Find the equation of the line passing through the point `(-4,5)` and the point of intersection of the lines `4x-3y+7=0` and `2x+3y+5=0`.

Answer

To find the equation of the line passing through the point \((-4, 5)\) and the point of intersection of the lines \(4x - 3y + 7 = 0\) and \(2x + 3y + 5 = 0\), we can follow these steps:

### Step 1: Find the point of intersection of the two lines.

We have the equations:
1. \(4x - 3y + 7 = 0\) (Equation 1)
2. \(2x + 3y + 5 = 0\) (Equation 2)

We can solve these equations simultaneously to find the point of intersection.

From Equation 1, we can express \(y\) in terms of \(x\):
\[ 3y = 4x + 7 \implies y = \frac{4x + 7}{3} \]

Now substitute this expression for \(y\) into Equation 2:
\[ 2x + 3\left(\frac{4x + 7}{3}\right) + 5 = 0 \]
\[ 2x + 4x + 7 + 5 = 0 \]
\[ 6x + 12 = 0 \implies 6x = -12 \implies x = -2 \]

Now substitute \(x = -2\) back into the expression for \(y\):
\[ y = \frac{4(-2) + 7}{3} = \frac{-8 + 7}{3} = \frac{-1}{3} \]

Thus, the point of intersection is \((-2, -\frac{1}{3})\).

### Step 2: Use the point-slope form of the line equation.

Now we have two points: \((-4, 5)\) and \((-2, -\frac{1}{3})\). We can find the slope \(m\) of the line that passes through these points.

The slope \(m\) is given by:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-\frac{1}{3} - 5}{-2 - (-4)} = \frac{-\frac{1}{3} - \frac{15}{3}}{-2 + 4} = \frac{-\frac{16}{3}}{2} = -\frac{8}{3} \]

### Step 3: Write the equation of the line.

Using the point-slope form of the line equation:
\[ y - y_1 = m(x - x_1) \]
Substituting \(m = -\frac{8}{3}\), \(x_1 = -4\), and \(y_1 = 5\):
\[ y - 5 = -\frac{8}{3}(x + 4) \]

### Step 4: Simplify the equation.

Distributing the slope:
\[ y - 5 = -\frac{8}{3}x - \frac{32}{3} \]
Adding 5 (which is \(\frac{15}{3}\)) to both sides:
\[ y = -\frac{8}{3}x - \frac{32}{3} + \frac{15}{3} \]
\[ y = -\frac{8}{3}x - \frac{17}{3} \]

### Final Equation

To express this in standard form, we can multiply through by 3 to eliminate the fractions:
\[ 3y = -8x - 17 \]
Rearranging gives:
\[ 8x + 3y + 17 = 0 \]

Thus, the equation of the line is:
\[ \boxed{8x + 3y + 17 = 0} \]