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標題:
a maths...equation of straight lines....
發問:
two points A (3,2) and B (-4,1) are given. (a) Find the equation of the perpendicular bisector ofAB. (b) Find a point on the line 2x-y+3=0 such that it is equidistant from A and B.
最佳解答:
其他解答:
(a) m = -(3+4)/(2-1) = -7 x1 = (3-4)/2 = -1/2 y1 = (2+1)/2 = 3/2 y-y1=m(x-x1) ans: y = -14x + 1 (b) y = 2x+3 ---(1) (x-3)^2+(y-2)^2 = (x+4)^2+(y-1)^2 ---(2) From (2) y = -7x-2 ---(3) (1) = (3) ans: (-5/9, 17/9)|||||(a) Let (x, y) be the mid point of A, B x = (3-(-4))/2 = 3.5 y =(2-1)/2 = 0.5 slope of AB = (2-1)/(3-(-4)) = 1/7 slope of the line perpendicular to AB = -1/slope of AB = -1/(1/7)= -7 The equation of the line: (y-0.5)/(x-3.5) = -7 y -0.5 = -7x +24.5 y+7x -25 = 0 (2) Let the point P(x, y) on the line 2x-y+3 =0 to produce PA= PB (x-3)^2 + (y-2)^2 = (x+4)^2 + (y-1)^2 x^2 - 6x + 9 + y^2 - 4y + 4 = x^2 + 8x + 16 + y^2 - 2y + 1 14x + 2y + 4 = 0 => 7x + y + 2 =0 -------(1) Sub 2x+3 = y into Eq(1): 7x + 2x+3 + 2 = 0 x = -5/9 sub into Eq(1): 7(-5/9) + y + 2 = 0 y = 17/9 The point P is (-5/9, 17/9)
a maths...equation of straight lines....
發問:
two points A (3,2) and B (-4,1) are given. (a) Find the equation of the perpendicular bisector ofAB. (b) Find a point on the line 2x-y+3=0 such that it is equidistant from A and B.
最佳解答:
此文章來自奇摩知識+如有不便請留言告知
a) The slope of AB: ( 2 - 1 ) / ( 3 + 4 ) = 1 / 7 So the slope of the perpendicular bisector of AB: -7 The mid-pt of AB: [ ( 3 - 4 ) / 2, ( 2 + 1 ) / 2 ] = ( - 1/2, 3/2 ) So, the equation: ( y - 3/2 ) / ( x + 1/2 ) = - 7 7x + y + 2 = 0 b) Let the point be ( x , y ). 2x - y + 3 = 0 y = 2x + 3 --- ( 1 ) √[ ( x - 3 )^2 + ( y - 2 )^2 ] = √[ ( x + 4 )^2 + ( y - 1 )^2 ] ( x - 3 )^2 + ( y - 2 )^2 = ( x + 4 )^2 + ( y - 1 )^2 ( x - 3 + x + 4 )( x - 3 - x - 4 ) = ( y - 1 + y - 2 )( y - 1 - y + 2 ) ( 2x + 1 )( - 7 ) = ( 2y - 3 )( 1 ) -14x - 7 = 2y - 3 14x + 2y + 4 = 0 7x + y + 2 = 0 --- ( 2 ) Put ( 1 ) into ( 2 ), 7x + 2x + 3 + 2 = 0 9x = -5 x = - 5 / 9 y = 2 ( - 5 / 9 ) + 3 = 17/9 So the pt. is ( - 5 / 9, 17 / 9 ).其他解答:
(a) m = -(3+4)/(2-1) = -7 x1 = (3-4)/2 = -1/2 y1 = (2+1)/2 = 3/2 y-y1=m(x-x1) ans: y = -14x + 1 (b) y = 2x+3 ---(1) (x-3)^2+(y-2)^2 = (x+4)^2+(y-1)^2 ---(2) From (2) y = -7x-2 ---(3) (1) = (3) ans: (-5/9, 17/9)|||||(a) Let (x, y) be the mid point of A, B x = (3-(-4))/2 = 3.5 y =(2-1)/2 = 0.5 slope of AB = (2-1)/(3-(-4)) = 1/7 slope of the line perpendicular to AB = -1/slope of AB = -1/(1/7)= -7 The equation of the line: (y-0.5)/(x-3.5) = -7 y -0.5 = -7x +24.5 y+7x -25 = 0 (2) Let the point P(x, y) on the line 2x-y+3 =0 to produce PA= PB (x-3)^2 + (y-2)^2 = (x+4)^2 + (y-1)^2 x^2 - 6x + 9 + y^2 - 4y + 4 = x^2 + 8x + 16 + y^2 - 2y + 1 14x + 2y + 4 = 0 => 7x + y + 2 =0 -------(1) Sub 2x+3 = y into Eq(1): 7x + 2x+3 + 2 = 0 x = -5/9 sub into Eq(1): 7(-5/9) + y + 2 = 0 y = 17/9 The point P is (-5/9, 17/9)
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