Find the equation of line passing through the point (6,-5) and perpendicular to the line 2x+3y+1=0. Show all work. Explain all steps.

1. Identify the slope of the line given by the equation: \(2x + 3y + 1 = 0\).
- Rewrite the equation in slope-intercept form (\(y = mx + b\)):
3y = -2x - 1 \Rightarrow y = -\frac{2}{3}x - \frac{1}{3}
- The slope (\(m_1\)) of the given line is \(-\frac{2}{3}\).

2. Find the slope of the line perpendicular to the given line.
- Perpendicular lines have slopes that are negative reciprocals:
m_2 = -\frac{1}{m_1} = -\left(-\frac{3}{2}\right) = \frac{3}{2}

3. Use the point-slope form of the line equation to find the equation of the perpendicular line that passes through the given point \((6, -5)\):
- Point-slope form:
y - y_1 = m(x - x_1)
- Substitute \(m = \frac{3}{2}\) and the point \((6, -5)\) into the equation:
y + 5 = \frac{3}{2}(x - 6)

4. Simplify to find the equation of the line:
- Distribute \(\frac{3}{2}\):
y + 5 = \frac{3}{2}x - 9
- Isolate \(y\):
y = \frac{3}{2}x - 9 - 5
y = \frac{3}{2}x - 14

5. Convert to standard form (\(Ax + By = C\)) with integers:
- Multiply everything by 2 to clear the fraction:
2y = 3x - 28
- Rearrange to standard form:
3x - 2y = 28

The equation of the line perpendicular to the given line and passing through (6, -5) is:

3x - 2y = 28