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 point through which the line represented by the equation \(2ax + by + 3c = 0\) always passes when \(a\), \(b\), 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 relationship mathematically:
\[ b = \frac{a + c}{2} \]
This implies:
\[ 2b = a + c \quad \text{or} \quad a - 2b + c = 0 \]

2. **Substituting into the Line Equation**:
The line equation given is:
\[ 2ax + by + 3c = 0 \]
We will substitute \(c\) in terms of \(a\) and \(b\) into this equation.

3. **Expressing \(c\)**:
From the A.P. condition, we can express \(c\) as:
\[ c = 2b - a \]

4. **Substituting \(c\) into the Line Equation**:
Substitute \(c\) into the line equation:
\[ 2ax + by + 3(2b - a) = 0 \]
Simplifying this gives:
\[ 2ax + by + 6b - 3a = 0 \]

5. **Rearranging the Equation**:
Rearranging the equation, we get:
\[ 2ax - 3a + by + 6b = 0 \]
This can be grouped as:
\[ 2a(x - \frac{3}{2}) + b(y + 6) = 0 \]

6. **Finding the Point**:
To find the point through which this line passes, we can set \(x = \frac{3}{2}\) and \(y = -6\):
\[ 2a\left(\frac{3}{2} - \frac{3}{2}\right) + b(-6 + 6) = 0 \]
This simplifies to \(0 = 0\), confirming that the line passes through the point \(\left(\frac{3}{2}, -6\right)\).

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