A) 2 sq unit
B) 3 sq unit
C) 1 sq unit
D) None of these
Correct Answer: D
Solution :
Since, 3x + 4y -1 =0 is the perpendicular bisector of AB, therefore B is the image of A in this .line. \[\therefore \]Coordinates of B are \[\left( 1-\frac{2.3(3.1+4.2-1)}{25},2-\frac{2.4(3.1+4.2-1)}{25} \right)\] \[=\left( -\frac{7}{5},-\frac{6}{5} \right)\] Similarly, coordinates of C are\[\left( -\frac{3}{5},\frac{4}{5} \right).\] \[\therefore \]Area of \[\Delta ABC=\frac{1}{2}\left| \left| \begin{matrix} 1 & 2 & 1 \\ -\frac{7}{3} & -\frac{6}{5} & 1 \\ -\frac{3}{5} & \frac{4}{5} & 1 \\ \end{matrix} \right| \right|\] \[=\frac{1}{2}\left| 1\left( -\frac{6}{5}-\frac{4}{5} \right)-2\left( -\frac{7}{5}+\frac{3}{5} \right)+1\left( -\frac{28}{25}-\frac{18}{25} \right) \right|\] \[=\frac{1}{2}\left| \left( -2+\frac{8}{5}-\frac{46}{25} \right) \right|\] \[=\frac{28}{25}\]sq unit
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