question_answer

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) \[b=a\]  

B)         \[b=-2\sqrt{a}\]  

C)         \[a=\pm \sqrt{2b}\]           

D)         \[b=\sqrt{2}\,a\]

Correct Answer: C

Solution :

Here, the vertices of a triangle are \[A(0,b),B(0,0)\]and \[C(a,0)\] Mid point of \[BC,D=\left( \frac{a+0}{2},\frac{0+0}{2} \right)\]                                 \[=\left( \frac{a}{2},0 \right)\] Mid point of AC, \[E=\left( \frac{a}{2},\frac{b}{2} \right)\] Slope of \[AD=\frac{b-0}{0-\frac{a}{2}}=-\frac{2b}{a}\] Slope of \[BE=\frac{\frac{b}{2}-0}{\frac{a}{2}-0}=\frac{b}{a}\] Since, AD is perpendicular to BE \[\Rightarrow \]    slope of \[AD\times \]slope of \[BE=-1\] \[\Rightarrow \]    \[-\frac{2b}{a}\times \frac{b}{a}=-1\] \[\Rightarrow \]    \[{{a}^{2}}=2{{b}^{2}}\Rightarrow a=\pm \sqrt{2}b\]