If O, G, S are respectively the orthocenter centroide circumcentre of a triangle whose vertices are A (2, 3), B(2, 4) and C (4, 3) then AO²+9BG²+4CS² =

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$\frac{77}{36}$

$\frac{8}{9}$

$\frac{5}{4}$

answer is B.

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Detailed Solution

Given A (2, 3) B ( 2, 4) C (4, 3)

$\begin{array}{l} S l o p e o f A B = \frac{4 - 3}{2 - 2} = \infty \\ S l o p e o f \bar{A C} = \frac{3 - 3}{4 - 2} = 0 \\ ∴ O = A (2, 3) \\ S = M i d p o int o f B C = (\frac{2 + 4}{2}, \frac{4 + 3}{2}) \Rightarrow (3, \frac{7}{2}) \\ G = (\frac{2 + 2 + 4}{3}, \frac{3 + 4 + 3}{3}) \Rightarrow (\frac{8}{3}, \frac{10}{3}) \end{array}$

$\begin{array}{l} O A = O \\ B G = \sqrt{{(\frac{8}{3} - 2)}^{2} + {(\frac{10}{3} - 4)}^{2}} = \sqrt{\frac{4}{9} + \frac{4}{9}} = \sqrt{\frac{8}{9}} \\ C S = \frac{1}{2} B C = \frac{1}{2} \sqrt{{(4 - 2)}^{2} + {(3 - 4)}^{2}} = \frac{1}{2} \sqrt{4 + 1} = \frac{\sqrt{5}}{2} \\ A O^{2} + 9 B G^{2} + 4 C S^{2} = 0 + 9 (\frac{8}{9}) + 4 (\frac{5}{4}) = 13 \end{array}$