Question 123193
Find a point on the y-axis that is equidistant from the points (8, -8) and (2, 2).
:
Since the point is on the y axis we know the x coordinate = 0
Call the point: 0,y and solve for y
:
Use the distance formula d = {{{sqrt((x2-x1)^2+(y2-y1)^2)}}}
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In the distance equation let x1 = 0 and y1 = y
:
dist from points 2,2 to 0,y  =  dist from points 8,-8 to 0,y
{{{sqrt((2-0)^2+(2-y)^2)}}}  = {{{sqrt((8-0)^2+(-8-y)^2)}}}
Square both sides and we can forget about the radicals
:
 4 + (4 - 4y + y^2) = 64 + (64 + 16y + y^2)
:
 8 - 4y + y^2 = 128 + 16y + y^2
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Group the y's on the left and numerical values on the right
y^2 - y^2 - 4y - 16y = 128 - 8
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Conveniently the y^2 cancel so we have:
-20y = 120
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Divide both sides by -20
y = {{{120/(-20)}}}
y = -6
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Equidistant point on the y axis = 0,-6
:
:
Rather than check our solution in the distance equations, I plotted these
points on piece of graph paper and, it indeed checks out! You can do the same.
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Hope I made this understandable. Any questions?