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Question 1122728: Verify that P = (1, -1) is the same distance from A = (5, 1) as it is from B = (-1, 3). Find three more points that are equidistant from A and B. Can points equidistant from A and B be found in every quadrant?
Answer by greenestamps(13203) (Show Source):
You can put this solution on YOUR website!
The points equidistant from A and B are the points on the perpendicular bisector of segment AB.
Segment AB has slope -1/3 and midpoint (2,2).
The perpendicular bisector of AB has slope 3 and passes through (2,2); its equation is y = 3x-4.
(a) Verify that (1,-1) is equidistant from A and B.
Yes; the point (1,-1) satisfies the equation of the perpendicular bisector of AB.
(b) Find three more points that are equidistant from A and B.
You can do this easily; choose any values of x and use the equation of the perpendicular bisector to find the corresponding y values.
(c) No. The perpendicular bisector has slope 3 and y-intercept -4; it passes through quadrants I, III, and IV but not through quadrant II.
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