Question

Find a relation between x and y such that the point (x, y) is equidistant from the points (3, 6)
2
and (-3, 4).
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Answer 1

Answer:

hope it is helpful to you ok

Answer 2

Answer:

The answer is 3x + y - 5 = 0 .

Step-by-step explanation:

Let the point P(x,y) be the equidistant from the points A(3,6) and B(-3,4).

therefore, PA = PB.

By using distance formula,

$\sqrt{(x2-x1)^{2}\:+\: (y2-y1)^{2} } \:we\:have,\\\\\sqrt{(x-3)^{2}\:+\: (y-6)^{2} }\:=\:\sqrt{(x+3)^{2}\:+\: (y-4)^{2} }\\\\{(x-3)^{2}\:+\: (y-6)^{2} }\:=\:{(x+3)^{2}\:+\: (y-4)^{2} }\\\\x^{2} \:-\:6x\:+\:9\:+\:y^{2} \:-\:12y\:+\:36\:=\:x^{2} \:+\:6x\:+\:9\:+\:y^{2} \:-\:8y\:+\:16\\\\-\:6x\:-\:6x\:-\:12y\:+\:8y\:+\:36\:-\:16\:=\:0\\\\-\:12x\:-\:4y\:+\:20\:=\:0\\\\therefore,\:3x\:+\:y\:-\:5\:=\:0$

Hope it helps you :)