Question

The family of straight lines \(4ax+3by+c = 0\) such that \(a+b+c = 0\) (where a, b, c are real constants) are concurrent at the point...

• A. \((4,3)\)
• B. \((\frac{1}{2},\frac{1}{3})\)
• C. \((\frac{1}{4},\frac{1}{3})\)
• D. \((\frac{1}{3},\frac{1}{2})\)

Hint:

Rewrite the family as a(4x - 1) + b(3y - 1) = 0 using c = -a - b.

Answer 1

The Correct Option is C

Step 1: Idea
The condition $a + b + c = 0$ says that $x = y = 1$ makes $ax + by + c$ vanish. Here the coefficients are $4a$ and $3b$.

Step 2: Rescale
Put $X = 4x$ and $Y = 3y$. The line becomes $aX + bY + c = 0$ with $a + b + c = 0$, which always passes through $X = 1$, $Y = 1$.

Step 3: Convert back
$4x = 1$ gives $x = \frac{1}{4}$ and $3y = 1$ gives $y = \frac{1}{3}$.

Step 4: Verify
Substitute $\left(\frac{1}{4}, \frac{1}{3}\right)$: $4a \cdot \frac{1}{4} + 3b \cdot \frac{1}{3} + c = a + b + c = 0$. For option (B), $(\frac{1}{2}, \frac{1}{3})$ gives $2a + b + c = a \neq 0$ in general.

Final Answer:
The common point is (1/4, 1/3). This is option (C). \[ \boxed{\text{(C) }\left(\frac{1}{4},\frac{1}{3}\right)} \]