Question

The ortho centre of the triangle with vertices $ (-2,-6),\,(-2,4) $ and $ (1,3) $ is

• A. $ (3,\,1) $
• B. $ (1,\,1/3) $
• C. $ (1,3) $
• D. None of these

Answer 1

Answer: (C)

Let the vertices of the triangle are
$A(-2,-6), B(-2,4)$ and $C(1,3)$
Now, $A B=\sqrt{(-2+2)^{2}+(-6-4)^{2}}=10$
$B C=\sqrt{(-2-1)^{2}+(4-3)^{2}}=\sqrt{10}$ and
$C A=\sqrt{(1+2)^{2}+(3+6)^{2}}=\sqrt{90}$
Here, $A B^{2}=B C^{2}+C A^{2}$
$\therefore $ Triangle is right angled triangle at $C$.
$ \therefore $ $\angle C=90^{\circ}$ So, orthocenter is $(1,3)$.