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

The correct option is B

$2-\sqrt{2}$

Explanation of the correct option:

Given : The midpoints of a triangle are $(0,1),(1,1)$ and $(1,0)$,

By mid point theroem : The line segment in a triangle joining the midpoint of two sides of the triangle is said to be parallel to its third side and is also half of the length of the third side.

When these points are plotted, the sides of the triangle obtained will be $2,2$and $2\sqrt{2}$.

Therefore $△EDF$,vertices are $D(0,2),E(2,0)andF(0,0).$by figure.

$x_1=0,a=2,x_2=0,b=2\sqrt{2},x_3=2,c=2$

Thus the $x$ coordinate of the incentre of a triangle$\mathbf{=}\frac{{\mathbf{x}}_{\mathbf{1}}\mathbf{\times }\mathbf{a}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{x}}_{\mathbf{2}}\mathbf{\times }\mathbf{b}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{x}}_{\mathbf{3}}\mathbf{\times }\mathbf{c}}{\mathbf{a}\mathbf{+}\mathbf{b}\mathbf{+}\mathbf{c}}$

$=\frac{2\left(0\right)+2\sqrt{2}\left(0\right)+2\left(2\right)}{(2+2+2\sqrt{2})}$

$$\begin{gathered} =\frac{4}{4+2\sqrt{2}} \\ =\frac{2}{2+\sqrt{2}}\times \frac{2-\sqrt{2}}{2-\sqrt{2}} \end{gathered}$$

$=2-\sqrt{2}$

Therefore, the $x$ coordinate of the incentre of a triangle is $2-\sqrt{2}$.

Hence option B is the correct option.