The x- coordinate of the incentre of the tria | vedclass

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(B) Let the vertices of the triangle be $A, B, C$. The midpoints of the sides are given as $M_1(0,1), M_2(1,1), M_3(1,0)$.Since the midpoints form a t

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$2 - \sqrt{2}$

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(B) Let the vertices of the triangle be $A, B, C$. The midpoints of the sides are given as $M_1(0,1), M_2(1,1), M_3(1,0)$.Since the midpoints form a triangle,the vertices of the original triangle are formed by lines passing through these midpoints parallel to the opposite sides.The vertices are found to be $B(0,0), A(2,0), C(0,2)$.The side lengths are $c = AB = 2$,$a = BC = 2$,and $b = AC = \sqrt{(2-0)^2 + (0-2)^2} = \sqrt{4+4} = 2\sqrt{2}$.The $x-$ coordinate of the incentre $(I_x, I_y)$ is given by $I_x = \frac{ax_A + bx_B + cx_C}{a+b+c}$.Substituting the values: $I_x = \frac{2(2) + 2\sqrt{2}(0) + 2(0)}{2 + 2\sqrt{2} + 2} = \frac{4}{4 + 2\sqrt{2}} = \frac{2}{2 + \sqrt{2}}$.Rationalizing the denominator: $I_x = \frac{2(2 - \sqrt{2})}{(2 + \sqrt{2})(2 - \sqrt{2})} = \frac{2(2 - \sqrt{2})}{4 - 2} = \frac{2(2 - \sqrt{2})}{2} = 2 - \sqrt{2}$.

The $x-$ coordinate of the incentre of the triangle that has the coordinates of midpoints of its sides as $(0,1), (1,1)$ and $(1,0)$ is

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