Write the nature of the equations `6x-2y+9 = 0 and 3x - y +12 = 0`

Options

A. coincident
B. parallel
C. intersecting exactly at one point
D. perpendicular to each other

Answer

To determine the nature of the equations \(6x - 2y + 9 = 0\) and \(3x - y + 12 = 0\), we will follow these steps:

### Step 1: Identify coefficients
We need to identify the coefficients \(a_1\), \(b_1\), and \(c_1\) from the first equation and \(a_2\), \(b_2\), and \(c_2\) from the second equation.

For the first equation \(6x - 2y + 9 = 0\):
- \(a_1 = 6\)
- \(b_1 = -2\)
- \(c_1 = 9\)

For the second equation \(3x - y + 12 = 0\):
- \(a_2 = 3\)
- \(b_2 = -1\)
- \(c_2 = 12\)

### Step 2: Calculate ratios
Next, we will calculate the ratios \( \frac{a_1}{a_2} \), \( \frac{b_1}{b_2} \), and \( \frac{c_1}{c_2} \).

1. Calculate \( \frac{a_1}{a_2} \):
\[ \frac{a_1}{a_2} = \frac{6}{3} = 2 \]

2. Calculate \( \frac{b_1}{b_2} \):
\[ \frac{b_1}{b_2} = \frac{-2}{-1} = 2 \]

3. Calculate \( \frac{c_1}{c_2} \):
\[ \frac{c_1}{c_2} = \frac{9}{12} = \frac{3}{4} \]

### Step 3: Analyze the ratios
Now we will analyze the ratios:
- \( \frac{a_1}{a_2} = 2 \)
- \( \frac{b_1}{b_2} = 2 \)
- \( \frac{c_1}{c_2} = \frac{3}{4} \)

### Step 4: Determine the nature of the equations
According to the conditions for the nature of the equations:
- If \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \) and \( \frac{c_1}{c_2} \) is not equal to \( \frac{a_1}{a_2} \) (or \( \frac{b_1}{b_2} \)), then the lines are **parallel**.

Here, we have:
- \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = 2 \)
- \( \frac{c_1}{c_2} = \frac{3}{4} \)

Since \( \frac{c_1}{c_2} \) is not equal to \( \frac{a_1}{a_2} \) (which is 2), the two equations are **parallel**.

### Conclusion
The nature of the equations \(6x - 2y + 9 = 0\) and \(3x - y + 12 = 0\) is that they are **parallel**.

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