pmatrix}0 &0.0&1\\-1&5.(-1)&1\\3&-2.3&1\end{pmatrix}+\text{Det}\begin{pmatrix}0&0^2&1\\5&5^2&1\\-2&(-2)^2&1\end{pmatrix}\right]=-7\\
H=(-4,-7)$$
