SOLUTION: Find an equation describing all points P(x,y) equidistant from Q(-3,4) and R(1,-3). (use the general equation of a line - input the numerical coefficient of each term of the equa

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Question 1118795: Find an equation describing all points P(x,y) equidistant from Q(-3,4) and R(1,-3).
(use the general equation of a line - input the numerical coefficient of each term of the equation below)
(Answer should be in this format:) _____x - _____y + _____ = _____
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39614) About Me  (Show Source):
You can put this solution on YOUR website!
slope -%28%28-3-1%29%2F%284-%28-3%29%29%29
-%28%28-4%29%2F%287%29%29
4%2F7

midpoint of QR
(-1, 1/2 )


All point equidistant from Q and R
y-1%2F2=%284%2F7%29%28x%2B1%29
2y-1=%288%2F7%29%28x%2B1%29
14y-7=8%28x%2B1%29
14y-7=8x%2B8
-7-8=8x-14y
highlight%288x-14y=-15%29

Answer by ikleyn(52756) About Me  (Show Source):
You can put this solution on YOUR website!
.
The line under the question is the perpendicular bisector to the segment, connecting given points.


The slope of the segment connecting the given points is m = %28%28-3%29-4%29%2F%281-%28-3%29%29 = %28-7%29%2F4 = -7%2F4.


Therefore, the slope of the perpendicular bisector to it is  4%2F7.


The midpoint of the given segment is  (-1,0.5).


The straight line through the point (1,-0.5) with the slope of 4%2F7 is


y - 0.5 = %284%2F7%29%2A%28x-%28-1%29%29,   or

y - 0.5 = %284%2F7%29%2A%28x%2B1%29.


Multiply both sides by 14 to get


14y - 7 = 8*(x + 1),   or   14y - 7 = 8x + 8,   or


8x - 14y + 15 = 0.


It is your final equation (=answer).