Question

the equation of locus of a point equidistant from A(2, 0) and B(0, 2) is emergency fast. thnx ra shivani​

Answer 1

Answer:

I hope my answer is helpful to you

Answer 2

$point \: (x \: \: y) is \: equidistant \: from \: (2 \: \: 0) \: and \: (0 \: \: 2) \\ distance \: of \: (2 \: \: 0) = \: distance \: of \: (0 \: \: 2) \\ \sqrt{ {(x - 2)}^{2} + {(y - 0)}^{2} } = \sqrt{ {(x - 0)}^{2} + {(y - 2)}^{2} } \\ \sqrt{ {(x - 2)}^{2} + {y}^{2} } = \sqrt{ {x}^{2} + {(y - 2)}^{2} } \\ sobs \\ {(x - 2)}^{2} + {y}^{2} = {x}^{2} + {(y - 2)}^{2} \\ {x}^{2} + 4 - 4x + {y}^{2} = {x}^{2} + {y}^{2} + 4 - 4y \\ 4 - 4x = 4 - 4y \\ 4 - 4x - 4 + 4y = 0 \\ - 4x + 4y = 0 \\ y - x = 0$

✫ ᕼOᑭᗴ TᕼIՏ ᗰᗩY ᕼᗴᒪᑭ YOᑌ ✫