To find the area of triangle PQR formed by the points where the line \(2x + 3y = 24\) meets the axes and the perpendicular bisector of segment PQ, we will follow these steps:
### Step 1: Find the coordinates of points P and Q
1. **Finding point P (x-intercept)**:
- Set \(y = 0\) in the equation \(2x + 3y = 24\).
- This gives us \(2x = 24\) or \(x = 12\).
- Thus, \(P = (12, 0)\).
2. **Finding point Q (y-intercept)**:
- Set \(x = 0\) in the equation \(2x + 3y = 24\).
- This gives us \(3y = 24\) or \(y = 8\).
- Thus, \(Q = (0, 8)\).
### Step 2: Find the midpoint S of segment PQ
- The coordinates of midpoint S can be calculated using the midpoint formula:
\[
S = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)
\]
where \(P = (12, 0)\) and \(Q = (0, 8)\).
- Therefore,
\[
S = \left(\frac{12 + 0}{2}, \frac{0 + 8}{2}\right) = (6, 4)
\]
### Step 3: Find the slope of line PQ
- The slope \(m_{PQ}\) of line PQ can be calculated as:
\[
m_{PQ} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{8 - 0}{0 - 12} = \frac{8}{-12} = -\frac{2}{3}
\]
### Step 4: Find the slope of the perpendicular bisector RS
- The slope of the perpendicular bisector \(m_{RS}\) is the negative reciprocal of the slope of PQ:
\[
m_{RS} = -\frac{1}{m_{PQ}} = \frac{3}{2}
\]
### Step 5: Find the equation of line RS
- Using point S (6, 4) and the slope \(m_{RS} = \frac{3}{2}\), we can write the equation in point-slope form:
\[
y - 4 = \frac{3}{2}(x - 6)
\]
- Simplifying this, we get:
\[
y - 4 = \frac{3}{2}x - 9 \implies y = \frac{3}{2}x - 5
\]
### Step 6: Find point R where line RS intersects the vertical line through (-2, 0)
- The vertical line through (-2, 0) has the equation \(x = -2\).
- Substituting \(x = -2\) into the equation of line RS:
\[
y = \frac{3}{2}(-2) - 5 = -3 - 5 = -8
\]
- Thus, \(R = (-2, -8)\).
### Step 7: Calculate the area of triangle PQR
- The area \(A\) of triangle formed by points \(P(x_1, y_1)\), \(Q(x_2, y_2)\), and \(R(x_3, y_3)\) can be calculated using the formula:
\[
A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
\]
- Substituting \(P(12, 0)\), \(Q(0, 8)\), and \(R(-2, -8)\):
\[
A = \frac{1}{2} \left| 12(8 - (-8)) + 0(-8 - 0) + (-2)(0 - 8) \right|
\]
\[
= \frac{1}{2} \left| 12(16) + 0 + 16 \right| = \frac{1}{2} \left| 192 + 16 \right| = \frac{1}{2} \times 208 = 104
\]
### Final Answer
The area of triangle PQR is \(104\) square units.
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