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(B) Let the vertices of the triangle be $A(x_1, y_1)$,$B(x_2, y_2)$,and $C(x_3, y_3)$.Given midpoints are $M_1(0, 1)$,$M_2(1, 1)$,and $M_3(1, 0)$.Usin

Vedclass · Accepted Answer

$2 - \sqrt{2}$

Vedclass · Answer

(B) Let the vertices of the triangle be $A(x_1, y_1)$,$B(x_2, y_2)$,and $C(x_3, y_3)$.Given midpoints are $M_1(0, 1)$,$M_2(1, 1)$,and $M_3(1, 0)$.Using the midpoint formula,we have:$\frac{x_1+x_2}{2} = 0, \frac{y_1+y_2}{2} = 1 \implies x_1+x_2 = 0, y_1+y_2 = 2$$\frac{x_2+x_3}{2} = 1, \frac{y_2+y_3}{2} = 1 \implies x_2+x_3 = 2, y_2+y_3 = 2$$\frac{x_3+x_1}{2} = 1, \frac{y_3+y_1}{2} = 0 \implies x_3+x_1 = 2, y_3+y_1 = 0$Solving these equations:For $x$: $x_1+x_2=0, x_2+x_3=2, x_3+x_1=2$. Adding them: $2(x_1+x_2+x_3) = 4 \implies x_1+x_2+x_3 = 2$.$x_3 = 2-0 = 2, x_1 = 2-2 = 0, x_2 = 2-2 = 0$.For $y$: $y_1+y_2=2, y_2+y_3=2, y_3+y_1=0$. Adding them: $2(y_1+y_2+y_3) = 4 \implies y_1+y_2+y_3 = 2$.$y_3 = 2-2 = 0, y_1 = 2-2 = 0, y_2 = 2-0 = 2$.Vertices are $A(0, 0)$,$B(0, 2)$,and $C(2, 0)$.Side lengths are $a = BC = \sqrt{(2-0)^2 + (0-2)^2} = \sqrt{4+4} = 2\sqrt{2}$,$b = AC = \sqrt{(2-0)^2 + (0-0)^2} = 2$,$c = AB = \sqrt{(0-0)^2 + (2-0)^2} = 2$.The $x$-coordinate of the incentre is $\frac{ax_1 + bx_2 + cx_3}{a+b+c} = \frac{2\sqrt{2}(0) + 2(0) + 2(2)}{2\sqrt{2} + 2 + 2} = \frac{4}{2\sqrt{2} + 4} = \frac{2}{\sqrt{2} + 2} = \frac{2(2-\sqrt{2})}{4-2} = 2 - \sqrt{2}$.

If the midpoints of the sides of a triangle are $(0, 1)$,$(1, 1)$,and $(1, 0)$,find the $x$-coordinate of its incentre.

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