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January 18th, 2014, 08:58 PM   #1
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Geometrical locus

I have changed the values so it's not solved for me but I have find the equation of the locus of point p which is equidistant from the y axis (point b)and point A (-3,1)

So
pa=pb
Pa^2=pb^2
Pb^2= (x--3)^2 +(y-1)^2
Pb^2=x^2-6x+10+y^2-2y
Then it equals the root of that? I am really banging my head against a wall on this one I have hunted back through my study notes and cannot even find a reference to solving a problem like this.
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January 19th, 2014, 04:29 AM   #2
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Re: Geometrical locus

i understand that whats its asking me for is an equation for a line thats equal distance from the 2 points and that ltiple points along that line.... aslong as it is equal distance from both A & B at any given point
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January 19th, 2014, 05:48 AM   #3
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Re: Geometrical locus

i think i nutted it out with my actual values
given point A (3,-1) the point on the Y axis must be at (0,-1)
PA=PB
PA^2=PB^2
(x-0)^2+(y--1)^2=(x-3)^2+(y--1)^2
x^2+Y^2+2y+1=x^2-6x+9+y^2+2y+1
-6x+9=0
-6x=-9
x=9/6
x=3/2
therefore the line x=3/2 is the required equation for the loci between the 2 points as all points along the line will be equidistant between the 2 points.

am i right? i still don't fully understand this topic given what i have been given to learn off
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