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:

Step-by-step explanation:

d =√(x2 -x1)2 + (y2 – y1)2

let P = (x, y) , Q = (3, 6) and R = (-3, 4).

distance between P and R =√(x + 3)2+ (y-4)2

Similarly, distance between P and Q =√(x- 3)2+ (y – 6)2

(x, y) is the equidistant from the points (3,6) and (-3,4) i.e.,PQ = PR

√(x -3)2+ (y – 6)2 =√(x + 3)2+ (y-4)2

squaring both sides we get,

⇒(x – 3)² + (y – 6)² = (x + 3)² + (y – 4)²

⇒x² + 9 – 6x + y² + 36 – 12y = x² + 9 + 6x + y² + 16 – 8y

⇒-6x + 45 – 12y = 6x – 8y + 25

⇒-6x – 6x – 12y + 8y = 25- 45

⇒-12x – 4y = -20

⇒3x + y = 5

Hence, the relation between x and y is 3x + y = 5

ANSWERED BY GAUTHMATH

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 :)