Question

find a relation between X and Y such that the point x, y is equidistant from the points 3, 6 and minus 3, 4 ​

Answer 1

A = x, y

B = 3,6

C = 3,4

AB = AC

$\sqrt{( {3 - x})^{2} + {(6 - y})^{2} } = ab$

AC =

$\sqrt{( {3 - x})^{2} + {(4 - y})^{2} }$

squaring both sides

(AB) ^2 = (BC) ^2

$( {3 - x})^{2} + {(6 - y})^{2} = \sqrt{ {(3 - x})^{2} } + \sqrt{ {(4 - y})^{2} }$

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

$36 - 16 = - 8y + 12y$

$20 = 4y$

$20 \div 4 = y$

$y = 5$

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 :)