If A(5, 2), B(2, -2) and C(-2, t) are the vertices of a right triangle with `angleB = 90^(@)` then find the value of t.

Explanation

To find the value of \( t \) such that triangle ABC is a right triangle with \( \angle B = 90^\circ \), we can use the property that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.

### Step-by-Step Solution:

1. **Identify the points:**
- \( A(5, 2) \)
- \( B(2, -2) \)
- \( C(-2, t) \)

2. **Calculate the distances:**
We will calculate the lengths of sides \( AB \), \( BC \), and \( AC \) using the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

- **Distance \( AB \):**
\[ AB = \sqrt{(2 - 5)^2 + (-2 - 2)^2} = \sqrt{(-3)^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

- **Distance \( BC \):**
\[ BC = \sqrt{(-2 - 2)^2 + (t + 2)^2} = \sqrt{(-4)^2 + (t + 2)^2} = \sqrt{16 + (t + 2)^2} \]

- **Distance \( AC \):**
\[ AC = \sqrt{(-2 - 5)^2 + (t - 2)^2} = \sqrt{(-7)^2 + (t - 2)^2} = \sqrt{49 + (t - 2)^2} \]

3. **Apply the Pythagorean theorem:**
Since \( \angle B \) is a right angle, we can apply the Pythagorean theorem:
\[ AB^2 + BC^2 = AC^2 \]
Plugging in the distances:
\[ 5^2 + \left(\sqrt{16 + (t + 2)^2}\right)^2 = \left(\sqrt{49 + (t - 2)^2}\right)^2 \]
Simplifying this gives:
\[ 25 + 16 + (t + 2)^2 = 49 + (t - 2)^2 \]
\[ 41 + (t + 2)^2 = 49 + (t - 2)^2 \]

4. **Expand the squares:**
\[ (t + 2)^2 = t^2 + 4t + 4 \]
\[ (t - 2)^2 = t^2 - 4t + 4 \]
Substituting these into the equation:
\[ 41 + t^2 + 4t + 4 = 49 + t^2 - 4t + 4 \]

5. **Simplify the equation:**
\[ 45 + 4t = 49 - 4t + 4 \]
\[ 45 + 4t = 53 - 4t \]
Adding \( 4t \) to both sides:
\[ 45 + 8t = 53 \]
Subtracting 45 from both sides:
\[ 8t = 8 \]
Dividing by 8:
\[ t = 1 \]

### Final Answer:
The value of \( t \) is \( 1 \).