Let A(h, 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 (1) `{1,""3}` (2) `{0,""2}` (3) `{-1,""3}` (4) `{-3,-2}`

Options

A. {1, 3}
B. {0, 2}
C. {-1, 3}
D. {-3, -2}

Answer

To solve the problem step by step, we need to determine the possible values of \( k \) for the triangle formed by points \( A(h, k) \), \( B(1, 1) \), and \( C(2, 1) \) with \( AC \) as the hypotenuse and the area of the triangle equal to 1.

### Step 1: Understand the triangle configuration
Given points:
- \( A(h, k) \)
- \( B(1, 1) \)
- \( C(2, 1) \)

Since \( AC \) is the hypotenuse, we can visualize the triangle with \( B \) and \( C \) lying on the same horizontal line (y = 1).

### Step 2: Calculate the lengths of the sides using the distance formula
Using the distance formula, we can find the lengths of the sides:
- Length \( AC \):
\[ AC = \sqrt{(h - 2)^2 + (k - 1)^2} \]
- Length \( AB \):
\[ AB = \sqrt{(h - 1)^2 + (k - 1)^2} \]
- Length \( BC \):
\[ BC = \sqrt{(2 - 1)^2 + (1 - 1)^2} = \sqrt{1} = 1 \]

### Step 3: Use the Pythagorean theorem
Since \( AC \) is the hypotenuse, we apply the Pythagorean theorem:
\[ AC^2 = AB^2 + BC^2 \]
Substituting the lengths:
\[ (h - 2)^2 + (k - 1)^2 = (h - 1)^2 + (k - 1)^2 + 1^2 \]

### Step 4: Simplify the equation
Expanding both sides:
\[ (h - 2)^2 + (k - 1)^2 = (h - 1)^2 + (k - 1)^2 + 1 \]
The \( (k - 1)^2 \) terms cancel out:
\[ (h - 2)^2 = (h - 1)^2 + 1 \]
Expanding:
\[ h^2 - 4h + 4 = h^2 - 2h + 1 + 1 \]
This simplifies to:
\[ h^2 - 4h + 4 = h^2 - 2h + 2 \]
Cancelling \( h^2 \) from both sides:
\[ -4h + 4 = -2h + 2 \]
Rearranging gives:
\[ -4h + 2h = 2 - 4 \]
\[ -2h = -2 \implies h = 1 \]

### Step 5: Calculate the area of the triangle
The area \( A \) of triangle \( ABC \) is given by:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, the base \( BC = 1 \) and the height is the vertical distance from point \( A \) to line \( BC \), which is \( |k - 1| \):
\[ A = \frac{1}{2} \times 1 \times |k - 1| = \frac{1}{2} |k - 1| \]
Setting the area equal to 1:
\[ \frac{1}{2} |k - 1| = 1 \]
Multiplying both sides by 2:
\[ |k - 1| = 2 \]

### Step 6: Solve for \( k \)
Removing the absolute value gives two cases:
1. \( k - 1 = 2 \) → \( k = 3 \)
2. \( k - 1 = -2 \) → \( k = -1 \)

### Conclusion
The possible values of \( k \) are \( -1 \) and \( 3 \).

### Final Answer
The set of values which \( k \) can take is:
\[ \{-1, 3\} \]