If O, G, S are respectively the orthocentre,c | vedclass

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(B) The vertices of the triangle are $A(2,3)$,$B(2,4)$,and $C(4,3)$.Since $AB$ is vertical (along $x=2$) and $AC$ is horizontal (along $y=3$),the tria

Vedclass · Accepted Answer

$13$

Vedclass · Answer

(B) The vertices of the triangle are $A(2,3)$,$B(2,4)$,and $C(4,3)$.Since $AB$ is vertical (along $x=2$) and $AC$ is horizontal (along $y=3$),the triangle is a right-angled triangle with the right angle at $A$.For a right-angled triangle,the orthocentre $O$ is the vertex at the right angle. Thus,$O = A = (2,3)$.Therefore,$AO^2 = (2-2)^2 + (3-3)^2 = 0$.The centroid $G$ is given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}ight) = \left(\frac{2+2+4}{3}, \frac{3+4+3}{3}ight) = \left(\frac{8}{3}, \frac{10}{3}ight)$.We calculate $9BG^2 = 9 	imes \left[ (2 - \frac{8}{3})^2 + (4 - \frac{10}{3})^2 ight] = 9 	imes [(-\frac{2}{3})^2 + (\frac{2}{3})^2] = 9 	imes [\frac{4}{9} + \frac{4}{9}] = 9 	imes \frac{8}{9} = 8$.The centroid $G$ divides the line segment joining the orthocentre $O$ and the circumcentre $S$ in the ratio $2:1$. Using the section formula,$G = \frac{1 \cdot O + 2 \cdot S}{3}$ $\Rightarrow 3G = O + 2S$ $\Rightarrow S = \frac{3G - O}{2}$.$S = \frac{3(\frac{8}{3}, \frac{10}{3}) - (2,3)}{2} = \frac{(8-2, 10-3)}{2} = \frac{(6, 7)}{2} = (3, \frac{7}{2})$.Now,$4CS^2 = 4 	imes [(4-3)^2 + (3-\frac{7}{2})^2] = 4 	imes [1^2 + (-\frac{1}{2})^2] = 4 	imes [1 + \frac{1}{4}] = 4 	imes \frac{5}{4} = 5$.Finally,$AO^2 + 9BG^2 + 4CS^2 = 0 + 8 + 5 = 13$.

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 $AO^2 + 9BG^2 + 4CS^2 =$

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