Question

The medians $AD$ and $BE$ of a triangle with vertices $A(0,b),B(0,0)$ & $C(a,0)$ 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

Answer: (D)

We have,
$BE$ and $AD$ are the medians. So, $E$ and $D$ are the mid points of $AC$ and $BC$ respectively.

$\therefore$ Coordinates of $E=\left(\frac{a}{2},\frac{b}{2}\right)$
and coordinates of $D=\left(\frac{a}{2},0\right)$
Now, slope of median $BE=m_1=\frac{b}{a}$
Also, slope of median $AD=m_2=\frac{-2b}{a}$
Now, $m_1$ & $m_2$ are perpendicular if $m_1{m}_2=-1$
$\Rightarrow \frac{b}{a}\times \frac{-2b}{a}=-1$
$\Rightarrow 2b^2=a^2\Rightarrow a=\pm \sqrt{2}b$