Math, asked by ritikachoudhary4474, 2 months ago

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

Answers

Answered by cuteprincess463
2

Answer:

let p(x, y) be equidistant point from the given points A(3, 6) and B(3, 4)

(AP) =(BP)

Step-by-step explanation:

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Attachments:
Answered by Anonymous
11

Answer:

Given.

  • points of line segment A(3,6) and B (-3,4).
  • AC = CB.

Find a relation between x and y such that the point (x, y) is equidistant from the point

Solution

_______________

A (3,6) C(x, y) B (-3,4)

By distance formula.

=> AC

=_/(x+3)^2+(y+6)^2.

=_/x^2+(3)^2+2*x*3+y^2+(6)^2+2*6*y.

=_/x^2+9+6x+y^2+36+12y.

then,

=>CB

=_/(-3+x)^2+(4+y)^2.

=_/(-3)^2+x^2+2*x*(-3)+(4)^2+y^2+2*4*y.

= _/9+x^2-6x+16+y^2+8y.

→AC= CB .

=>_/x^2+9+6x+y^2+36+12y=

_/9+x^2-6x+16+y^2+8y.

Squaring both sides.

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

=>6x+6x+12y-8y = 16 - 36.

=>12x+4y = -20.

=> 4 (3x+y) = -20.

=> 3x+y = -20/4.

=> 3x+y = -3.

Thus, the relation between x and y such that the point (x, y) is equidistant from the point is 3x+y = -3.

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