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

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

Answer 1

Answer: (B)

Coordinate of vertices of triangle
$A(2,3),B(2,4),C(4,3)$
$\therefore AB=1,BC=\sqrt{5},CA=2$
So, $\Delta ABC$ ' 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 $2:1$

$\frac{8}{3}=\frac{2x+2}{3}$
$\Rightarrow x=3$
$\frac{10}{3}=\frac{2y+3}{3}$
$\Rightarrow y=\frac{7}{2}$
So, $A{O}^2+9B{G}^2+4C{S}^2$
$A{O}^2=(2-2{)}^2+(3-3{)}^2=0$
$\Rightarrow A{O}^2=0$
$B{G}^2={\left(2-\frac{8}{3}\right)}^2+{\left(4-\frac{10}{3}\right)}^2=\frac{8}{9}$
$\Rightarrow 9B{G}^2=8$
$C{S}^2=(4-3{)}^2+{\left(3-\frac{7}{2}\right)}^2=\frac{5}{4}$
$\Rightarrow 4C{S}^2=5$
$\therefore A{O}^2+9B{G}^2+4C{S}^2=13$