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
(a) $a=\frac{b}{2}$

(b) $b=\frac{a}{2}$

(c) ab = 1

(d) $a=\pm \sqrt{2}b$

Answer 1

(d) $a=\pm \sqrt{2}b$

The midpoints of BC and AC are $D\left(\frac{a}{2},0\right)\mathrm{and}E\left(\frac{a}{2},\frac{b}{2}\right)$.
Slope of AD= $\frac{0-b}{{\displaystyle \frac{a}{2}}-0}$
Slope of BE = $\frac{-{\displaystyle \frac{b}{2}}}{{\displaystyle \frac{-a}{2}}}$

It is given that the medians are perpendicular to each other.

$$\begin{gathered} \frac{0-b}{{\displaystyle \frac{a}{2}-0}}\times \frac{-{\displaystyle \frac{b}{2}}}{-{\displaystyle \frac{a}{2}}}=-1 \\ \Rightarrow a=\pm \sqrt{2}b \end{gathered}$$