If `a + 2b + 3c = 0 " then " a/3+(2b)/3+c=0` and comparing with line ax + by + c, we get `x = 1/3 & y = 2/ 3` so there will be a point `(1/3,2/3)` from where each of the lines of the form ax + by + c = 0 will pass for the given relation between a,b,c . We can say if there exists a linear relation between a,b,c then the family of straight lines of the form of `ax + by +c ` pass through a fixed point .
If a , b, c are in H.P . then the line acx + bcy + ab = 0 passes through,

Options

A. (1,-2)
B. (1,2)
C. (-2,1)
D. (2,-1)

Answer

To solve the problem, we need to analyze the given conditions and derive the necessary equations step by step.

### Step 1: Understand the Given Condition
We start with the condition:
\[ a + 2b + 3c = 0 \]

### Step 2: Transform the Equation
We can divide the entire equation by 3 to simplify it:
\[ \frac{a}{3} + \frac{2b}{3} + c = 0 \]

### Step 3: Compare with the General Line Equation
Now, we compare this equation with the general line equation:
\[ ax + by + c = 0 \]
From our equation, we can identify:
- \( a = 1 \)
- \( b = 2 \)
- \( c = 3 \)

This gives us the coordinates:
- \( x = \frac{1}{3} \)
- \( y = \frac{2}{3} \)

Thus, we find the point:
\[ \left( \frac{1}{3}, \frac{2}{3} \right) \]

### Step 4: Conclusion from the First Part
From the above, we conclude that if there exists a linear relation between \( a, b, c \), then the family of straight lines of the form \( ax + by + c = 0 \) passes through the fixed point \( \left( \frac{1}{3}, \frac{2}{3} \right) \).

### Step 5: Analyze the Second Part of the Problem
Now, we need to analyze the case when \( a, b, c \) are in Harmonic Progression (HP). If \( a, b, c \) are in HP, then their reciprocals \( \frac{1}{a}, \frac{1}{b}, \frac{1}{c} \) are in Arithmetic Progression (AP).

### Step 6: Set Up the Equation
From the properties of AP, we know:
\[ 2\frac{1}{b} = \frac{1}{a} + \frac{1}{c} \]

### Step 7: Substitute into the Line Equation
We need to find the line \( acx + bcy + ab = 0 \). Dividing the entire equation by \( abc \), we can rewrite it as:
\[ \frac{x}{b} + \frac{y}{a} + \frac{1}{c} = 0 \]

### Step 8: Substitute \( \frac{1}{c} \)
Using the relationship from the AP condition:
\[ \frac{1}{c} = 2\frac{1}{b} - \frac{1}{a} \]
Substituting this into the line equation gives:
\[ \frac{x}{b} + \frac{y}{a} + \left( 2\frac{1}{b} - \frac{1}{a} \right) = 0 \]

### Step 9: Rearranging the Equation
Rearranging the equation, we have:
\[ \frac{x + 2}{b} + \frac{y - 1}{a} = 0 \]

### Step 10: Identify the Fixed Point
From this equation, we can identify the fixed point where the line passes through:
- The coefficient of \( \frac{1}{b} \) gives \( x + 2 = 0 \) → \( x = -2 \)
- The coefficient of \( \frac{1}{a} \) gives \( y - 1 = 0 \) → \( y = 1 \)

Thus, the point through which the line passes is:
\[ (-2, 1) \]

### Final Answer
The line \( acx + bcy + ab = 0 \) passes through the point:
\[ (-2, 1) \]

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