If a , b , c are in A.P ., then the equation `2ax+by +3c = 0 ` always passes through

Options

A. `(-(3)/2 , - 6 )`
B. `(3/2,-6)`
C. `(-3/2,6)`
D. `(3/2 , 6)`

Answer

To solve the problem, we need to determine the coordinates (x, y) through which the equation \(2ax + by + 3c = 0\) passes, given that \(a\), \(b\), and \(c\) are in arithmetic progression (A.P.).

### Step-by-Step Solution:

1. **Understanding A.P.**: Since \(a\), \(b\), and \(c\) are in A.P., we can express this condition mathematically:
\[ b = \frac{a + c}{2} \]
This implies:
\[ a + c = 2b \]

2. **Rearranging the A.P. Condition**: We can rearrange the equation \(a + c = 2b\) to form:
\[ a - 2b + c = 0 \tag{1} \]

3. **Substituting in the Given Equation**: The given equation is:
\[ 2ax + by + 3c = 0 \]
We want to compare this with equation (1). To facilitate this, we can multiply equation (1) by 3:
\[ 3a - 6b + 3c = 0 \tag{2} \]

4. **Comparing Coefficients**: Now we have two equations:
- From equation (2): \(3a - 6b + 3c = 0\)
- From the given equation: \(2ax + by + 3c = 0\)

We can compare the coefficients of \(a\), \(b\), and \(c\) in both equations.

5. **Setting Up the Coefficient Comparisons**:
- Coefficient of \(a\): \(2x = 3\)
- Coefficient of \(b\): \(y = -6\)
- Coefficient of \(c\) is already equal in both equations.

6. **Solving for x and y**:
- From \(2x = 3\):
\[ x = \frac{3}{2} \]
- From \(y = -6\):
\[ y = -6 \]

7. **Conclusion**: Therefore, the coordinates through which the equation \(2ax + by + 3c = 0\) always passes is:
\[ (x, y) = \left(\frac{3}{2}, -6\right) \]

### Final Answer:
The equation \(2ax + by + 3c = 0\) always passes through the point \(\left(\frac{3}{2}, -6\right)\).

---