Slope and Equation of stright line

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.

I don't kone