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

Options

A. (-5,4)
B. (5 , -4)
C. (4 , -5)
D. none of these

Answer

To find the limiting position of the point of intersection of the lines \(3x + 4y = 1\) and \((1+c)x + 3c^2 y = 2\) as \(c\) tends to 1, we can follow these steps:

### Step 1: Express \(x\) in terms of \(y\) from the first equation
The first equation is:
\[ 3x + 4y = 1 \]
We can solve for \(x\):
\[ 3x = 1 - 4y \implies x = \frac{1 - 4y}{3} \]

**Hint:** Rearranging the equation helps isolate one variable, making substitution easier.

### Step 2: Substitute \(x\) in the second equation
Now, substitute \(x\) into the second equation:
\[ (1+c)x + 3c^2 y = 2 \]
Substituting for \(x\):
\[ (1+c)\left(\frac{1 - 4y}{3}\right) + 3c^2 y = 2 \]

**Hint:** When substituting, ensure to distribute correctly and maintain the equation's structure.

### Step 3: Simplify the equation
Multiply through by 3 to eliminate the fraction:
\[ (1+c)(1 - 4y) + 9c^2 y = 6 \]
Expanding this:
\[ 1 + c - 4(1+c)y + 9c^2 y = 6 \]
Combine like terms:
\[ 1 + c - 4y - 4cy + 9c^2 y = 6 \]
Rearranging gives:
\[ -4y + (9c^2 - 4c)y = 5 - c \]

**Hint:** Always combine like terms carefully to simplify the equation.

### Step 4: Factor out \(y\)
Rearranging the equation:
\[ y(-4 + (9c^2 - 4c)) = 5 - c \]
Thus,
\[ y = \frac{5 - c}{-4 + (9c^2 - 4c)} \]

**Hint:** Isolate \(y\) to understand how it behaves as \(c\) approaches 1.

### Step 5: Take the limit as \(c\) approaches 1
Now, we find the limit of \(y\) as \(c\) approaches 1:
\[ y = \frac{5 - 1}{-4 + (9(1)^2 - 4(1))} = \frac{4}{-4 + 9 - 4} = \frac{4}{1} = 4 \]

**Hint:** Substituting the limit directly into the simplified equation gives the value of \(y\).

### Step 6: Find \(x\) using the value of \(y\)
Now that we have \(y = 4\), substitute back to find \(x\):
\[ x = \frac{1 - 4(4)}{3} = \frac{1 - 16}{3} = \frac{-15}{3} = -5 \]

**Hint:** Always substitute back into the original equation to find the corresponding variable.

### Conclusion
The limiting position of the point of intersection as \(c\) tends to 1 is:
\[ \boxed{(-5, 4)} \]