A) \[2Tx+{{a}^{2}}y+2aT=0\]
B) \[2Tx-{{a}^{2}}y+2aT=0\]
C) \[2Tx-{{a}^{2}}y-2aT=0\]
D) None of these
Correct Answer: B
Solution :
If the line cuts off the axes at A and B, then area of triangle is \[\frac{1}{2}\times OA\times OB=T\] Þ \[\frac{1}{2}.a.OB=T\Rightarrow OB=\frac{2T}{a}\] Hence the equation of line is \[\frac{x}{-a}+\frac{y}{2T/a}=1\] \[\Rightarrow 2Tx-{{a}^{2}}y+2aT=0\].
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