To prove that the area of the triangle with vertices at \((p-4, p+5)\), \((p+3, p-2)\), and \((p, p)\) remains constant as \(p\) varies, we will use the formula for the area of a triangle given its vertices.
### Step 1: Identify the vertices
Let the vertices of the triangle be:
- \(A = (p-4, p+5)\)
- \(B = (p+3, p-2)\)
- \(C = (p, p)\)
### Step 2: Use the area formula for a triangle
The area \(A\) of a triangle with vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) is given by the formula:
\[
\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
\]
### Step 3: Substitute the coordinates into the formula
Substituting the coordinates of points \(A\), \(B\), and \(C\):
- \(x_1 = p - 4\), \(y_1 = p + 5\)
- \(x_2 = p + 3\), \(y_2 = p - 2\)
- \(x_3 = p\), \(y_3 = p\)
The area becomes:
\[
\text{Area} = \frac{1}{2} \left| (p-4)((p-2) - p) + (p+3)(p - (p+5)) + p((p+5) - (p-2)) \right|
\]
### Step 4: Simplify the expression
Calculating each term:
1. First term:
\[
(p-4)((p-2) - p) = (p-4)(-2) = -2p + 8
\]
2. Second term:
\[
(p+3)(p - (p+5)) = (p+3)(-5) = -5p - 15
\]
3. Third term:
\[
p((p+5) - (p-2)) = p(7) = 7p
\]
Now, combine these terms:
\[
\text{Area} = \frac{1}{2} \left| -2p + 8 - 5p - 15 + 7p \right|
\]
Combine like terms:
\[
\text{Area} = \frac{1}{2} \left| (7p - 2p - 5p) + (8 - 15) \right| = \frac{1}{2} \left| 0p - 7 \right| = \frac{1}{2} \cdot 7 = \frac{7}{2}
\]
### Step 5: Conclusion
The area of the triangle is \(\frac{7}{2}\) square units, which is a constant value independent of \(p\).
### Explanation of the result
This result indicates that the area of the triangle formed by the given vertices does not change as \(p\) varies. This is because the vertices are linearly dependent on \(p\), meaning that they move in a way that maintains the same relative distances and angles, thus keeping the area constant.
