Question

The limiting position of the point of intersection of the straight lines $3x+5y=1$ and $(2+c)x+5c^{2}y$ = 1 as $c$ tends to one is

• A. $(\frac{2}{5},-\frac{1}{25})$
• B. $(\frac{1}{2},-\frac{1}{10})$
• C. $(\frac{3}{8},-\frac{1}{40})$
• D. none

Answer 1

The correct option is A $(\frac{2}{5},-\frac{1}{25})$

GIven: $3x+5y=1,(2+c)x+{5c}^{2}y=1$

To find: Limiting position of point of intersection as c tends to be one.

Sol: $3x+5y=1⟹x=\frac{1-5y}{3}$

$(2+c)x+{5c}^{2}y=1⟹x=\frac{1-{5c}^{2}y}{(2+c)}$

$\frac{1-5y}{3}=\frac{1-{5c}^{2}y}{(2+c)}⟹y={lim}_{c\to 1}\frac{c-1}{10+5c-{15c}^2}$

$⟹y={lim}_{c\to 1}\frac{1}{5}\frac{(c-1)}{(3c+2)(1-c)}⟹y=\frac{-1}{25}$

$⟹x=\frac{2}{5}$

Hence correct answer is $(\frac{2}{5},\frac{-1}{25})$