Maths(Coordinate Treatment of Simple Locus Problems)
2017/07/04 02:00
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發問:
In the figure,Q is a point on the straight line L:3x+15y-53=0 such that PQ⊥L.THe coordinates of P are (5,6). (a)Find the equation of PQ. (b)Find the coordinates of Q. (c)Find the length of PQ,leave your answer in surd form.
最佳解答:
(a)∵PQ⊥L,設PQ的斜率為k,L的斜率k1=-3/15=-1/5 k=-1/k1=5 又因P的坐標為(5,6). 由點斜式得 y-6=5(x-5) y-6=5x-25 ∴5x-y-19=0 (b)因為Q在L上,且Q在PQ上,則Q為L和PQ的交點 {3x+15y-53=0-----------(1) {5x-y-19=0---------------(2) 由(2)得y=5x-19 ---------(3) (3)代入(1) 3x+15(5x-19)-53=0 3x+75x-285-53=0 78x=338 x=169/39 代入(3) y=5*169/39-19 y=104/39 即Q的坐標是(169/39 ,104/39) (c)由兩點距離公式得: │PQ│=√[(5-169/39)^2+(6-104/39)^2] =√[(26/39)^2+(130/39)^2] =[√(676+16900)]/39 =(√17576)/39 =(26√26)/39
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Maths(Coordinate Treatment of Simple Locus Problems)發問:
In the figure,Q is a point on the straight line L:3x+15y-53=0 such that PQ⊥L.THe coordinates of P are (5,6). (a)Find the equation of PQ. (b)Find the coordinates of Q. (c)Find the length of PQ,leave your answer in surd form.
最佳解答:
(a)∵PQ⊥L,設PQ的斜率為k,L的斜率k1=-3/15=-1/5 k=-1/k1=5 又因P的坐標為(5,6). 由點斜式得 y-6=5(x-5) y-6=5x-25 ∴5x-y-19=0 (b)因為Q在L上,且Q在PQ上,則Q為L和PQ的交點 {3x+15y-53=0-----------(1) {5x-y-19=0---------------(2) 由(2)得y=5x-19 ---------(3) (3)代入(1) 3x+15(5x-19)-53=0 3x+75x-285-53=0 78x=338 x=169/39 代入(3) y=5*169/39-19 y=104/39 即Q的坐標是(169/39 ,104/39) (c)由兩點距離公式得: │PQ│=√[(5-169/39)^2+(6-104/39)^2] =√[(26/39)^2+(130/39)^2] =[√(676+16900)]/39 =(√17576)/39 =(26√26)/39
其他解答:
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