STATEMENT-1: The line `2009x+2010y+2011=0` where `2009a+2011c=0` passes through the point `((a)/(b),(b)/(c )),` where `abc ne 0.`
STATEMENT-2: If `2009a+2010+2011c=0,` then the straight line `ax+by+c=0, abc ne 0` passes through `((2009)/(2011),(2010)/(2011)).`

Options

A. Statement-1 is True, Statemetn-2 is True, Statemetn-2 is a correct explanation for Statement-5
B. Statement-1 is True, Statemetn-2 is True, Statement-2 is NOT a correct explanation for Statement-5
C. Statement-1 is True, Statement-2 is False
D. Statement-1 is False, Statement-2 is True

Answer

To solve the given problem, we will analyze both statements step by step.

### Step 1: Analyze Statement 1
The first statement is:
- The line \( 2009x + 2010y + 2011 = 0 \) passes through the point \( \left(\frac{a}{b}, \frac{b}{c}\right) \) where \( 2009a + 2011c = 0 \) and \( abc \neq 0 \).

#### Check if the point satisfies the line equation:
1. Substitute \( x = \frac{a}{b} \) and \( y = \frac{b}{c} \) into the line equation:
\[ 2009\left(\frac{a}{b}\right) + 2010\left(\frac{b}{c}\right) + 2011 = 0 \]
This simplifies to:
\[ \frac{2009a}{b} + \frac{2010b}{c} + 2011 = 0 \]
To eliminate the fractions, multiply through by \( bc \):
\[ 2009ac + 2010b^2 + 2011bc = 0 \]

2. From the condition \( 2009a + 2011c = 0 \), we can express \( c \) in terms of \( a \):
\[ c = -\frac{2009a}{2011} \]

3. Substitute \( c \) back into the equation:
\[ 2009a\left(-\frac{2009a}{2011}\right) + 2010b^2 + 2011b\left(-\frac{2009a}{2011}\right) = 0 \]
Simplifying gives:
\[ -\frac{2009^2 a^2}{2011} + 2010b^2 - \frac{2011 \cdot 2009ab}{2011} = 0 \]
\[ -2009^2 a^2 + 2010b^2 - 2009ab = 0 \]

4. This equation does not hold for all values of \( a, b, c \) under the condition \( abc \neq 0 \). Thus, Statement 1 is **false**.

### Step 2: Analyze Statement 2
The second statement is:
- If \( 2009a + 2010 + 2011c = 0 \), then the line \( ax + by + c = 0 \) passes through \( \left(\frac{2009}{2011}, \frac{2010}{2011}\right) \).

#### Check if the point satisfies the line equation:
1. Substitute \( x = \frac{2009}{2011} \) and \( y = \frac{2010}{2011} \) into the line equation:
\[ a\left(\frac{2009}{2011}\right) + b\left(\frac{2010}{2011}\right) + c = 0 \]
This simplifies to:
\[ \frac{2009a + 2010b + 2011c}{2011} = 0 \]
Thus:
\[ 2009a + 2010b + 2011c = 0 \]

2. Since we have the condition \( 2009a + 2010 + 2011c = 0 \), we can see that the equation holds true if we set \( b = 1 \).

3. Therefore, Statement 2 is **true**.

### Final Conclusion
- **Statement 1** is **false**.
- **Statement 2** is **true**.