Question
The limiting position of the point of intersection of the lines $$3x+4y=1$$ and $$\left( {1 + c} \right)x + 3{c^2}y = 2$$ as $$c$$ tends to 1 is :
A.
$$\left( { - 5,\,4} \right)$$
B.
$$\left( {5,\, - 4} \right)$$
C.
$$\left( {4,\, - 5} \right)$$
D.
none of these
Answer :
$$\left( { - 5,\,4} \right)$$
Solution :
$$\eqalign{
& {\text{Solving, }}x = \frac{{3{c^2} - 8}}{{9{c^2} - 4c - 4}},\,y = \frac{{5 - c}}{{9{c^2} - 4c - 4}} \cr
& {\text{Now,}}\,\,\mathop {\lim }\limits_{c \to 1} \frac{{3{c^2} - 8}}{{9{c^2} - 4c - 4}} = \frac{{ - 5}}{1} = - 5\,\,{\text{and}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\mathop {\lim }\limits_{c \to 1} \frac{{5 - c}}{{9{c^2} - 4c - 4}} = \frac{4}{1} = 4 \cr} $$