The medians AD and BE of a triangle with vert | vedclass

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(C) The vertices of the triangle are $A(0, b)$,$B(0, 0)$,and $C(a, 0)$.$D$ is the midpoint of $BC$,so $D = (\frac{0+a}{2}, \frac{0+0}{2}) = (\frac{a}{

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Both $(a)$ and $(b)$

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(C) The vertices of the triangle are $A(0, b)$,$B(0, 0)$,and $C(a, 0)$.$D$ is the midpoint of $BC$,so $D = (\frac{0+a}{2}, \frac{0+0}{2}) = (\frac{a}{2}, 0)$.$E$ is the midpoint of $AC$,so $E = (\frac{0+a}{2}, \frac{b+0}{2}) = (\frac{a}{2}, \frac{b}{2})$.The slope of median $AD$ is $m_1 = \frac{0-b}{a/2 - 0} = \frac{-b}{a/2} = -\frac{2b}{a}$.The slope of median $BE$ is $m_2 = \frac{b/2 - 0}{a/2 - 0} = \frac{b/2}{a/2} = \frac{b}{a}$.Since the medians are perpendicular,$m_1 	imes m_2 = -1$.$(-\frac{2b}{a}) 	imes (\frac{b}{a}) = -1$.$-\frac{2b^2}{a^2} = -1$.$a^2 = 2b^2$.$a = \pm \sqrt{2}b$.

The medians $AD$ and $BE$ of a triangle with vertices $A(0, b)$,$B(0, 0)$,and $C(a, 0)$ are perpendicular to each other,if

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