• # question_answer The limiting position of the point of intersection of the straight lines, $3x+5y=1$ and $(2+c)\,x+5{{c}^{2}}y=1$ as c tends to one is: A) $\left( \frac{2}{5},\,\,-\frac{1}{25} \right)$ B) $\left( \frac{1}{2},\,\,-\frac{1}{10} \right)$ C) $\left( \frac{3}{8},\,\,-\frac{1}{40} \right)$ D) none of these

 $x=\underset{c\,\to \,1}{\mathop{Limit}}\,\frac{5\,(c-1)\,\,(c+1)}{5\,(c-1)\,\,(3c-2)}=\frac{2}{5};$ $y=\underset{c\,\to \,1}{\mathop{Limit}}\,\frac{1-c}{5\,(c-1)\,\,(3c+2)}=-\frac{1}{25}$