Question

Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (– 3, 4).

Answer 1

Answer:

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Answer 2

Answer:

3x + y = 5

Step-by-step explanation:

the distance of point (x1,y1) from point (x2,y2) is given by the distance formula

{(x2-x1)^2 + (y2-y1)^2}^(1/2)

applying the same formula, the distance between (x,y) and (3,6) is given by

{(x-3)^2+(y-6)^2}^(1/2)

distance between (x,y) and (-3,4) is

{(x+3)^2+(y-4)^2}^(1/2)

since points are equidistant, the distance is equal, Equating the square of the distance, we have

(x-3)^2 + (y-6)^2 = (x+3)^2 + (y-4)^2

(x-3)^2 - (x+3)^2 = (y-4)^2 - (y-6)^2

x^2 -6x +9 - x^2 -6x -9 = y^2 - 8y + 16 - y^2 + 12y - 36

-12x = 4y - 20

12x + 4y = 20

3x + y = 5 is the required relation