A)distance and internal distance
1.Given two points A(-2,1) and B(4,2), the line segment AB cuts the y-axis at Q(0,k),
find the value of k.
B)Slope,equation....
2.Find the equation of the perpendicular bisector of A(-2,1) and B(4,-5).
3.ON is perpendicular to the line l and N=(1,-2), find the equation of the line l.
4.Find the point of intersection of the lines L1 :2x+3y+4=0 and L2 :x-y-8=0.
5.Find the point of intersection of the lines L1 :x-2y+6=0 and L2 :3x+y+4=0.
Slope and Equation of stright line
2007-01-31 5:12 am
回答 (2)
2007-01-31 9:03 am
✔ 最佳答案
Q.1(k - 2)/(2 - 1) = (0 - 4)/(4 - (-2))
k - 2 = -4/6
= -2/3
k = 4/3.
Q.2
Let P(x,y) be any point on the line.
PA = PB
PA² = PB²
(x+2)² + (y-1)² = (x-4)² + (y+5)²
12x - 12y - 36 = 0
x - y - 3 = 0.
Q.3
Slope of ON = (-2-0)/(1-0) = -2
So, slope of line L = -1/(-2) = 0.5
Equation of line L:
y = 0.5x + c, for any real number c.
(as it is not defined that N is on the line L)
If N(1,-2) is on line L,
-2 = 0.5 + c , i.e. c = -2.5
Then, the line is y = 0.5x - 2.5
i.e. x - 2y - 5 = 0
2007-01-31 01:07:50 補充：
Q.4L1 : 2x 3y 4=0 ................. (a)L2 : x-y-8=0 ...................... (b)(a) 3 (b): 5x - 20 = 0 x = 4.Putting into (b), y = x - 8 = -4.Thus, the intersection point = (4, -4)
2007-01-31 01:08:03 補充：
Q.5L1 : x-2y 6=0 ................. (a)L2 :3x y 4=0 ................. (b)(a) 2 (b): 7x 14 = 0 x = -2.Putting into (b), y = -3x - 4 = 2.Thus, the intersection point = (-2, 2)
2007-01-31 01:08:51 補充：
So many questions... I suppose 10 points should be more reasonable than just 5 points.
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