Question

If \( (x, y) \) is equidistant from \( (a+b, b-a) \) and \( (a-b, a+b) \), then:

• A. \( x+y = 0 \)
• B. \( bx-ay = 0 \)
• C. \( ax-by = 0 \)
• D. \( bx+ay = 0 \)
• E. \( ax+by = 0 \)

Hint:

The locus of points equidistant from two points is the perpendicular bisector of the segment joining them. The equation \( bx-ay=0 \) represents a straight line passing through the origin.

Answer 1

The Correct Option is B

Concept: The distance formula between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is \( \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \). Since the point is equidistant, the squares of the distances must also be equal, which allows us to remove the square root and simplify the resulting quadratic equation into a linear one.

Step 1: Set up the distance equation.
Let \( P(x, y) \), \( A(a+b, b-a) \), and \( B(a-b, a+b) \). Given \( PA^2 = PB^2 \): \[ [x-(a+b)]^2 + [y-(b-a)]^2 = [x-(a-b)]^2 + [y-(a+b)]^2 \] Expanding both sides: \[ x^2 - 2x(a+b) + (a+b)^2 + y^2 - 2y(b-a) + (b-a)^2 = x^2 - 2x(a-b) + (a-b)^2 + y^2 - 2y(a+b) + (a+b)^2 \]

Step 2: Simplify and solve.
Cancel common terms (\( x^2, y^2, (a+b)^2 \)) and rearrange: \[ -2x(a+b) - 2y(b-a) + (b-a)^2 = -2x(a-b) - 2y(a+b) + (a-b)^2 \] Note that \( (b-a)^2 = (a-b)^2 \), so they cancel as well: \[ -2x(a+b) - 2y(b-a) = -2x(a-b) - 2y(a+b) \] Divide by \(-2\) and expand: \[ ax + bx + by - ay = ax - bx + ay + by \] \[ bx - ay = -bx + ay \implies 2bx - 2ay = 0 \implies bx - ay = 0 \]