Prove that the circumcenter, orthocenter, incenter and centroid of the triangle formed by the points $A (- 1, 11)$ $B (- 9, - 8)$ and $C (15, - 2)$ are collinear, without actually finding any of them.

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Solution

Distance of AB, BC and CA
$\begin{matrix} \Rightarrow A B = \sqrt{{(- 8)}^{2} + {(19)}^{2}} \Rightarrow A B = \sqrt{64 + 361} = \sqrt{425} u n i t \Rightarrow B C = \sqrt{{(15 + 9)}^{2} + {(- 6)}^{2}} = \sqrt{{(24)}^{2} + 36} = \sqrt{612} u n i t \Rightarrow a n d A C = \sqrt{{(16)}^{2} + {(13)}^{2}} = \sqrt{256 + 109} = \sqrt{425} u n i t \end{matrix}$
$∴ Δ A B C$ are Isosceles triangle
Hence, centroid, orthocentre, incentre and circumcentre lie on same line.

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Which of the following statements are correct?

1. For triangle orthocenter,circumcenter and

centroid are collinear.

2. Centroid divides the line joining the circumcenter $&$ orthocenter

in the ratio of $2 : 1$ i.e., $\frac{C S}{O C} = \frac{2}{1}$

Where C-Coordinate of centroid

O-Coordinate of orthocenter

S--Coordinate of circumcenter

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