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

Answer 1

Answer: (B)

Given mid-points of a triangle are $(0,1),(1,1)$ and $(1,0)$. Plotting these points on a graph paper and make a triangle.
So, the sides of a triangle will be $2,2$ and $\sqrt{2^2+2^2}$ i.e. $2\sqrt{2}$

$x$ -coordinate of incentre $=\frac{2\times 0+2\sqrt{2}\cdot 0+2\cdot 2}{2+2+2\sqrt{2}}$
$=\frac{2}{2+\sqrt{2}}\times \frac{2-\sqrt{2}}{2-\sqrt{2}}=2-\sqrt{2}$