4 A.maths Q.(Equation of Straight Line and Circle) 20POINTS.－hzb53jl55v的部落格

標題:

4 A.maths Q.(Equation of Straight Line and Circle) 20POINTS.

發問:

1, The angle between a pair of straight lines with slopes 7 and m is 45 degree Find the values of m2, Find the equation of the perpendicular bisector of AB where coordinates of A and B are (-1,4) and (7,8) repectively.3, A straight line passes through (1,2) and cuts the axes at two distinct points P... 顯示更多 1, The angle between a pair of straight lines with slopes 7 and m is 45 degree Find the values of m 2, Find the equation of the perpendicular bisector of AB where coordinates of A and B are (-1,4) and (7,8) repectively. 3, A straight line passes through (1,2) and cuts the axes at two distinct points P and Q. If PO=PQ, where O is the origin, find the equations of the line. 4, Find the equation of circle with its centre at(2,4) which touches the line x-y-3=0 THANK YOU VERY MUCH

最佳解答:

1. Let tan a = 7 and tan b = m. so a + 45 = b.............(1) or b + 45 = a.................(2) [ext. angle of triangle.] From (1) tan (a + 45) = tan b (tan a + 1)/[1 - (tan a)(tan 45)] = m (7 + 1)/( 1 - 7) = m = 8/(-6) = -4/3. From (2) tan (b + 45) = tan a (tan b + tan 45)/[1 - (tan b)(tan 45)] = 7 (m + 1)/(1 - m) = 7 m + 1 = -7m + 7 6 = 8m m = 6/8 = 3/4. 2. Slope of AB = (4-8)/(-1-7) = -4/(-8) = 1/2. so slope of perpendicular bisector = -2. Mid-point of A and B is (3,6). so equation of perpendicular bisector is y - 6 = -2(x - 3) y - 6 = -2x + 6 2x + y = 12. 3. I think it should be OP = OQ, not PQ. If this is correct, then slope of line = -1 or 1. So equations of the line are: a) y - 2 = x - 1 y = x + 1. b) y - 2 = -(x - 1) y - 2 = -x + 1 y + x = 3. 4. Radius of circle = distance from centre to line = abs[(2 - 4 - 3)/sqrt(1^2 + 1^2)] = abs[-5/sqrt2] = 5/sqrt2. so equation of circle is (x - 2)^2 + (y-4)^2 = (5/sqrt2)^2 x^2 + 4 - 4x + y^2 + 16 - 8y - 25/2 = 0 2x^2 + 2y^2 - 8x - 16y + 15 = 0