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

• \[a = \pm \sqrt{2}b\]

Answer 1

\[a = \pm \sqrt{2}b\]

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.

\[\frac{0 - b}{\frac{a}{2} - 0} \times \frac{- \frac{b}{2}}{- \frac{a}{2}} = - 1\]

\[ \Rightarrow a = \pm \sqrt{2}b\]