Find the equation of the line through the intersection of `y+x=9 and 2x-3y+7=0`, and perpendicular to the line `2y-3x-5=0`.

Answer

To find the equation of the line through the intersection of the lines \(y + x = 9\) and \(2x - 3y + 7 = 0\), and perpendicular to the line \(2y - 3x - 5 = 0\), we can follow these steps:

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

The equations of the lines are:
1. \(y + x = 9\) (Let's call this Line 1)
2. \(2x - 3y + 7 = 0\) (Let's call this Line 2)

We can express \(y\) from Line 1:
\[ y = 9 - x \]

Now, substitute this expression for \(y\) into Line 2:
\[ 2x - 3(9 - x) + 7 = 0 \]
Expanding this:
\[ 2x - 27 + 3x + 7 = 0 \]
Combine like terms:
\[ 5x - 20 = 0 \]
Solving for \(x\):
\[ 5x = 20 \implies x = 4 \]

Now substitute \(x = 4\) back into the equation for \(y\):
\[ y = 9 - 4 = 5 \]

Thus, the point of intersection is \((4, 5)\).

### Step 2: Find the slope of the line perpendicular to \(2y - 3x - 5 = 0\)

First, we need to find the slope of the line given by \(2y - 3x - 5 = 0\). Rewriting it in slope-intercept form:
\[ 2y = 3x + 5 \implies y = \frac{3}{2}x + \frac{5}{2} \]
The slope \(m_1\) of this line is \(\frac{3}{2}\).

The slope \(m_2\) of the line we want (which is perpendicular) is given by:
\[ m_1 \cdot m_2 = -1 \implies m_2 = -\frac{2}{3} \]

### Step 3: Write the equation of the line through the intersection point with the found slope

Using the point-slope form of the equation of a line:
\[ y - y_1 = m(x - x_1) \]
Where \((x_1, y_1) = (4, 5)\) and \(m = -\frac{2}{3}\):
\[ y - 5 = -\frac{2}{3}(x - 4) \]

### Step 4: Simplify the equation

Distributing the slope:
\[ y - 5 = -\frac{2}{3}x + \frac{8}{3} \]
Adding 5 to both sides:
\[ y = -\frac{2}{3}x + \frac{8}{3} + 5 \]
Converting 5 to a fraction:
\[ 5 = \frac{15}{3} \]
So:
\[ y = -\frac{2}{3}x + \frac{8}{3} + \frac{15}{3} = -\frac{2}{3}x + \frac{23}{3} \]

### Step 5: Convert to standard form

To convert to standard form \(Ax + By + C = 0\):
\[ \frac{2}{3}x + y - \frac{23}{3} = 0 \]
Multiplying through by 3 to eliminate the fraction:
\[ 2x + 3y - 23 = 0 \]

Thus, the final equation of the line is:
\[ \boxed{2x + 3y - 23 = 0} \]