Find the co-ordinates of intersection

The straight line of graph y=3x-6 cuts the x axis at A, and the y axis at B.

a) Find the co-ordinates of A and B. (write down forumla, if there's any)

b) Calculate the length of AB (please write the formula used)

c) M is the mid-point of AB.
Find the co-ordinates of M (please write the forumla used)

Please show all your working out.
Thanks, Best answer will be chosen.

a) y = 3x - 6

When it cuts the x-axis, let y=0
0 = 3x - 6
6 = 3x
3x = 6
x = 2

When it cuts the y-axis, let x=0
y = 3 (0) - 6
= -6

Therefore, the co-ordinate of A is (2) and B is (-6) AB (2,-6)

b) For the length we must use the distance formula:
d = √(x2 - x1)^2 + (y2-y1)^2

Information we have: x1 = 0 x2 = 2 y1 = -6 y2 = 0

Therefore,
d = √(2-0)^2 + (0-(-6))^2
= √ (4) + (6)^2
= √40

Therefore, the length of AB is √40 units

c) Midpoint Formula: ((x1+x2)/2) , (y1+y2)/2))

From the values of x1, x2, y1 and y2 we are going to use it again.

Midpoint of AB = ((0+2)/2) , ((-6+0)/2)
= 1 , -3

Therefore, the co-ordinates of M is (1, -3)

a/ This line cuts the x axis at A
y = 3x - 6 (1)
y = 0 (2) --> x axis
From (1)(2) -> 3x - 6 = 0 x = 2 -> A(2,0)
This line cuts the y axis at B
y = 3x - 6 (3)
x = 0 (4) -> y axis
From (1)(2) -> y = 3*0 -6 = -3 -> B(0,-3)

b/
AB = sqrt[(0-2)^2 + (-3-0)^2]= sqrt(4+9)=sqrt(13)

c/
M = ((0+2)/2,(-3+0)/2) = (1,-3/2)

ok all you have to do is go to this website right here its
http://www.wolframalpha.com/widgets/gallery/
and then you eneter what kind of caculator you want
and it will do it for you
and show you step by step :)