Question

find the relation between x and y such that the point x y is equal distance from point 7, 1 3, 5​

Answer 1

Answer:

x=y+2

Step-by-step explanation:

Given, AP = BP

AP2 = BP2

By distance formula,

(x - 7)2 + (y - 1)2 = (x - 3)2 + (y - 5)2

x2 + 49 - 14x + y2 + 1 - 2y = x2 + 9 - 6x + y2 + 25 - 10y

-7x + 25 - y = -3x + 17 - 5y

-4x + 8 + 4y = 0

x - y = 2

Answer 2

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Answer: Let P(x, y) be the point equidistant from the points A(7, 1) and B(3, 5).

Given, AP = BP

AP^2 = BP^2

By distance formula,

(x - 7)^2 + (y - 1)^2 = (x - 3)^2 + (y - 5)^2

x2 + 49 - 14x + y2 + 1 - 2y = x2 + 9 - 6x + y2 + 25 - 10y

-7x + 25 - y = -3x + 17 - 5y

-4x + 8 + 4y = 0

x - y = 2

Hence ,  x  = 2 + y is the relation.

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