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