The median BE and AD of triangle with vertices \( \Large A \left(0,\ b\right), B \left(0,\ 0\right)\ and\ C \left(a,\ 0\right) \) are perpendicular to each other if;
A) \( \Large a=\frac{b}{2} \)
B) \( \Large b=\frac{a}{2} \)
C) ab = 1
D) \( \Large a\ =\ \pm \sqrt{2}b \)
Correct Answer:
D) \( \Large a\ =\ \pm \sqrt{2}b \)
Description for Correct answer: The vertices of triangle ABC are \( \Large A \left(0, b\right),\ B \left(0,\ 0\right)\ and\ C \left(a,\ 0\right) \)
mid points of E and D are \( \Large \left(\frac{a}{2},\ \frac{b}{2}\right)\ and\ \left(\frac{a}{2},\ 0\right) \)
The slope of median BE, \( \Large m_{1} = \frac{b}{a}\ and \)
Slope AD, \( \Large m_{2} = -\frac{2b}{a} \)
Since, the medians are perpendicular to each other.