ISI2015-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. $b=\pm \sqrt{2}b$
• C. $b= – \sqrt{2}a$
• D. $b=a$

Answer 1

Answer: $\mathbf B$

Explanation:

$\mathrm {AD}$ and $\mathrm {BE}$ are $\perp$s

$\mathrm D$ and $\mathrm E$ are the mid-points of $\mathrm {BC}$ and $\mathrm {AC}$ respectively.

So, coordinates of $\mathrm {D = (0+a/2, 0+0/2)$ = $(a/2,0)}$

$\mathrm {AD \perp BE}$

$\therefore\; \text{Slope of AD} \times \;\text{Slope of BE} = -1$ tex

$\mathrm {= \frac{(0-b)}{(a/2-0)} *\frac{ (b/2-0)}{(a/2-0)} = -1}$

$\mathrm{= \frac{b^2}{2} = \frac{a^2}{4} \\\Rightarrow a^2 = 2b^2}$

$\mathrm {\Rightarrow a = \pm\sqrt{(2b)}}$

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