To solve the problem, we need to find the value of \( \lambda \) such that the area of triangle \( PQR \) is 2 square units, where point \( R \) divides the line segment joining points \( A(-5, 1) \) and \( B(3, 5) \) in the ratio \( \lambda:1 \).
### Step 1: Find the coordinates of point R using the section formula.
The coordinates of point \( R \) that divides the line segment \( AB \) in the ratio \( \lambda:1 \) can be calculated using the section formula:
\[
R\left( \frac{\lambda x_2 + x_1}{\lambda + 1}, \frac{\lambda y_2 + y_1}{\lambda + 1} \right)
\]
Here, \( A(-5, 1) \) and \( B(3, 5) \) give us \( x_1 = -5, y_1 = 1, x_2 = 3, y_2 = 5 \).
Thus, the coordinates of \( R \) are:
\[
R\left( \frac{\lambda \cdot 3 + (-5)}{\lambda + 1}, \frac{\lambda \cdot 5 + 1}{\lambda + 1} \right)
\]
### Step 2: Substitute the coordinates of points P and Q.
The coordinates of points \( P \) and \( Q \) are given as \( P(1, 5) \) and \( Q(7, 2) \).
### Step 3: Use the formula for the area of triangle PQR.
The area \( A \) of triangle \( PQR \) can be calculated using the formula:
\[
\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
\]
Substituting \( P(1, 5) \), \( Q(7, 2) \), and \( R\left( \frac{3\lambda - 5}{\lambda + 1}, \frac{5\lambda + 1}{\lambda + 1} \right) \):
\[
\text{Area} = \frac{1}{2} \left| 1(2 - \frac{5\lambda + 1}{\lambda + 1}) + 7\left(\frac{5\lambda + 1}{\lambda + 1} - 5\right) + \frac{3\lambda - 5}{\lambda + 1}(5 - 2) \right|
\]
### Step 4: Simplify the expression.
1. Calculate each term:
- First term:
\[
1 \left( 2 - \frac{5\lambda + 1}{\lambda + 1} \right) = 2 - \frac{5\lambda + 1}{\lambda + 1} = \frac{2(\lambda + 1) - (5\lambda + 1)}{\lambda + 1} = \frac{2\lambda + 2 - 5\lambda - 1}{\lambda + 1} = \frac{-3\lambda + 1}{\lambda + 1}
\]
- Second term:
\[
7\left(\frac{5\lambda + 1}{\lambda + 1} - 5\right) = 7\left(\frac{5\lambda + 1 - 5(\lambda + 1)}{\lambda + 1}\right) = 7\left(\frac{5\lambda + 1 - 5\lambda - 5}{\lambda + 1}\right) = 7\left(\frac{-4}{\lambda + 1}\right) = \frac{-28}{\lambda + 1}
\]
- Third term:
\[
\frac{3\lambda - 5}{\lambda + 1}(5 - 2) = \frac{3\lambda - 5}{\lambda + 1} \cdot 3 = \frac{9\lambda - 15}{\lambda + 1}
\]
2. Combine all terms:
\[
\text{Area} = \frac{1}{2} \left| \frac{-3\lambda + 1 - 28 + 9\lambda - 15}{\lambda + 1} \right| = \frac{1}{2} \left| \frac{6\lambda - 42}{\lambda + 1} \right|
\]
### Step 5: Set the area equal to 2 and solve for \( \lambda \).
Setting the area equal to 2:
\[
\frac{1}{2} \left| \frac{6\lambda - 42}{\lambda + 1} \right| = 2
\]
Multiplying both sides by 2:
\[
\left| \frac{6\lambda - 42}{\lambda + 1} \right| = 4
\]
This gives us two equations to solve:
1. \( \frac{6\lambda - 42}{\lambda + 1} = 4 \)
2. \( \frac{6\lambda - 42}{\lambda + 1} = -4 \)
### Step 6: Solve each equation.
**For the first equation:**
\[
6\lambda - 42 = 4(\lambda + 1)
\]
\[
6\lambda - 42 = 4\lambda + 4
\]
\[
2\lambda = 46 \implies \lambda = 23
\]
**For the second equation:**
\[
6\lambda - 42 = -4(\lambda + 1)
\]
\[
6\lambda - 42 = -4\lambda - 4
\]
\[
10\lambda = 38 \implies \lambda = \frac{19}{5}
\]
### Final Answer
The possible values of \( \lambda \) are \( 23 \) and \( \frac{19}{5} \).
