Is the line `3x + 4y + 7 = 0` perpendicular to the line `28x - 21y + 50 = 0 `? 

Answer

To determine if the lines represented by the equations \(3x + 4y + 7 = 0\) and \(28x - 21y + 50 = 0\) are perpendicular, we will follow these steps:

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

**For the first line:**
Starting with the equation:
\[ 3x + 4y + 7 = 0 \]
We can isolate \(y\):
\[ 4y = -3x - 7 \]
Now, divide by 4:
\[ y = -\frac{3}{4}x - \frac{7}{4} \]
Thus, the slope \(m_1\) of the first line is:
\[ m_1 = -\frac{3}{4} \]

**For the second line:**
Starting with the equation:
\[ 28x - 21y + 50 = 0 \]
We can isolate \(y\):
\[ -21y = -28x - 50 \]
Now, divide by -21:
\[ y = \frac{28}{21}x + \frac{50}{21} \]
Simplifying \(\frac{28}{21}\):
\[ y = \frac{4}{3}x + \frac{50}{21} \]
Thus, the slope \(m_2\) of the second line is:
\[ m_2 = \frac{4}{3} \]

### Step 2: Check if the slopes are negative reciprocals
For two lines to be perpendicular, the product of their slopes must equal \(-1\):
\[ m_1 \cdot m_2 = -1 \]
Substituting the values we found:
\[ -\frac{3}{4} \cdot \frac{4}{3} \]
Calculating the product:
\[ = -\frac{3 \cdot 4}{4 \cdot 3} = -1 \]

### Conclusion
Since the product of the slopes \(m_1\) and \(m_2\) is \(-1\), the lines are indeed perpendicular.

### Final Answer:
Yes, the lines \(3x + 4y + 7 = 0\) and \(28x - 21y + 50 = 0\) are perpendicular.

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