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 \[BE=-1\] \[\Rightarrow \] \[-\frac{2b}{a}\times \frac{b}{a}=-1\] \[\Rightarrow \] \[{{a}^{2}}=2{{b}^{2}}\Rightarrow a=\pm \sqrt{2}b\]You need to login to perform this action.
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