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locus Figure shows a circle with centre O.AB is a chord.P is point on AB such that OP I AB (a) Find the equation of OP (b) Fing the coordinates of P (C) Hence prove that AP=PB http://i129.photobucket.com/albums/p203/yu123_2006/maths2.jpg?t=1165917047 thx!!! 更新: Figure shows a circle with centre O.AB is a chord.P is point on AB such that OP 垂直AB. The figure is put on the coordinate plane, such that the coordinates of O, A and B are (4.5),(1,1) and (8,2) respectively.
最佳解答:
a. slope of AB = (2-1)/(8-2) = 1/7 slope of OP = -7 (since AB perpendicular PO) (y-5)/(x-4) = -7 y-5 = -7x+28 7x+y-33=0 b. P lies on both OP and AB equation of AB: (y-1)/(x-1) = 1/7 7y-7 = x-1 x-7y+6=0 x-7y+6=0 7x+y-33=0 upon solving, we have x=9/2,y=3/2 so P=(9/2,3/2) c. method 1 AP = sqrt((9/2-1)^2 + (3/2-1)^2) = 5/sqrt2 PB = sqrt((9/2-8)^2 + (3/2-2)^2) = 5/sqrt2 method 2 angleOPA = angleOPB OP = OP OA = OB = radius therefore, OAP congruent OBP(RHS) so AP=BP(corresponding sides, congruent triangles)
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發問:locus Figure shows a circle with centre O.AB is a chord.P is point on AB such that OP I AB (a) Find the equation of OP (b) Fing the coordinates of P (C) Hence prove that AP=PB http://i129.photobucket.com/albums/p203/yu123_2006/maths2.jpg?t=1165917047 thx!!! 更新: Figure shows a circle with centre O.AB is a chord.P is point on AB such that OP 垂直AB. The figure is put on the coordinate plane, such that the coordinates of O, A and B are (4.5),(1,1) and (8,2) respectively.
最佳解答:
a. slope of AB = (2-1)/(8-2) = 1/7 slope of OP = -7 (since AB perpendicular PO) (y-5)/(x-4) = -7 y-5 = -7x+28 7x+y-33=0 b. P lies on both OP and AB equation of AB: (y-1)/(x-1) = 1/7 7y-7 = x-1 x-7y+6=0 x-7y+6=0 7x+y-33=0 upon solving, we have x=9/2,y=3/2 so P=(9/2,3/2) c. method 1 AP = sqrt((9/2-1)^2 + (3/2-1)^2) = 5/sqrt2 PB = sqrt((9/2-8)^2 + (3/2-2)^2) = 5/sqrt2 method 2 angleOPA = angleOPB OP = OP OA = OB = radius therefore, OAP congruent OBP(RHS) so AP=BP(corresponding sides, congruent triangles)
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