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What part of the coordinate plane is equidistant from the points A(-3,2) and B(3,2) explain?
Let point A = (3,2) and point B = (-3,2). Now imagine a point P with coordinates (x,y) in the coordinate plane. The distance from point A to P is:
dA2 = (y-2)2 + (x-3)2
The distance from from point B to point P is:
dB2 = (y-2)2 + (x-(-3))2 = (y-2)2 + (x+3)2
Point P is equidistant from points A and B if dA = dB (or dA2 = dB2). So let’s set them equal and solve for x and y:
(y-2)2 + (x-3)2 = (y-2)2 + (x+3)2
(x-3)2 = (x+3)2
x2 – 6x + 9 = x2 + 6x + 9
0 = 12x
x = 0
This is the equation of a vertical line through the point x=0. In other words, it is the y-axis. So the part of the plane equidistant from points A and B are all points along the y-axis.
If you plot points A and B on graph, you can see that point B is a reflection of point A across the y-axis. The y-axis is thus the axis of symmetry and all points along it are equidistant from points A and B. If you draw a line from point A and point B to any point P on the y-axis, you can see that the distances are equal.
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