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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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