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Find the relation between x and y such that the point (x,y) is equidistant from the points (7,1) and (3,5)?

Solution

✅ Correct Option: 3

Equate the squared distances: (x−7)2+(y−1)2=(x−3)2+(y−5)2(x-7)^2 + (y-1)^2 = (x-3)^2 + (y-5)^2.

Expanding, −14x+50−2y=−6x+34−10y-14x + 50 - 2y = -6x + 34 - 10y.

This gives −8x+8y+16=0-8x + 8y + 16 = 0, that is x−y=2x - y = 2.

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