Let A(1,k),B(1,1) and C(2,1) be the vertices of a right angled triangle with AC as its hypotenuse. If the area of the triangle is 1, then the set of values which k can take is given by:

Options

A. (1,3)
B. (0,2)
C. (-1,3)
D. (-3,-2)

Answer

To solve the problem step by step, we need to find the values of \( k \) such that the area of triangle \( ABC \) is equal to 1, with \( AC \) as the hypotenuse.

### Step 1: Identify the coordinates of the points
The vertices of the triangle are given as:
- \( A(1, k) \)
- \( B(1, 1) \)
- \( C(2, 1) \)

### Step 2: Calculate the length of side \( BC \)
The length of side \( BC \) can be calculated using the distance formula:
\[ BC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting the coordinates of points \( B \) and \( C \):
\[ BC = \sqrt{(2 - 1)^2 + (1 - 1)^2} = \sqrt{1^2 + 0^2} = \sqrt{1} = 1 \]

### Step 3: Calculate the length of side \( AB \)
The length of side \( AB \) is:
\[ AB = \sqrt{(1 - 1)^2 + (k - 1)^2} = \sqrt{0^2 + (k - 1)^2} = |k - 1| \]

### Step 4: Use the area formula of the triangle
The area \( A \) of triangle \( ABC \) can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, we can take \( BC \) as the base and \( AB \) as the height:
\[ 1 = \frac{1}{2} \times |k - 1| \times 1 \]
This simplifies to:
\[ 1 = \frac{1}{2} |k - 1| \]

### Step 5: Solve for \( |k - 1| \)
Multiplying both sides by 2 gives:
\[ 2 = |k - 1| \]

### Step 6: Set up cases for \( |k - 1| = 2 \)
This absolute value equation leads to two cases:
1. \( k - 1 = 2 \)
2. \( k - 1 = -2 \)

### Step 7: Solve each case
**Case 1:**
\[ k - 1 = 2 \implies k = 3 \]

**Case 2:**
\[ k - 1 = -2 \implies k = -1 \]

### Step 8: Conclusion
The possible values of \( k \) are:
\[ k = -1 \quad \text{and} \quad k = 3 \]
Thus, the set of values which \( k \) can take is \( \{-1, 3\} \).