Question

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

• A. $(4,-5)$
• B. $(-5,4)$
• C. $(5,-4)$
• D. $(-4,5)$

Answer 1

The correct option is B $(-5,4)$

$3x+4y=1$ ---(1)

And $(1+c)x+3c^{2}y=2$ ---(2)

Limiting position of intersection of this lines as c tends to 1

So, equation 2 is

$2x+3y=2$

And $3x+4y=1$ ---(1)

From 2 & 1,

$\Rightarrow y=4$ and $x=-5$