Find the formula that expresses the arbitrary point P(x,y) is on the perpendicular bisector I of segment AB.?

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Find the formula that expresses the arbitrary point P(x,y) is on the perpendicular bisector I of segment AB.?

A(-10,2), B (12,-4)

2007-10-01 21:44:14 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics

3 answers

slope of AB is (- 4 - 2)/(12 + 10) = - 6/22 = - 3/11
midpoint of AB is (1,- 1)
slope of bisector is 11/3
y + 1 = (11/3)(x - 1)
3y + 3 = 11x - 11
3y = 11x - 14
11x - 3y - 14 = 0

2007-10-01 22:34:20 · answer #1 · answered by Helmut 7 · 0⤊ 1⤋

ill try

first find the midpoint of segment AB
which is
(1,-1)
therefore the arbitrary point P passes to this point

because a bisector divides a segment equally..

dont know how to continue hehehe

2007-10-01 21:57:02 · answer #2 · answered by Jr Levesque 2 · 0⤊ 1⤋

I fly under the radar around here, so I'm happy to say that I feel under-appreciated. I guess the economy has hit Yahoo points, too, huh?

2016-05-18 23:48:17 · answer #3 · answered by ? 3 · 0⤊ 0⤋