- #1
offtheleft
- 131
- 1
Homework Statement
having a little bit of trouble. some of the information is new to me. ill point out because I am copying this down from my notes.
let A & B define some linear function, [tex]y=f(x)[/tex].
A = (1, 2), B = (6, 6)
C = (x, y), D = (x, y)
let (x, y) represent some arbitrary points on the line defined by f but, not on points A or B.
C is between A and B and, D is somewhere past B.
so far, i know what's goin on. i have the slope which is [tex]\frac{4}{5}[/tex].
here's where I'm lost, i might have zoned out because of my ADHD but this is what's going on
1, from A [tex]\rightarrow[/tex] C, [tex]\frac{y-2}{x-1}[/tex]
2, from B [tex]\rightarrow[/tex] D, [tex]\frac{y-6}{x-6}[/tex]
3, from B [tex]\rightarrow[/tex] C, [tex]\frac{-(6-y)}{-(6-x)}[/tex]
4, and from A [tex]\rightarrow[/tex] D is the same as A [tex]\rightarrow[/tex] C
I have no idea how all that was figured out ( from 1 - 4 ) someone please explain that to me?
and it continues into the next part
[tex]\frac{y-2}{x-1}[/tex] = [tex]\frac{y-6}{x-6}[/tex]
i can comprehend that.
[tex](x-1)(x-6)[/tex] [tex]\frac{y-2}{x-1}[/tex] = [tex](x-1)(x-6)[/tex] [tex]\frac{y-6}{x-6}[/tex]
not sure how that all happened.
than it goes too..
[tex]\frac{x-1}{x-1}[/tex] [tex][(x-6)(y-2)][/tex] = [tex]\frac{x-6}{x-6}[/tex] [tex][(x-1)(y-6)][/tex]
(this is where i started paying attention again)
cancel stuff out and do some fun algebraic gymnastics and i came up with this;
[tex]xy-2x-6y+12=xy-6x-y+6[/tex]
more fun algebra stuff and i came up with;
[tex]y=\frac{4}{5}x+\frac{6}{5}[/tex]
i have no problem doing certain things. but that whole middle piece just lost me until it ws put into some fun polynomial expression. i need to know how i got that