If (0, 1), (1, 1) and (1, 0) are the midpoint | vedclass

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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 $(0, 1), (1, 1), (1, 0)$.For $x$-coordinates:

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$(2 - \sqrt{2}, 2 - \sqrt{2})$

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 $(0, 1), (1, 1), (1, 0)$.For $x$-coordinates: $x_1+x_2=0, x_2+x_3=2, x_3+x_1=2$. Solving these gives $x_1=0, x_2=0, x_3=2$.For $y$-coordinates: $y_1+y_2=2, y_2+y_3=2, y_3+y_1=0$. Solving these gives $y_1=0, y_2=2, y_3=0$.Thus,the vertices are $A(0, 0), B(0, 2)$ 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 incentre $I$ is given by $(\frac{ax_1+bx_2+cx_3}{a+b+c}, \frac{ay_1+by_2+cy_3}{a+b+c})$.$I = (\frac{2\sqrt{2}(0) + 2(0) + 2(2)}{2\sqrt{2}+2+2}, \frac{2\sqrt{2}(0) + 2(2) + 2(0)}{2\sqrt{2}+2+2}) = (\frac{4}{4+2\sqrt{2}}, \frac{4}{4+2\sqrt{2}})$.Rationalizing: $\frac{4}{4+2\sqrt{2}} = \frac{2}{2+\sqrt{2}} = \frac{2(2-\sqrt{2})}{4-2} = 2-\sqrt{2}$.So,the incentre is $(2-\sqrt{2}, 2-\sqrt{2})$.

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

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