Perpendicular Bisector of 2 Points

[yourmom98]

is the locus of points equidistant from the two given points on the same line as the perpendicular bisector of the 2 points?

 

[bomba923]

Welcome to PF
*In 2D,
Yes. If we draw a line between any two given points, the locus of points equidistant from those two given points will lie on a unique perpendicular bisector that intersects exactly between the two points on the line (i.e., intersects at a point belonging to the line equidistant from the two given points). (In 2D all is coplanar)
*In space (3D that is),
Yes. If we draw a line between any two given points, the locus of points equidistant from those two given points will lie on a unique plane containing every perpendicular bisector of this line that intersects at a point belonging to the line equidistant from the two given points.

 

[yourmom98]

thanks i am asked to draw a diagram that represent the locus in each of the following situations also i am asked "what is the locus of each?"

a) Going down on a elevator
b)sitting in a seat on a ferris wheel as it rotates
c)all the points that are 2cm from a parabola
d)all the point that are 5 cm above a line




a) the diagram would be a vertical line
b) the diagram would be a sinusoidal function
c) a parabola
d) horizontal line

are these correct and what does it mean "what is the locus of each?" am i supposed to give an equation?


EDIT: also how would i find the equation of this
locus where point such that the sum of whoose distances from (0,-2) and (0,2) is 8 cause. well its not that i CANT find the equation its just that i have to draw and ellipse to figure it out i wonder if there is an more accuate way? so far my answer is 16=x^2+y^2 is this rite?

 

[EnumaElish]

a) the diagram would be a vertical line
b) the diagram would be a sinusoidal function

If the answer to (a) is a vertical line why isn't the answer to (b) a circle?

c) a parabola

Two parabolas (in 2D)? A cylinder (in 3D)?

d) horizontal line

I guess you meant a parallel line.

am i supposed to give an equation?

You can only give an equation if you have an equation to begin with. If the question didn't give you an equation, do you really want to be the one who starts it?

how would i find the equation of this locus where point such that the sum of whoose distances from (0,-2) and (0,2) is 8 cause. well its not that i CANT find the equation its just that i have to draw and ellipse to figure it out i wonder if there is an more accuate way? so far my answer is 16=x^2+y^2 is this rite?

Let a = (0,-2) and b = (0,2); and c = (x,y) is such a point that d(a,c) + d(b,c) = 8 where d is the (Euclidian) distance function. For any two points u = (u₁,u₂) and v = (v₁,v₂), d is defined as d(u,v) = \sqrt{(v_1-u_1)^2+(v_2-u_2)^2}. So the locus that the question is asking is "the set of all (x,y) points in \mathbb R^2 such that \sqrt{(x-0)^2+(y+2)^2} + \sqrt{(x-0)^2+(y-2)^2} = 8."

P.S. Function d is symmetric: d(u,v) = d(v,u) for any two poins u and v. You can verify this if you apply the definition of d once to d(u,v) and once for d(v,u).