Question

If the medians $AD$ and $BE$ of the triangle with vertices $A(0, b), B(0, 0), C(a, 0)$ are mutually perpendicular, then

• A. $b = \sqrt{2a}$
• B. $a = \pm \sqrt{2b}$
• C. $b = - \sqrt{2a}$
• D. $b = a$

Answer 1

The Correct Option is B

$AD$ and $BE$ are perpendicular.

$D$ and $E$ are the mid points of $BC$ and $AC$ respectively.
$ \therefore$ Coordinates of
$D = \left( \frac{0+a}{2} , \frac{0+a}{2} \right) = \left( \frac{a}{2} , 0\right)$
$AD \bot BE$
$ \therefore$ Slope of AD $\times$ Slope of BE = -1
$\left(\frac{0-b}{\frac{a}{2} -0}\right)\times\left(\frac{\frac{b}{2} -0}{\frac{a}{2} -0}\right) = -1 $
$\Rightarrow \frac{b^{2}}{2} = \frac{a^{2}}{4} \Rightarrow a^{2} = 2b^{2} \Rightarrow a = \pm\sqrt{2} b.$