The straight lines `4a x+3b y+c=0` passes through? , where `a+b+c`=0 (a)`(4,3)` (b) `(1/4,1/3)` `(1/2,1/3)` (d) none of these

Options

A. (4,3)
B. (1/4,1/3)
C. (1/2,1/3)
D. none of these

Explanation

To determine through which point the straight line \(4ax + 3by + c = 0\) passes, given that \(a + b + c = 0\), we can follow these steps:

### Step 1: Substitute \(c\) in the equation
From the condition \(a + b + c = 0\), we can express \(c\) as:
\[ c = -a - b \]
Now, substitute this value of \(c\) into the line equation:
\[ 4ax + 3by + c = 0 \implies 4ax + 3by - a - b = 0 \]

### Step 2: Rearranging the equation
Rearranging the equation gives us:
\[ 4ax + 3by = a + b \]

### Step 3: Factor out common terms
We can factor out \(a\) and \(b\) from the left side:
\[ a(4x - 1) + b(3y - 1) = 0 \]
Since \(a\) and \(b\) are constants, for this equation to hold true, both terms must be equal to zero:
\[ 4x - 1 = 0 \quad \text{and} \quad 3y - 1 = 0 \]

### Step 4: Solve for \(x\) and \(y\)
Solving these equations gives us:
1. From \(4x - 1 = 0\):
\[ 4x = 1 \implies x = \frac{1}{4} \]
2. From \(3y - 1 = 0\):
\[ 3y = 1 \implies y = \frac{1}{3} \]

### Step 5: Conclusion
Thus, the point through which the line passes is:
\[ \left(\frac{1}{4}, \frac{1}{3}\right) \]
This corresponds to option (b).