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標題:

F5 math problems ( Locus )

發問:

1.Passing through ( 5 , 7 ) and with inclination 45°, find the equation of straight line. 2.Let L be a straight line joining A( 2 , 5 ) and B( 6 , 3 ). Find the equation of the line perpendicular to L at A.

最佳解答:

1. slope = tan 45 = 1 If the equation is y = mx + c, therefore y = x + c as m = 1 Put (5, 7) into the equation, 7 = 5 + c c = 2 so, the equation is y = x + 2 2. slope of AB = (5 - 3) / (2 - 6) = - 0.5 L is perpendicular to AB, therefore slope of L * slope of AB = -1 slope of L = -1 / slope of AB = 2 therefore the equation of L is y = 2x + c Put A(2, 5) into L, 5 = 2(2) + c 5 - 4 = c c = 1 the equation of c is y = 2x + 1

其他解答:

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• F.1 Maths_4
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1. For an inclination 45°, the slope of the line is tan 45° = 1. We use point-slope form. (y - 7) / (x - 5) = 1 y - 7 = x - 5 Equation: x - y = -2 2. Slope of L: (5 - 3) / (2 - 6) = -2 / -4 = 1/2 Slope of line perpendicular to L = -1 / (1/2) = -2 We apply point-slope form here. (y - 5) / (x - 2) = -2 y - 5 = -2x + 4 Equation: 2x + y = 9 2007-10-27 01:09:28 補充： Some amendments to answer 2:Slope of L: (5 - 3) / (2 - 6) = 2/ -4 = -1/2Slope of line perpendicular to L: -1 / (-1/2) = 2Apply point-slope form.(y - 5) / (x - 2) = 2y - 5 = 2x - 4Equation: 2x - y = -1