The centroid of a triangle is at $G(0,6)$ and the orthocenter is at $H(0,8)$. Knowing that one of the vertices of this triangle is at the origin, find its area.
I deduced that this triangle must be isosceles. I also know that the centroid divides the median in $2k$ to $k$ ratio, and it seems that we can draw the other altitudes to construct a solution using similarity. Is there an analytic approach or an alternative approach without using similarity?
On
Since the centroid $G = (0, 6)$ and vertex $A = (0, 0)$ , then $M = (0, 9) $ lies on $BC$. Since the orthocenter is $H(0, 8)$ , the line $AH$ is orthogonal to $BC$ , so $BC$ must be horizontal. Thus, taking into account that the $x$-coordinate of $G$ is 0, it follows that,
$B = (-a, 9), C= (a, 9)$
The normal vector to AC is along (9, -a) and passes through $B$, hence it parametric equation is
$p(t) = B + t (9, -a) = (-a, 9) + t (9, -a) $
At $x=0$, $t = a/9$, which in turn implies that,
$y = 9 - a^2/9$
But at the orthocenter we have $y = 8$
Hence, $8 = 9 - a^2/9$, from what $a = 3$
Hence the area of the triangle = $\frac{1}{2} (2 a)(9) = (3)(9) = 27 $
We know that the centroid of a triangle divides the join of it's orthocentre and circumcentre in ratio $2 : 1$. Hence, the circumcentre is (0, 5).
As illustrated in the image, the co-ordinates of vertices are $(a,b)$ and $(-a,b)$. Here we have uses that the triangle is isoceles.
Now, as the $y$- coordinate of centroid=$6$, $b=9$.
As the circumcentre is equidistant from the vertices, a=3.