JEE MAIN - Mathematics (2013 (Offline) - No. 11)
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 :
$$2 + \sqrt 2 $$
$$2 - \sqrt 2 $$
$$1 + \sqrt 2 $$
$$1 - \sqrt 2 $$
Explanation
From the figure, we have
$$a = 2,b = 2\sqrt 2 ,c = 2$$
$${x_1} = 0,\,{x^2} = 0,\,{x_3} = 2$$
_en_11_1.png)
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
$$ = {{2 \times 0 + 2\sqrt 2 .0 + 2.2} \over {2 + 2 + 2\sqrt 2 }}$$
$$=$$ $${2 \over {2 + \sqrt 2 }} = 2 - \sqrt 2 $$
$$a = 2,b = 2\sqrt 2 ,c = 2$$
$${x_1} = 0,\,{x^2} = 0,\,{x_3} = 2$$
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
$$ = {{2 \times 0 + 2\sqrt 2 .0 + 2.2} \over {2 + 2 + 2\sqrt 2 }}$$
$$=$$ $${2 \over {2 + \sqrt 2 }} = 2 - \sqrt 2 $$
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