The limiting position of the point of intersection of the lines $3 x + 4 y = 1$ and $(1 + c) x + 3 c^{2} y = 2$ as $c$ tends to $1$ is

(- 5, 4)

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(5, - 4)

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(4, - 5)

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none of these

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Solution

The correct option is A $(- 5, 4)$
$3 x + 4 y = 1$
$x = \frac{1 - 4 y}{3}$
Substituting value of $x$ in other equation, we get
$(1 + c) (\frac{1 - 4 y}{3}) + 3 c^{2} y = 2$
$y = \frac{5 - c}{9 c^{2} - 4 c - 4}$
As $c$ tending to $1.$
Therefore, $y = 4$ and $x = - 5$
Hence, option 'A' is correct.

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