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Q.
In an acute triangle ABC, if the coordinates of orthocentre 'H' are (4, b), centroid 'G' are
(b, 2b - 8) and circumcentre 'S' are (- 4, 8), then 'b' can not be
Straight Lines
Solution:
As H, G and C are collinear
$\therefore \begin{vmatrix}4 & b & 1 \\ b & 2 b -8 & 1 \\ -4 & 8 & 1\end{vmatrix}=0 \Rightarrow\begin{vmatrix}4 & b & 1 \\ b -4 & b -8 & 0 \\ -( b +4) & 16-2 b & 0\end{vmatrix}=0$
$\Rightarrow (b - 4)(16 - 2b) + (b + 4)(b - 8) = 0$
$\Rightarrow 2(b - 4)(8 - b) + (b + 4)(b - 8) = 0$
$\Rightarrow (8 - b)[(2b - 8) - (b + 4)] = 0$
$\Rightarrow (8 - b)(b - 12) = 0$
Hence b = 8 or 12, which is wrong because collinearity does not explain centroid, orthocentre and circumcentre
$\therefore \frac{-8+4}{3}=b \Rightarrow b=\frac{-4}{3}$
and $ \frac{16+b}{3}=2 b-8 \Rightarrow b=8$
But no common value of ' $b$ ' is possible $\Rightarrow A , B , C , D$