question_answer

The x−coordinate of the in centre of the triangle that has the coordinates of mid points of its sides as (0, 1) (1, 1) and (1, 0) is:     AIEEE Solevd Paper-2013

A) \[2+\sqrt{2}\]   

B)        \[2-\sqrt{2}\]

C) \[1+\sqrt{2}\]   

D)        \[1-\sqrt{2}\]

Correct Answer: B

Solution :

On solving a = 0 and b = 2 \[{{I}_{x}}=\frac{0\times 2+0\times \sqrt{8}+2\times 2}{2+2+2\sqrt{2}}\] \[{{I}_{x}}=\frac{4}{4+2\sqrt{2}}\] \[{{I}_{x}}=\frac{2}{2+\sqrt{2}}\times \frac{2-\sqrt{2}}{2-\sqrt{2}}\] \[{{I}_{x}}=\frac{2\left( 2-\sqrt{2} \right)}{2}\]\[\Rightarrow \]\[2-\sqrt{2}\]