ISI2016-DCG-39

Question

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

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

Answer 1

Answer: $\mathbf B$

The mid-points of $BC$ and $AC$ are: $D\Big(\frac{a}{2}, 0\Big)$ and $E\Big(\frac{a}{2}\frac{b}{2}\Big)$

$$\;\text{Slope of AD}  =m_1= \frac{y_2-y_1}{x_2-x_1} =\Bigg(\frac{0-b}{\frac{a}{2}-0}\Bigg)$$

Similarly, $$\text{Slope of BE } =m_2=\frac{y_2-y_1}{x_2-x_1} = \Bigg(\frac{ \frac{-b}{2}} {\frac{-a}{2}}\Bigg)$$

It is given that the medians are mutually perpendicular to each other.

$$\therefore m_1\times m_2 = -1$$

$$\implies  \Bigg(\frac{0-b}{\frac{a}{2}-0}\Bigg)*\Bigg(\frac{ \frac{-b}{2}} {\frac{-a}{2}}\Bigg) = -1$$

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

$\therefore$ $\mathbf B$ is the correct answer.