To determine the value of \( p \) for which the points \( (p+1, 1) \), \( (2p+1, 3) \), and \( (2p+2, 2p) \) are collinear, we will use the concept of the area of a triangle formed by these three points. If the area is zero, the points are collinear.
### Step 1: Set up the area formula
The area \( A \) of a triangle formed by three points \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) can be calculated using the determinant formula:
\[
A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
\]
For our points:
- \( (x_1, y_1) = (p+1, 1) \)
- \( (x_2, y_2) = (2p+1, 3) \)
- \( (x_3, y_3) = (2p+2, 2p) \)
### Step 2: Substitute the points into the formula
Substituting the points into the area formula gives:
\[
A = \frac{1}{2} \left| (p+1)(3 - 2p) + (2p+1)(2p - 1) + (2p+2)(1 - 3) \right|
\]
### Step 3: Simplify the expression
Now we will simplify the expression inside the absolute value:
1. Calculate \( (p+1)(3 - 2p) \):
\[
= 3p + 3 - 2p^2 - 2p = -2p^2 + p + 3
\]
2. Calculate \( (2p+1)(2p - 1) \):
\[
= 4p^2 - 2p + 2p - 1 = 4p^2 - 1
\]
3. Calculate \( (2p+2)(1 - 3) \):
\[
= (2p+2)(-2) = -4p - 4
\]
Now combine these results:
\[
-2p^2 + p + 3 + 4p^2 - 1 - 4p - 4 = 2p^2 - 3p - 2
\]
### Step 4: Set the area to zero
Since the points are collinear, the area must be zero:
\[
\frac{1}{2} | 2p^2 - 3p - 2 | = 0
\]
This implies:
\[
2p^2 - 3p - 2 = 0
\]
### Step 5: Solve the quadratic equation
Now we will solve the quadratic equation \( 2p^2 - 3p - 2 = 0 \) using the factorization method or the quadratic formula.
Using the quadratic formula:
\[
p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 2 \), \( b = -3 \), and \( c = -2 \):
\[
p = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot (-2)}}{2 \cdot 2}
\]
\[
= \frac{3 \pm \sqrt{9 + 16}}{4}
\]
\[
= \frac{3 \pm \sqrt{25}}{4}
\]
\[
= \frac{3 \pm 5}{4}
\]
This gives us two solutions:
1. \( p = \frac{8}{4} = 2 \)
2. \( p = \frac{-2}{4} = -\frac{1}{2} \)
### Final Answer
Thus, the values of \( p \) for which the points are collinear are:
\[
p = 2 \quad \text{and} \quad p = -\frac{1}{2}
\]
