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

CASE I

(x,y) is equidistant from (3,6)

Let the distance between them be d1

Using distance formula ,

d1 = (3-x)^2 + (6-y)^2

= x^2 + y^2 - 6x - 12y + 45

CASE II

(x,y) is equidistant from (-3,4)

Let the distance between them be d2

Using distance formula,

d2 = (-3-x)^2 + (4-y)^2

= x^2 + y^2 + 6x - 8y + 19

Now , according to question ,

d1 = d2

= x^2 + y^2 - 6x - 12y + 45 = x^2 + y^2 + 6x - 8y + 19

On solving ,

3x + 5y = 9

HOPE THIS HELPS YOU

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