If the points `(1, 2)` and `(3, 4)` were to be on the opposite side of the `3x - 5y + alpha = 0`, then:

Options

A. `7 lt a lt 11`
B. a=7
C. a=1
D. `a lt 7 " or " a gt 11`

Answer

To determine the values of \(\alpha\) such that the points \((1, 2)\) and \((3, 4)\) lie on opposite sides of the line defined by the equation \(3x - 5y + \alpha = 0\), we can follow these steps:

### Step 1: Substitute the points into the line equation
We will evaluate the line equation \(3x - 5y + \alpha = 0\) for both points.

1. For the point \((1, 2)\):
\[ L_1 = 3(1) - 5(2) + \alpha = 3 - 10 + \alpha = -7 + \alpha \]

2. For the point \((3, 4)\):
\[ L_2 = 3(3) - 5(4) + \alpha = 9 - 20 + \alpha = -11 + \alpha \]

### Step 2: Determine the conditions for opposite sides
For the points to be on opposite sides of the line, the products of \(L_1\) and \(L_2\) must be negative:
\[ L_1 \cdot L_2 < 0 \]

Substituting the expressions we found:
\[ (-7 + \alpha)(-11 + \alpha) < 0 \]

### Step 3: Solve the inequality
To solve the inequality, we first find the roots of the equation:
\[ (-7 + \alpha) = 0 \quad \Rightarrow \quad \alpha = 7 \]
\[ (-11 + \alpha) = 0 \quad \Rightarrow \quad \alpha = 11 \]

Now we have the critical points \(\alpha = 7\) and \(\alpha = 11\). We will test the intervals defined by these points:

1. **Interval 1:** \(\alpha < 7\)
- Choose \(\alpha = 0\):
\[ (-7 + 0)(-11 + 0) = 7 \cdot 11 > 0 \quad \text{(not valid)} \]

2. **Interval 2:** \(7 < \alpha < 11\)
- Choose \(\alpha = 9\):
\[ (-7 + 9)(-11 + 9) = 2 \cdot (-2) < 0 \quad \text{(valid)} \]

3. **Interval 3:** \(\alpha > 11\)
- Choose \(\alpha = 12\):
\[ (-7 + 12)(-11 + 12) = 5 \cdot 1 > 0 \quad \text{(not valid)} \]

### Step 4: Conclusion
The values of \(\alpha\) for which the points \((1, 2)\) and \((3, 4)\) lie on opposite sides of the line \(3x - 5y + \alpha = 0\) are:
\[ \alpha \in (7, 11) \]