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

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Question

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

Options

  • 3x – 7y = 0

  • x – 3y = – 5

  • x – 3y = 5

  • 3x – y = – 9

MCQ
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Solution

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

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Equation of Locus
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