SOLUTION: 2)PQR is a straight line PQ=4PR P(-4,-1) Q(6,11) find the co-ordinate of R?
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Question 935497: 2)PQR is a straight line PQ=4PR P(-4,-1) Q(6,11) find the co-ordinate of R?
Found 2 solutions by josgarithmetic, stanbon:
Answer by josgarithmetic(39617) (Show Source): You can put this solution on YOUR website!
Fix the description. PQ is less than PR because Q is between P and R. Either PQR is the line using Q between P and R; or PQ is some fraction of PR. Both cannot be true at the same time.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
PQR is a straight line PQ=4PR P(-4,-1) Q(6,11) find the co-ordinate of R
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length of PQ = sqrt[(11--1)^2+(6--4)^2] = sqrt[144+100)
= sqrt(244) = 2sqrt(61)
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PR = (1/4)PQ = (1/2)sqrt(61)
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Comment: since PQ is larger than PR, R must be between P and Q
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Ratio of PQ to PR = 2sqrt(61)/(1/2)sqrt(61) = 4:1
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So R is 1/4 of the way from P to Q
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Let the coordinates of R be (x,y)
x = (1/4)(6--4) = (1/4)(10) = 2.5
y = (1/4)(11--1) = (1/4)(12) = 3
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Ans:: R = (2.5,3)
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Cheers,
Stan H.
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