The point P is equidistant from A(1, 3), B(- 3, 5) and C(5, - 1), then PA is equal to
Mathematics · JEE Main · NTA Exams — Co-ordinate Geometry
The point P is equidistant from A(1, 3), B(- 3, 5) and C(5, - 1), then PA is equal to
- 5
- \(5 \sqrt{5}\)
- 25
- \(5 \sqrt{10}\)
Answer
(D) 5 10
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