A) \[5{{a}^{2}}sq\]. units
B) \[\frac{5}{2}{{a}^{2}}sq.\]units
C) \[\frac{25{{a}^{2}}}{2}sq.\]units
D) None of these
Correct Answer: B
Solution :
Let the co-ordinates of the third vertex be \[(2a,\,\,t)\]. \[AC=BC\Rightarrow t=\sqrt{4{{a}^{2}}+{{(a-t)}^{2}}}\Rightarrow \]\[t=\frac{5a}{2}\] So the coordinates of third vertex C are \[\left( 2a,\frac{5a}{2} \right)\] Therefore area of the triangle \[=\pm \frac{1}{2}\left| \,\begin{matrix} 2a & \frac{5a}{2} & 1 \\ 2a & 0 & 1 \\ 0 & a & 1 \\ \end{matrix}\, \right|=\left| \,\begin{matrix} a & \frac{5a}{2} & 1 \\ 0 & -\frac{5a}{2} & 0 \\ 0 & a & 1 \\ \end{matrix}\, \right|=\frac{5{{a}^{2}}}{2}sq.\]units.
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