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Algebra 2 help

  1. Sep 10, 2008 #1
    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
     
  2. jcsd
  3. Sep 10, 2008 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Yes, and each of those is equal to 4/5. you don't really need the "-" in (3) because -1/-1= 1. [tex]\frac{6- y}{6- x}[/itex] is sufficent. Of course that is exactly the same as [tex]\frac{y- 6}{x- 6}[/tex].

    Well yes, because C= D= (x,y)!


    [/quote]I have no idea how all that was figured out ( from 1 - 4 ) someone please explain that to me?[/quote]
    There wasn't anything "figured out" until you add the "= 4/5" part! What you wrote in 1- 4 is just [tex]\frac{y_1- y_0}{x_1- x_0}[/tex] for two points [tex](x_0, y_0)[/tex] and [tex](x_1, y_1)[/tex]. Those are all from the definition of "slope" of a line which you had already figured out was 4/5.

    Do you know the definition of "slope" of a line? Do you know how [tex]tan(\theta)[/tex] is defined where [itex]\theta[/tex] is an angle of right triangle? Without knowing those, you can't comprehend it.

    You are clearing the fractions by multiplying both sides by the denominators of the fractions.

    Okay, just to make it clear what is happening they move the corresponding numerator and denominator together so you can see they cancel. That was what I said before: they are clearing the fractions by multiplying by the denominators so the denominators cancel.

    So you weren't paying attention before? No wonder you didn't understand!

    Sometimes you can do mathematics by memorizing formulas. Sometimes you actually have to think. If you get to a part you don't understand, DON'T stop "paying attention". Concentrate on it and look at every detail. There was nothing there that you hadn't seen before.
     
  4. Sep 10, 2008 #3
    i didnt mean to stop paying attention. i have REALLY bad ADHD.

    so that whole section where i had 1 - 4 was just a more complicated or elaborate way to find the slope?

    after you clarified a few things i understand it more except for the [tex]
    tan(\theta)[tex]. when it comes to sin, cos, tan, etc... im lost. i had geom in highschool, three years later i went from algebra 1 straight to algebra 2.
     
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