Question

Orthocentre of the triangle whose vertices are $(2,-\frac{1}{2}),(\frac{1}{2},-\frac{1}{2}),(2,\frac{\sqrt{3}-1}{2})$ is

• A. $(2,-\frac{1}{2})$
• B. $(\frac{1}{2},-\frac{1}{2})$
• C. $(2,-\frac{\sqrt{3}-1}{2})$
• D. $(\frac{1}{2},-\frac{\sqrt{3}-1}{2})$

Answer 1

The correct option is A $(2,-\frac{1}{2})$

Let the points be $A(2,-\frac{1}{2}),B(\frac{1}{2},-\frac{1}{2})$ and $C(2,\frac{\sqrt{3-1}}{2})$ and let $AD,BE,CF$ be the altitudes From $A,B,C$respectively.

Slope of $AB=\frac{y_2-y_1}{x_2-x_1}=0$

So slope of $AF=-infinity$.So,AF is perpendicular to $X$ axis.It's equation is $x=k$\\ ie.$x=2⟹\left(1\right)$

Slope of $AC$=infinity

So slope of $BE$ is $0$

So it is parallel to x-axis.

It's equation is $y=k.$\\ ie.$y=-\frac{1}{2}⟹\left(2\right)$

Solving $\left(1\right)$ and $\left(2\right)$ we get the ortho centre as $(2,-\frac{1}{2})$.