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

用戶238057

2008-11-12 6:39 pm

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

回答 (1)

用戶10012

2008-11-12 8:19 pm

✔ 最佳答案

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

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