Question 1128278: find the equation satisfied by the set of points such that each is equidistant from (2,-1) and 3x+y-7=0
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! find the equation satisfied by the set of points such that each is equidistant from P(2,-1) and 3x+y-7=0
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It's a parabola. The focus is (2,-1) and the directrix is 3x+y-7 = 0
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Find the equation of the axis of the parabola, the line thru (2,-1) and perpendicular to (call it L1) 3x+y-7=0.
Slope of L1 = -3
Slope of perpendicular line = 1/3
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y+1= (1/3)*(x-2)
The intersection of the 2 lines is D(13/5,-4/5)
The midpoint of PD is M(23/10,-9/10), a point on the parabola.
The distance from M to the line L1 is MD = PM =
= 
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The distance from a point (m,n) to a line Ax + By + C = 0 is:
(Am + Bn + C)/sqrt(A^2 + B^2)
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The distance from P to the line L is:
|2*3 -1*1 -7|/sqrt(9+1) = 2/sqrt(10)
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Line L2 parallel to L1 thru (2,-1) is y+1 = -3(x-2)
The 2 points on L2 at a distance of sqrt(10)/10 from P are Q(2.1,-1.3) and
R(1.9,-0.7)
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Find the parabola thru the 3 points about the axis.
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More later.
My Excel sheet always finds a parabola with its axis parallel to the y-axis.
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Use this method:
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For any point (x,y) the distance to (2,-1) is sqrt((x-2)^2 + (y+1)^2)
The distance to the line L1 is |3x + y - 7|/sqrt(10)
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