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Question
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
Options
\[a = \frac{b}{2}\]
\[b = \frac{a}{2}\]
ab = 1
\[a = \pm \sqrt{2}b\]
Solution
\[a = \pm \sqrt{2}b\]
The midpoints of BC and AC are \[D\left( \frac{a}{2}, 0 \right) \text { and } E\left( \frac{a}{2}, \frac{b}{2} \right)\].
Slope of AD= \[\frac{0 - b}{\frac{a}{2} - 0}\]
Slope of BE = \[\frac{- \frac{b}{2}}{\frac{- a}{2}}\]
It is given that the medians are perpendicular to each other.
\[\frac{0 - b}{\frac{a}{2} - 0} \times \frac{- \frac{b}{2}}{- \frac{a}{2}} = - 1\]
\[ \Rightarrow a = \pm \sqrt{2}b\]
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