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

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$2 - \sqrt{2}$

Vedclass · Answer

(C) Let the vertices of the triangle be $A(x_1, y_1), B(x_2, y_2),$ and $C(x_3, y_3)$.The 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$$\frac{x_2+x_3}{2} = 1, \frac{y_2+y_3}{2} = 1$$\frac{x_3+x_1}{2} = 1, \frac{y_3+y_1}{2} = 0$Solving these,we get the vertices as $A(0, 0), B(0, 2),$ and $C(2, 0)$.The side lengths are $a = BC = \sqrt{(2-0)^2 + (0-2)^2} = \sqrt{4+4} = 2\sqrt{2}$,$b = AC = \sqrt{(2-0)^2 + (0-0)^2} = 2$,and $c = AB = \sqrt{(0-0)^2 + (2-0)^2} = 2$.The $x$-coordinate of the incenter $I$ is given by $x_I = \frac{ax_1 + bx_2 + cx_3}{a+b+c}$.$x_I = \frac{(2\sqrt{2})(0) + (2)(0) + (2)(2)}{2\sqrt{2} + 2 + 2} = \frac{4}{4 + 2\sqrt{2}} = \frac{2}{2 + \sqrt{2}}$.Rationalizing the denominator: $x_I = \frac{2(2 - \sqrt{2})}{(2 + \sqrt{2})(2 - \sqrt{2})} = \frac{2(2 - \sqrt{2})}{4 - 2} = \frac{2(2 - \sqrt{2})}{2} = 2 - \sqrt{2}$.

$i$. $P$	$A$. $(0,0)$
$ii$. $Q$	$B$. $(6,0)$
$iii$. $R$	$C$. $(-2,1)$
$iv$. $S$	$D$. $(-6,0)$
	$E$. $(-6,-3)$
	$F$. $(-6,3)$

If the midpoints of the sides of a triangle are $(0, 1), (1, 1),$ and $(1, 0)$,what is the $x$-coordinate of the incenter of the triangle?

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