Question

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 $A O^{2}+9 B G^{2}+4 C S^{2}=$

• A. $\frac{77}{36}$
• B. $13$
• C. $\frac{8}{9}$
• D. $\frac{5}{4}$

Answer 1

Answer: (B)

Coordinate of vertices of triangle
$A(2,3), B(2,4), C(4,3)$
$\therefore \, A B=1,\, B C=\sqrt{5}, \,C A=2$
So, $\Delta \,A B C$ ' is right angle triangle where right angle at $A$ that is orthocentre also. Coordinate of orthocentre is $O(2,3)$.
Coordinates of centroid
$=\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}\right)$
$G=\left(\frac{8}{3}, \frac{10}{3}\right)$
$G$ divide the line joining $O$ and $S$ in the ratio $2: 1$

$\frac{8}{3}=\frac{2 x+2}{3}$
$ \Rightarrow \, x=3$
$\frac{10}{3}=\frac{2 y+3}{3} $
$\Rightarrow \, y=\frac{7}{2}$
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$
$C S^{2}=(4-3)^{2}+\left(3-\frac{7}{2}\right)^{2}=\frac{5}{4} $
$\Rightarrow \,4 C S^{2}=5$
$\therefore \, A O^{2}+9 B G^{2}+4 C S^{2}=13 $