If `A(9,-9),B(1,3)` are the ends of a right angled isosceles triangle then the third vertex is

Options

A. (8,-2)
B. (-8,2)
C. (8,-8)
D. None of these

Answer

To find the third vertex \( C(h, k) \) of the right-angled isosceles triangle with vertices \( A(9, -9) \) and \( B(1, 3) \), we will use the properties of the triangle and the distance formula.

### Step 1: Understand the Properties of the Triangle
Since \( A \) and \( B \) are the ends of a right-angled isosceles triangle, the lengths of the sides \( AC \) and \( BC \) must be equal, i.e., \( AC = BC \).

### Step 2: Apply the Distance Formula
Using the distance formula, we can express the lengths \( AC \) and \( BC \):
- The distance \( AC \) is given by:
\[ AC = \sqrt{(h - 9)^2 + (k + 9)^2} \]
- The distance \( BC \) is given by:
\[ BC = \sqrt{(h - 1)^2 + (k - 3)^2} \]

### Step 3: Set the Distances Equal
Since \( AC = BC \), we can set the two equations equal to each other:
\[ \sqrt{(h - 9)^2 + (k + 9)^2} = \sqrt{(h - 1)^2 + (k - 3)^2} \]

### Step 4: Square Both Sides
To eliminate the square roots, we square both sides:
\[ (h - 9)^2 + (k + 9)^2 = (h - 1)^2 + (k - 3)^2 \]

### Step 5: Expand Both Sides
Expanding both sides, we get:
\[ (h^2 - 18h + 81) + (k^2 + 18k + 81) = (h^2 - 2h + 1) + (k^2 - 6k + 9) \]

### Step 6: Simplify the Equation
Combining like terms:
\[ h^2 + k^2 - 18h + 18k + 162 = h^2 + k^2 - 2h - 6k + 10 \]
Cancelling \( h^2 \) and \( k^2 \) from both sides:
\[ -18h + 18k + 162 = -2h - 6k + 10 \]

### Step 7: Rearrange the Equation
Rearranging gives:
\[ -18h + 2h + 18k + 6k + 162 - 10 = 0 \]
This simplifies to:
\[ -16h + 24k + 152 = 0 \]

### Step 8: Solve for \( h \)
We can express \( h \) in terms of \( k \):
\[ 16h = 24k + 152 \]
\[ h = \frac{24k + 152}{16} \]
\[ h = \frac{3k + 19}{2} \]

### Step 9: Find Possible Values for \( C \)
To find integer coordinates for \( C(h, k) \), we can substitute integer values for \( k \) and calculate \( h \).

### Step 10: Check Possible Points
We will check the integer points that satisfy the triangle's properties.

Let’s check some integer values for \( k \):
1. If \( k = 0 \):
\[ h = \frac{3(0) + 19}{2} = \frac{19}{2} \quad \text{(not an integer)} \]
2. If \( k = 1 \):
\[ h = \frac{3(1) + 19}{2} = \frac{22}{2} = 11 \quad \text{(point (11, 1))} \]
3. If \( k = 2 \):
\[ h = \frac{3(2) + 19}{2} = \frac{25}{2} \quad \text{(not an integer)} \]
4. If \( k = 3 \):
\[ h = \frac{3(3) + 19}{2} = \frac{28}{2} = 14 \quad \text{(point (14, 3))} \]

Continuing this process will yield possible integer coordinates for \( C \).

### Final Answer
After checking various integer values, we find that the third vertex \( C \) can be \( (11, 1) \) or \( (14, 3) \) depending on the orientation of the triangle.