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 Equidistant points on the Axes (Posted on 2003-03-28)
Two points, A and B, are on the Cartesian plane at (-1,4) and (9,6).

A. What point on the x-axis is equidistant from each of these two points?

B. What point on the y-axis is equidistant from these two points?

 Submitted by Charlie Rating: 3.2000 (10 votes) Solution: (Hide) The locus of all points equidistant from the two points is the perpendicular bisector of the line segment bounded by the two points. The perpendicular bisector goes through the point midway between them, (4,5), and has a slope that is the negative of the reciprocal of the slope of the segment. The segment's slope is 1/5, so the slope of the perpendicular bisector is -5. Its equation is y=-5x+b and b has to satisfy 5=-5(4)+b, so b=25 and the line sought is y=-5x+25. Part A: That linear function intersects the x-axis when y=0, so 0=-5x+25, and x=5. Alternatively, the Pythagorean relationship would give (x+1)^2+4^2 = (9-x)^2+6^2, which also solves to x=5. For part B, since the equation of the perpendicular bisector is -5x+25, if we substitute x=0, we get y=25. Or, (-1)^2+(y-4)^2 = 9^2+(y-6)^2, which also solves to y=25.

 Subject Author Date Sorry, my last Y point was wrong... Antonio 2003-09-03 10:24:00 Solution Antonio 2003-09-03 09:34:09 Solution Trevor John Streeton 2003-03-28 22:05:40 re: pretty easy :easier Jer 2003-03-28 07:33:59 re: oops Little Different jude 2003-03-28 07:14:08 pretty easy DuCk 2003-03-28 07:10:45 oops Little Different jude 2003-03-28 06:53:24 Little different jude 2003-03-28 06:50:07 solution Christian Perfect 2003-03-28 06:48:59

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