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

The correct option is D

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

The midpoints of BC and AC are $D(\frac{a}{2},0)$ and $E(\frac{a}{2},\frac{b}{2})$

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.

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

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