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 and explain the result.

Answer

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-by-Step Solution:

1. **Identify the vertices**:
Let the vertices of the triangle be:
- \(A = (p-4, p+5)\)
- \(B = (p+3, p-2)\)
- \(C = (p, p)\)

2. **Use the area formula**:
The area \(A\) of a triangle with vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) is given by:
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]

3. **Substituting the coordinates**:
Substitute the coordinates of points \(A\), \(B\), and \(C\) into the area formula:
\[ \text{Area} = \frac{1}{2} \left| (p-4)((p-2) - p) + (p+3)(p - (p+5)) + (p)( (p+5) - (p-2)) \right| \]

4. **Simplifying the expression**:
Calculate each term:
- First term:
\[ (p-4)((p-2) - p) = (p-4)(-2) = -2p + 8 \]
- Second term:
\[ (p+3)(p - (p+5)) = (p+3)(-5) = -5p - 15 \]
- Third term:
\[ p((p+5) - (p-2)) = p(7) = 7p \]

5. **Combining the terms**:
Combine the results:
\[ \text{Area} = \frac{1}{2} \left| (-2p + 8) + (-5p - 15) + 7p \right| \]
Simplifying further:
\[ = \frac{1}{2} \left| 0p + (8 - 15) \right| = \frac{1}{2} \left| -7 \right| = \frac{7}{2} \]

6. **Conclusion**:
The area of the triangle is \(\frac{7}{2}\), which is a constant value regardless of the value of \(p\).

### Explanation of the Result:
The area of the triangle remains constant as \(p\) varies because the terms involving \(p\) cancel each other out in the area formula. This indicates that the shape and size of the triangle do not change as we move the vertices along the line defined by their coordinates.