# Find the orthocenter of the triangle with vertices $(2, \frac {\sqrt {3}-1}{2})$, $(\frac {1}{2}, -\frac {1}{2})$ and $(2, -\frac {1}{2})$

Find the orthocenter of the triangle with vertices $(2, \frac {\sqrt {3}-1}{2})$, $(\frac {1}{2}, -\frac {1}{2})$ and $(2, -\frac {1}{2})$

As orthocentre is the point of intersection of the perpendiculars drawn from vertices to the opposite sides. Thinking straight forward, we can make the equations of the sides and find the equations of perpendiculars to those sides and then finally solve those equations of perpendiculars to get the orthocentre. But this much of calculation becomes quite cumbersome for objective type of questions. Isn't there any other approach to this?

• This triangle has the right angle at $(2,-1/2)$, right? It is the orthocenter. – SMM Jul 8 '18 at 12:24
• @SMM, How do you know that it is right angled at $(2,-1/2)$?? – pi-π Jul 8 '18 at 12:28
• Just look at the coordinates. One side is parallel to $x$-axis, and one to the $y$-axis. – SMM Jul 8 '18 at 12:29

The last point shares its $x$-coordinate with the first point and its $y$-coordinate with the second. Therefore the triangle is right, and right triangles have their orthocentres at the right angle – $\left(2,-\frac12\right)$ here.
• @blue_eyed_... You would use the dot product for this in general. The vector from point 1 to 3 is $(0,-\sqrt3/2)$ and that from 2 to 3 is $(3/2,0)$. Their dot product is zero, showing that the angle at point 3 is right and thus that point 3 is the orthocentre. – Parcly Taxel Jul 8 '18 at 12:33
Let respectively your points be $A,B$ and $C$. Notice $AC$ and $BC$ are perpendicular, from the sketch of points. In general we compute:$$\text{slope of AC}\times\text{slope of BC}=\cdots=-1$$ to conclude that two lines are perpendicular, but in this case $\text{slope of } AC$ isn't defined. We immediately know that $BC$ and $AC$ pass through the orthocentre of the triangle, since they are altitudes. In other words, the 2 sides, that are not the hypotenuse, of a right-angled triangle are altitudes. So, point $C$ is the desired orthocentre (the intersection of altitudes).