1.Find the equation of a straight line passing through(3,2)such that
a)it is perpendicular to the line X=6
b)it is parallel to the line X=6
2.Find the equation of the perpendicular bisector of the line segment with endpoints A(6,-3)and B(2,3).
F.5Locus
2007-11-18 6:29 am
回答 (2)
2007-11-18 6:37 am
✔ 最佳答案
1 (a) When the line is perpendicular to x = 6, the line must in the form y = k.Since the line passes through (3, 2), 2 = k
Hence the equation required is y = 2.
(b) When the line is parallel to x = 6, the line must in the form x = k.
Since the line passes through (3, 2), 3 = k
Hence the equation required is x = 3.
2. Mid-point of AB = ((6+2)/2, (-3+3)/2) = (4, 0)
Slope of AB = (3+3)/(2-6) = 6/(-4) = -3/2
The slope of perpendicular bisector = -1/(-3/2) = 2/3
Equation required: y - 0 = (2/3)(x-4)
3y = 2x-8
2x-3y-8 = 0
2007-11-18 6:44 am
1.A. Y=2
B. X=3
2.The slope of AB: 3-(-3)/2-6=-3/2
So:the slope of required equation :-1/ (-3/2)=2/3
The point it pass through:X=(6+2)/2=4
Y=(3-3)/2=o
So:the equation : y=2/3X-8/3
2x-3y-8=0
B. X=3
2.The slope of AB: 3-(-3)/2-6=-3/2
So:the slope of required equation :-1/ (-3/2)=2/3
The point it pass through:X=(6+2)/2=4
Y=(3-3)/2=o
So:the equation : y=2/3X-8/3
2x-3y-8=0
收錄日期: 2021-04-13 18:12:33
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20071117000051KK04728