oh yah i saw that question in the text book...didn't know how to solve so i skipped
but a point can have 3 coordinate, if you're considering the z axis, but then again we haven't been taught into such details
im sure its something simple, that we're just not getting it.
anyways i have one more question
could someone plz solve C2 question 8 Jun05,
8a) x^2+y^2-10x+9=0
x^2-10x+y^2+9=0
(x-5)^2-25+y^2+9=0
(x-5)^2+y^2+9=0
hence
(x-5)^2+y^2-16=0
center: (5,0)
8b.) (x-5)^2+y^2=16
square root 16 =4=radius
8c) radius= 4 center : (5,0)
substitute y in the f(x) equation with 0
therefore:
x^2-10x+9=0
(x-9)(x-1)=0
x=9
x=1
hence
(9,0) and (1,0)
8d) gradiant of AT
m= -2/7
y-y=m(x-x)
y-0=-2/7(x-5)
hence
y=-2/7(x-5)
