Find the equation of the line which passes through the point (-2, 3) and is perpendicular to the line `2x + 3y + 4 = 0`

Answer

To find the equation of the line that passes through the point (-2, 3) and is perpendicular to the line given by the equation \(2x + 3y + 4 = 0\), we can follow these steps:

### Step 1: Determine the slope of the given line
The equation of the line is in the form \(Ax + By + C = 0\). Here, \(A = 2\), \(B = 3\), and \(C = 4\).

The slope \(m_1\) of the line can be calculated using the formula:
\[ m_1 = -\frac{A}{B} = -\frac{2}{3} \]

### Step 2: Find the slope of the perpendicular line
If two lines are perpendicular, the product of their slopes is \(-1\). Let \(m_2\) be the slope of the line we want to find. Therefore, we have:
\[ m_1 \cdot m_2 = -1 \]
Substituting \(m_1\):
\[ -\frac{2}{3} \cdot m_2 = -1 \]
Solving for \(m_2\):
\[ m_2 = \frac{3}{2} \]

### Step 3: Use the point-slope form to find the equation of the line
We know that the line passes through the point \((-2, 3)\) and has a slope of \(\frac{3}{2}\). We can use the point-slope form of the equation of a line:
\[ y - y_1 = m(x - x_1) \]
Substituting \(y_1 = 3\), \(m = \frac{3}{2}\), and \(x_1 = -2\):
\[ y - 3 = \frac{3}{2}(x + 2) \]

### Step 4: Simplify the equation
Now, we will simplify the equation:
\[ y - 3 = \frac{3}{2}x + 3 \]
Adding 3 to both sides:
\[ y = \frac{3}{2}x + 6 \]

### Step 5: Convert to standard form
To convert this equation to standard form \(Ax + By + C = 0\), we rearrange it:
\[ -\frac{3}{2}x + y - 6 = 0 \]
Multiplying through by 2 to eliminate the fraction:
\[ -3x + 2y - 12 = 0 \]
Rearranging gives:
\[ 3x - 2y + 12 = 0 \]

### Final Equation
Thus, the equation of the line that passes through the point (-2, 3) and is perpendicular to the line \(2x + 3y + 4 = 0\) is:
\[ 3x - 2y + 12 = 0 \]