question_answer

The \[x-\]co-ordinate of the incentre of the triangle where the mid points of the sides are (0, 1), (1,1) and (1, 0), is:

A)  \[2+\sqrt{2}\]  

B)         \[1+\sqrt{2}\]

C)  \[2-\sqrt{2}\]   

D)         \[1-\sqrt{2}\]

E)  \[3-\sqrt{2}\]

Correct Answer: C

Solution :

Since, (0, 1), (1, 1) and (1, 0) are the mid points of sides AB, BC and CA respectively. \[\therefore \]Co-ordinates of A, B and C are (0, 0), (0, 2) and (2, 0) respectively. Now,\[AB=2,\text{ }BC=2\sqrt{2},\text{ }CA=2\] \[\therefore \] \[x\]co-ordinate of incentre \[=\frac{0+0+2.2}{2+2\sqrt{2}+2}\]                          \[\left( \because \frac{a{{x}_{1}}+b{{y}_{1}}+c{{z}_{1}}}{a+b+c} \right)\] \[=\frac{2}{2+\sqrt{2}}=2-\sqrt{2}\]