Determine the angle between the lines whose equation are
`3x+y-7=0 and x+2y+9=0`,

Answer

To determine the angle between the lines given by the equations \(3x + y - 7 = 0\) and \(x + 2y + 9 = 0\), we will follow these steps:

### Step 1: Convert the equations to slope-intercept form
We need to rewrite both equations in the form \(y = mx + c\), where \(m\) is the slope.

**For the first line:**
\[ 3x + y - 7 = 0 \]
Rearranging gives:
\[ y = -3x + 7 \]
Thus, the slope \(m_1 = -3\).

**For the second line:**
\[ x + 2y + 9 = 0 \]
Rearranging gives:
\[ 2y = -x - 9 \]
\[ y = -\frac{1}{2}x - \frac{9}{2} \]
Thus, the slope \(m_2 = -\frac{1}{2}\).

### Step 2: Use the formula for the angle between two lines
The formula for the tangent of the angle \(\theta\) between two lines with slopes \(m_1\) and \(m_2\) is given by:
\[ \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]

### Step 3: Substitute the slopes into the formula
Substituting \(m_1 = -3\) and \(m_2 = -\frac{1}{2}\):
\[ \tan \theta = \left| \frac{-3 - \left(-\frac{1}{2}\right)}{1 + (-3) \left(-\frac{1}{2}\right)} \right| \]
This simplifies to:
\[ \tan \theta = \left| \frac{-3 + \frac{1}{2}}{1 + \frac{3}{2}} \right| \]

### Step 4: Simplify the expression
Calculating the numerator:
\[ -3 + \frac{1}{2} = -\frac{6}{2} + \frac{1}{2} = -\frac{5}{2} \]
Calculating the denominator:
\[ 1 + \frac{3}{2} = \frac{2}{2} + \frac{3}{2} = \frac{5}{2} \]
Thus, we have:
\[ \tan \theta = \left| \frac{-\frac{5}{2}}{\frac{5}{2}} \right| = \left| -1 \right| = 1 \]

### Step 5: Find the angle \(\theta\)
Since \(\tan \theta = 1\), we find:
\[ \theta = \tan^{-1}(1) = 45^\circ \]

### Conclusion
The angle between the lines is \(45^\circ\).

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