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Question 548497: Find the distance betweeen the points of intersection of the curves 3x - 2y + 6 = 0 and x^2 + y^2 = 9.
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Find the distance betweeen the points of intersection of the curves 3x - 2y + 6 = 0 and x^2 + y^2 = 9
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x^2+y^2=9
y^2=9-x^2
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3x-2y+6=0
2y=3x+6
y=3x/2+3
y^2=9x^2/4+18x/2+9
y^2=(9x^2/4)+9x+9
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equate y^2 derived from each curve
9-x^2=(9x^2/4)+9x+9
(9x^2/4)+x^2+9x=0
(9/4)x^2+(4/4)x^2+9x=0
13x^2/4+9x=0
x(13x/4+9)=0
x=0
y=3x/2+3=3
one point of intersection: (0,3)
or
13x/4+9=0
13x/4=-9
13x=-36
x=-36/13≈-2.7692≈-2.8
y=3x/2+3=(3*(-36/13)/2)+3≈1.1538≈-1.2
second point of intersection:(-2.8,-1.2)
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using distance formula:
d^2=(0-(-2.8))^2+(3-(-1.2))^2
d^2=(0+2.8)^2+(3+1.2)^2
d^2=(2.8)^2+(4.2)^2
d^2=(2.8)^2+(4.2)^2=25.48
d=√25.48≈5.1
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