If $O, G, S$ are respectively the orthocentre, centroid and circumcentre of a triangle whose vertices are…
If $O, G, S$ are respectively the orthocentre, centroid and circumcentre of a triangle whose vertices are $A(2,3), B(2,4)$ and $C(4,3)$, then $A O^2+9 B G^2+4 C S^2=$
$\frac{77}{36}$
13
$\frac{8}{9}$
$\frac{5}{4}$
Solution
Coordinate of vertices of triangle
$
\begin{aligned}
& A(2,3), B(2,4), C(4,3) \\
& \therefore \quad A B=1, B C=\sqrt{5}, C A=2
\end{aligned}
$
So, $\triangle A B C$ is right angle triangle where right angle at $A$ that is orthocentre also.
Coordinate of orthocentre is $O(2,3)$.
Coordinates of centroid
$
\begin{aligned}
& =\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right) \\
G & =\left(\frac{8}{3}, \frac{10}{3}\right)
\end{aligned}
$
$G$ divide the line joining $O$ and $S$ in the ratio $2: 1$
$
\begin{gathered}
\frac{8}{3}=\frac{2 x+2}{3} \Rightarrow x=3 \\
\frac{10}{3}=\frac{2 y+3}{3} \Rightarrow y=\frac{7}{2} \\
\Rightarrow \quad S\left(3, \frac{7}{2}\right) \\
\text { So, } A O^2+9 B G^2+4 C S^2 \\
A O^2=(2-2)^2+(3-3)^2=0 \Rightarrow A O^2=0 \\
B G^2=\left(2-\frac{8}{3}\right)^2+\left(4-\frac{10}{3}\right)^2=\frac{8}{9} \Rightarrow 9 B G^2=8
\end{gathered}
$
$\begin{aligned} & C S^2=(4-3)^2+\left(3-\frac{7}{2}\right)^2=\frac{5}{4} \Rightarrow 4 C S^2=5 \\ & \therefore \quad A O^2+9 B G^2+4 C S^2=13 . \\ & \end{aligned}$
