Question 130118: Hi! i have a question on my math homework i have to find the circumcenter of a triangle whose points are L(83,15) S(196,123) and B(22,123). i have found the midpoints which are on line: LS is (139.5,69), SB is (109,123), and LB is 52.5,69). then i found the slope of each line which is: LS 108/113, SB 0, and LB -108/61. then i have tried to find the perpendicular bisector of each but every time i do the same problem i have came out with a different number that equals b. i have tried for a few hours on this problem and i would aprieciate if you could help me find what X (of the circumcenter) would be. Thank you
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! The circumcenter is the center of a triangle's circumcircle. It can be found as the intersection of the perpendicular bisectors.
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i have to find the circumcenter of a triangle whose points are
L(83,15), S(196,123) and B(22,123).
i have found the midpoints which are on line:
LS is (139.5,69), SB is (109,123), and LB is (52.5,69).
then i found the slope of each line which is:
LS 108/113, SB 0, and LB -108/61
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Equations:
Perpendicular bisector of LS:
Finding the y-intercept.
69 = (-113/108)139.5 + b
b = 214.9583333
Equation of the perp/bisector of LS: y = (-113/108)x-214.959333
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Perpendicular bisector of SB
Since SB has slope = 0, the perpendicular is a vertical line thru the
midpoint (109,123)
Equation of the perp/bisector of SB is x=109
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Find the intersection of the two perp/bisectors:
Substitute x=109 into the perp/bisector of LS:
y = (-113/108)(109)-214.959333
y = -329.0056295
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The circumcenter is (109,-329.0056295)
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Cheers,
Stan H.
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