Question

Let $A(1,1), B(1,-1), C(-1,1)$ be the vertices of $\triangle A B C$. Let $S$ be the circum-centre, $O$ be the orthocentre and $I$ be the incentre of the $\triangle ABC$. Then $IS + OS =$

• A. $4$
• B. $2$
• C. $2+2 \sqrt{2}$
• D. $2+\sqrt{2}$

Answer 1

Answer: (B)

It is clear that $\triangle A B C$ is a right angled triangle at $A$ .
$\therefore $ Circum-centre, $S=$ Mid-point of hypotenuse $B C$
$=\left(\frac{-1+1}{2}, \frac{1-1}{2}\right)=(0,0)$
Orthocentre, $O=$ Coordinate of vertices at which right angle is there
$=A=(1,1)$
and Incentre
$I=\begin{pmatrix}\frac{2 \sqrt{2} \times 1+2 \times 1+2 \times(-1)}{2 \sqrt{2}+2+2} \\\frac{2 \sqrt{2} \times 1+2 \times 1+2 \times-1}{2 \sqrt{2}+2+2}
\end{pmatrix} $
$=\left(\frac{2 \sqrt{2}}{2 \sqrt{2}+4}, \frac{2 \sqrt{2}}{2 \sqrt{2}+4}\right)=(\sqrt{2}-1, \sqrt{2}-1)$
$I S+O S=\sqrt{(\sqrt{2}-1)^{2}+(\sqrt{2}-1)^{2}}+\sqrt{(1)^{2}+(1)^{2}} $
$=\sqrt{2}(\sqrt{2}-1)+\sqrt{2}=2-\sqrt{2}+\sqrt{2}=2$