Question

The orthocentre of the triangle formed by the vertices $(5,0),(0,0)$ and $(\frac{5}{2},\frac{5\sqrt{3}}{2})$ is

• A. $(2,3)$
• B. $(\frac{5}{2},\frac{5}{2\sqrt{3}})$
• C. $(\frac{5}{6},\frac{5}{2\sqrt{3}})$
• D. $(\frac{5}{2},\frac{5}{\sqrt{3}})$

Answer 1

The correct option is B $(\frac{5}{2},\frac{5}{2\sqrt{3}})$

Given coordinates of the vertices of triangle are $(5,0),(0,0)$ and $(\frac{5}{2},\frac{5\sqrt{3}}{2})$
Assuming $A=(5,0),B=(0,0),C=(\frac{5}{2},\frac{5\sqrt{3}}{2})$

Now, finding the sidelengths
$AB=5\text{units}BC=\sqrt{{\left(\frac{5}{2}\right)}^2+{\left(\frac{5\sqrt{3}}{2}\right)}^2}=5\text{units}AC=\sqrt{{\left(\frac{5}{2}\right)}^2+{\left(\frac{5\sqrt{3}}{2}\right)}^2}=5\text{units}$

So, the triangle is equilateral.
Orthocentre
$H\equiv (\frac{x_1\tan A+x_2\tan B+x_3\tan C}{\tan A+\tan B+\tan C},\frac{y_1\tan A+y_2\tan B+y_3\tan C}{\tan A+\tan B+\tan C})$
$\because \tan A=\tan B=\tan C=\sqrt{3}$
$\Rightarrow H\equiv (\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})$
$\Rightarrow H\equiv (\frac{5+\frac{5}{2}}{3},\frac{\frac{5\sqrt{3}}{2}}{3})\therefore H\equiv (\frac{5}{2},\frac{5}{2\sqrt{3}})$