Question

The $x$-coordinate of the incentre of the triangle that has the coordinates of mid points of its sides as $(0,1),(1,1)$ and $(1,0)$ is

• A. $2-\sqrt{2}$
• B. $1+\sqrt{2}$
• C. $1-\sqrt{2}$
• D. $2+\sqrt{2}$

Answer 1

The correct option is A $2-\sqrt{2}$

$x$-coordinate $=\frac{a{x}_1+b{x}_2+c{x}_3}{a+b+c}$

$=\frac{2\times 2+2\sqrt{2}\times 0+2\times 0}{2+2+2\sqrt{2}}$

$=\frac{4}{4+2\sqrt{2}}=\frac{2}{2+\sqrt{2}}=2-\sqrt{2}.$

Alternate Solution:

$x$-coordinate $=r=(s-a)\tan \frac{A}{2}$

$=(\frac{4+2\sqrt{2}}{2}-2\sqrt{2})\tan \frac{\pi }{4}$

$=2-\sqrt{2}$