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 :
$$2 - \sqrt 2 $$
Solution :
From the figure, we have
$$\eqalign{ & a = 2,\,\,b = 2\sqrt 2 ,\,\,c = 2 \cr & {x_1} = 0,\,{x_2} = 0,\,{x_3} = 2 \cr} $$
Now, $$x$$-co-ordinate of incentre is given as $$\frac{{a{x_1} + b{x_2} + c{x_3}}}{{a + b + c}}$$
$$ \Rightarrow \,x$$ -co-ordinate of incentre
$$\eqalign{ & = \frac{{2 \times 0 + 2\sqrt 2 .0 + 2.2}}{{2 + 2 + 2\sqrt 2 }} \cr & = \frac{2}{{2 + \sqrt 2 }} \cr & = 2 - \sqrt 2 \cr} $$
From the figure, we have
$$\eqalign{ & a = 2,\,\,b = 2\sqrt 2 ,\,\,c = 2 \cr & {x_1} = 0,\,{x_2} = 0,\,{x_3} = 2 \cr} $$
Now, $$x$$-co-ordinate of incentre is given as $$\frac{{a{x_1} + b{x_2} + c{x_3}}}{{a + b + c}}$$
$$ \Rightarrow \,x$$ -co-ordinate of incentre
$$\eqalign{ & = \frac{{2 \times 0 + 2\sqrt 2 .0 + 2.2}}{{2 + 2 + 2\sqrt 2 }} \cr & = \frac{2}{{2 + \sqrt 2 }} \cr & = 2 - \sqrt 2 \cr} $$