Question

Equation of locus of a point which is equidistant from (1, 2) and (– 5, 4) is ______.

विकल्प

• 3x – 7y = 0

• x – 3y = – 5

• x – 3y = 5

• 3x – y = – 9

Answer 1

Equation of locus of a point which is equidistant from (1, 2) and (– 5, 4) is 3x – y = – 9.

Explanation:

(x – 1)² + (y – 2)² = (x + 5)² + (y – 4)²

⇒ x² – 2x + 1 + y² – 4y + 4 = x² + 10x + 25 + y² – 8y + 16

⇒ – 12x + 4y = 36

⇒ 3x – y = – 9