• # question_answer The equation of straight line passing through $(-a,\ 0)$ and making the triangle with axes of area ?T? is      [RPET 1987] 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

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 $\frac{x}{-a}+\frac{y}{2T/a}=1$                    $\Rightarrow 2Tx-{{a}^{2}}y+2aT=0$.