retagged by
302 views

1 Answer

0 votes
0 votes

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.

edited by

Related questions

0 votes
0 votes
0 answers
1
gatecse asked Sep 18, 2019
339 views
If in a $\triangle ABC,\angle B=\dfrac{2\pi}{3},$ then $\cos A+\cos C$ lies in$\left[\:-\sqrt{3},\sqrt{3}\:\right]$$\left(\:-\sqrt{3},\sqrt{3}\:\right]$$\left(\:\frac{3}{...
0 votes
0 votes
0 answers
2
gatecse asked Sep 18, 2019
352 views
Which of the following relations is true for the following figure?$b^{2}=c(c+a)$$c^{2}=a(a+b)$$a^{2}=b(b+c)$All of these
0 votes
0 votes
2 answers
3
gatecse asked Sep 18, 2019
435 views
If $\tan\: x=p+1$ and $\tan\; y=p-1,$ then the value of $2\:\cot\:(x-y)$ is$2p$$p^{2}$$(p+1)(p-1)$$\frac{2p}{p^{2}-1}$
0 votes
0 votes
0 answers
4
gatecse asked Sep 18, 2019
362 views
The shaded region in the following diagram represents the relation$y\:\leq\: x$$\mid \:y\mid \:\leq\: \mid x\:\mid $$y\:\leq\: \mid x\:\mid$$\mid \:y\mid\: \leq\: x$