How to find the orthocentre of a triangle whose coordinates and area are given

Question

I have a triangle in $ xy $ plane, whose vertices being

$$ A(k, –3k)$$ $$ B(5, k) $$ $$ C(–k, 2) $$

Where $k$ is an integer.

And area is $28$ sq. units.

How to find the coordinates of its orthocentre?

What I've done

By using area formula, I got $k=2$

Answer 1

Hint

Since the area is given as well as the coordinates, this can help you determine the value of $k$ from the following: $$\frac{1}{2}\begin{vmatrix}k&-3k&1\\5&k&1\\-k&2&1\end{vmatrix}=28.$$

Once you have the $k$, it is a matter of determining the intersection of two altitudes of the triangle.

For example, to get the altitude from $C$ onto $AB$, you want to find a line that passes through $C$ and is perpendicular to $AB$. Slope of $AB$ is $m_{AB}=\frac{4k}{5-k}$. So your line must of the form $$y-2=-\frac{5-k}{4k}(x+k).$$ Similarly you can get the other altitude (say from $B$ onto $AC$) and then find the intersection of these two altitudes. By definition, the intersection is the orthocenter.

Answer 2

Once you get $k=2$ from the shoelace formula, you also get that the $BC$ side is horizontal, hence the abscissa of the orthocenter is the abscissa of $A$, i.e. $2$. Since the slope of the $CA$ side is $-2$, the orthocenter $H$ lies on the line through $B$ with slope $\frac{1}{2}$. From $y=\frac{1}{2}(x-5)+2$ you get $H=\left(2,\frac{1}{2}\right)$. Anyway there is another solution associated with $k=-\frac{23}{5}$.