The x-coordinate of the incentre of the triangle that has the coordinates of mid points of…

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:

Accepted Answer

From the figure, we have

$$a = 2,b = 2\sqrt 2 ,c = 2$$

$${x_1} = 0,\,{x^2} = 0,\,{x_3} = 2$$

Now, $x$-co-ordinate of incenter is given as

$${{a{x_1} + b{x_2} + c{x_3}} \over {a + b + c}}$$

$$ \Rightarrow x$$-coordinate of incentre

$$ = {{2 \times 0 + 2\sqrt 2 .0 + 2.2} \over {2 + 2 + 2\sqrt 2 }}$$

$=$ $${2 \over {2 + \sqrt 2 }} = 2 - \sqrt 2 $$

Solutions · 1