The set of lines `ax+by+c=0` where `3a+2b+4c=0` are concurrent at the point

Options

A. (3,2)
B. (2,4)
C. `(3//4,1//2)`
D. None

Answer

To solve the problem, we need to find the point at which the set of lines given by the equation \( ax + by + c = 0 \) is concurrent, given the condition \( 3a + 2b + 4c = 0 \).

### Step-by-Step Solution:

1. **Understand the Condition**: We have the equation of lines in the form \( ax + by + c = 0 \) and a condition \( 3a + 2b + 4c = 0 \). This condition implies that the coefficients \( a, b, c \) are related in such a way that the lines are concurrent.

2. **Rearranging the Condition**: We can rewrite the condition \( 3a + 2b + 4c = 0 \) by dividing the entire equation by 4:
\[ \frac{3}{4}a + \frac{1}{2}b + c = 0 \]
This gives us a new equation that we can compare with the original line equation.

3. **Comparing with the Line Equation**: The original line equation is \( ax + by + c = 0 \). We can denote this as Equation (1). The rearranged condition can be denoted as Equation (2):
\[ \frac{3}{4}a + \frac{1}{2}b + c = 0 \quad \text{(Equation 2)} \]

4. **Identifying the Coefficients**: From Equation (1) and Equation (2), we can identify:
- For \( a \): \( x = \frac{3}{4} \)
- For \( b \): \( y = \frac{1}{2} \)

5. **Finding the Concurrent Point**: The point of concurrency of the lines can be expressed as:
\[ \left( \frac{3}{4}, \frac{1}{2} \right) \]

6. **Final Answer**: Therefore, the point at which the set of lines is concurrent is:
\[ \boxed{\left( \frac{3}{4}, \frac{1}{2} \right)} \]