Question

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

• A. 4
• B. 8
• C. 12
• D. - 12

Answer 1

Answer: (A), (B), (C), (D)

As H, G and C are collinear
$\therefore \left|\begin{array}{ccc}4& b& 1\\ b& 2b-8& 1\\ -4& 8& 1\end{array}\right|=0\Rightarrow \left|\begin{array}{ccc}4& b& 1\\ b-4& b-8& 0\\ -(b+4)& 16-2b& 0\end{array}\right|=0$
$\Rightarrow (b-4)(16-2b)+(b+4)(b-8)=0$
$\Rightarrow 2(b-4)(8-b)+(b+4)(b-8)=0$
$\Rightarrow (8-b)[(2b-8)-(b+4)]=0$
$\Rightarrow (8-b)(b-12)=0$
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 $\frac{16+b}{3}=2b-8\Rightarrow b=8$
But no common value of ' $b$ ' is possible $\Rightarrow A,B,C,D$