Question

Let \( A(6, -1) \), \( B(1, 3) \) and \( C(x, 8) \) be three points such that \( AB = BC \). The values of \( x \) are:

• A. \( 3, 5 \)
• B. \( -3, 5 \)
• C. \( 3, -5 \)
• D. \( 4, 5 \)
• E. \( -3, -5 \)

Hint:

Always equate squared distances (\( AB^2 = BC^2 \)) instead of the distances themselves. This removes the square roots immediately and simplifies the algebraic manipulation.

Answer 1

The Correct Option is B

Concept: The distance between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by the distance formula: \( \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \). Given that \( AB = BC \), we set the squares of their distances equal to each other to solve for the unknown variable \( x \).

Step 1: Calculate the length of \( AB^2 \).
Using points \( A(6, -1) \) and \( B(1, 3) \): \[ AB^2 = (1 - 6)^2 + (3 - (-1))^2 \] \[ AB^2 = (-5)^2 + (4)^2 = 25 + 16 = 41 \]

Step 2: Calculate the length of \( BC^2 \) in terms of \( x \).
Using points \( B(1, 3) \) and \( C(x, 8) \): \[ BC^2 = (x - 1)^2 + (8 - 3)^2 \] \[ BC^2 = (x - 1)^2 + 5^2 = (x - 1)^2 + 25 \]

Step 3: Solve the equation \( AB^2 = BC^2 \).
\[ 41 = (x - 1)^2 + 25 \] \[ (x - 1)^2 = 41 - 25 = 16 \] Taking the square root of both sides: \[ x - 1 = \pm 4 \] This gives two cases: 1. \( x - 1 = 4 \implies x = 5 \) 2. \( x - 1 = -4 \implies x = -3 \) The values of \( x \) are \( -3, 5 \).