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is the locus of points equidistant from the two given points on the same line as the perpendicular bisector of the 2 points?
If the answer to (a) is a vertical line why isn't the answer to (b) a circle?yourmom98 said:a) the diagram would be a vertical line
b) the diagram would be a sinusoidal function
Two parabolas (in 2D)? A cylinder (in 3D)?c) a parabola
I guess you meant a parallel line.d) horizontal line
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?am i supposed to give an equation?
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 = (u_{1},u_{2}) and v = (v_{1},v_{2}), d is defined as d(u,v) = [itex]\sqrt{(v_1-u_1)^2+(v_2-u_2)^2}[/itex]. So the locus that the question is asking is "the set of all (x,y) points in [itex]\mathbb R^2[/itex] such that [itex]\sqrt{(x-0)^2+(y+2)^2}[/itex] + [itex]\sqrt{(x-0)^2+(y-2)^2} = 8.[/itex]"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?