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)

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

Answer 2

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 $m_1\, m_2 = - 1$
$\Rightarrow \:\:\: \frac{b}{a} \times \frac{-2b}{a} = - 1$
$\Rightarrow \:\:\: 2b^2 = a^2 \:\:\: \Rightarrow \:\: a = \pm \sqrt{2} b$