(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

having a little bit of trouble. some of the information is new to me. ill point out because im 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 whats 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 whats 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

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# Homework Help: Algebra 2 help

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