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The medians AD and BE of the triangle with vertices A (0, b), B(0, 0) and C (a, 0) are mutually perpendicular if

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$b = \sqrt{2} a$

$a = \pm \sqrt{2} b$

$b = - \sqrt{2} a$

b =a

answer is B.

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Detailed Solution

Midpoint D of BC is (a/2, 0) and midpoint E of CA is (a/2, b/2)

Now slope of $A D = - 2 \frac{b}{a},$ slope of $B E = \frac{b}{a}$

$A D ⊥ B E \Rightarrow - \frac{2 b}{a} . \frac{b}{a} = - 1 \Rightarrow 2 b^{2} = a^{2} \Rightarrow a = \pm \sqrt{2} b$

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