The lines `3x+4y+7=0` and `4x+3y+5=0` are perpendicular.

Answer

To determine if the lines \(3x + 4y + 7 = 0\) and \(4x + 3y + 5 = 0\) are perpendicular, we will follow these steps:

### Step 1: Write the equations of the lines in slope-intercept form
We need to convert both equations into the form \(y = mx + b\), where \(m\) is the slope.

**For the first line \(3x + 4y + 7 = 0\):**
1. Rearrange the equation to isolate \(y\):
\[ 4y = -3x - 7 \]
2. Divide by 4:
\[ y = -\frac{3}{4}x - \frac{7}{4} \]
Thus, the slope \(m_1 = -\frac{3}{4}\).

**For the second line \(4x + 3y + 5 = 0\):**
1. Rearrange the equation to isolate \(y\):
\[ 3y = -4x - 5 \]
2. Divide by 3:
\[ y = -\frac{4}{3}x - \frac{5}{3} \]
Thus, the slope \(m_2 = -\frac{4}{3}\).

### Step 2: Check the product of the slopes
To check if the lines are perpendicular, we need to verify if the product of their slopes \(m_1\) and \(m_2\) equals \(-1\).

Calculate the product:
\[ m_1 \cdot m_2 = \left(-\frac{3}{4}\right) \cdot \left(-\frac{4}{3}\right) \]
This simplifies to:
\[ m_1 \cdot m_2 = \frac{3 \cdot 4}{4 \cdot 3} = 1 \]

### Step 3: Conclusion
Since \(m_1 \cdot m_2 = 1\) and not \(-1\), we conclude that the lines are **not perpendicular**.

### Summary of Steps:
1. Convert both line equations to slope-intercept form to find the slopes.
2. Calculate the product of the slopes.
3. Determine if the lines are perpendicular based on the product of the slopes.

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