Let A(6, -1), B (1, 3) and C (x, 8) be three points such that AB = BC then the value of x are

Options

A. 3, 5
B. `-3, 5`
C. `3, -5`
D. `-3, -5`

Answer

To solve the problem, we need to find the value of \( x \) such that the distances \( AB \) and \( BC \) are equal. We will use the distance formula to calculate these distances.

### Step-by-step Solution:

1. **Identify the Points**:
- Let \( A(6, -1) \), \( B(1, 3) \), and \( C(x, 8) \).

2. **Use the Distance Formula**:
The distance formula between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

3. **Calculate Distance \( AB \)**:
- Using points \( A(6, -1) \) and \( B(1, 3) \):
\[ AB = \sqrt{(1 - 6)^2 + (3 - (-1))^2} \]
\[ = \sqrt{(-5)^2 + (3 + 1)^2} \]
\[ = \sqrt{25 + 16} = \sqrt{41} \]

4. **Calculate Distance \( BC \)**:
- Using points \( B(1, 3) \) and \( C(x, 8) \):
\[ BC = \sqrt{(x - 1)^2 + (8 - 3)^2} \]
\[ = \sqrt{(x - 1)^2 + 5^2} \]
\[ = \sqrt{(x - 1)^2 + 25} \]

5. **Set the Distances Equal**:
Since \( AB = BC \):
\[ \sqrt{41} = \sqrt{(x - 1)^2 + 25} \]

6. **Square Both Sides**:
To eliminate the square roots, square both sides:
\[ 41 = (x - 1)^2 + 25 \]

7. **Simplify the Equation**:
\[ 41 - 25 = (x - 1)^2 \]
\[ 16 = (x - 1)^2 \]

8. **Take the Square Root**:
Taking the square root of both sides gives:
\[ x - 1 = 4 \quad \text{or} \quad x - 1 = -4 \]

9. **Solve for \( x \)**:
- From \( x - 1 = 4 \):
\[ x = 5 \]
- From \( x - 1 = -4 \):
\[ x = -3 \]

10. **Final Values of \( x \)**:
Thus, the values of \( x \) are:
\[ x = 5 \quad \text{or} \quad x = -3 \]