Question

Ortho centre of the triangle formed by the vertices $(\sqrt{13},\sqrt{5})$, $(\sqrt{7},-\sqrt{11})$ and $(-\sqrt{18},0)$ is

• A. $(0,0)$
• B. $(-\sqrt{11},5)$
• C. $(\sqrt{7},-\sqrt{11}-\sqrt{18},\sqrt{5}-\sqrt{11})$
• D. $(\sqrt{17}+\sqrt{7},\sqrt{5}-5)$

Answer 1

The correct option is C $(\sqrt{7},-\sqrt{11}-\sqrt{18},\sqrt{5}-\sqrt{11})$

Clearly given points lie on the circle $x^2+y^2=18$
$\therefore$ Circum center(s) = (0,0)
Centroid (G) $=(\frac{\sum x}{3},\frac{\sum y}{3})$
Orthocenter divides line joining ‘G’ and ‘S’ in the ratio 2:3 external
$\therefore G=(\frac{\sum x}{3},\frac{\sum y}{3})s=(0,0)$
$0=(\sum X,\sum Y)$
$(\sqrt{13}+\sqrt{7}-\sqrt{18},\sqrt{5}-\sqrt{11})$