posted by isaiah

sorry, but one more question...
there are a couple questions that say something like write the equation 2x+3y-5=0 in normal form... but isnt it already in normal form?

  1. Reiny

    Good grief, haven't seen that type of question in about 50 years.

    didn't know they still used it .
    I will describe the method and then attempt to explain what happened.

    They are using the word "normal" to mean perpendicular.
    e.g. in your equation, 2x+3y - 5 = 0, the slope is -2/3.
    So the slope of a normal would be +3/2,
    (recall the negative reciprocal stuff.)

    1. change the equation so that the constant is on the right side as a positive number.
    ----> 2x + 3y = 5
    2. if the equation is in the form ax + by = c
    find √(a^2 + b^2)
    --- √(2^2 + 3^2) = √13

    3. divide each term by √13

    Normal form equivalent to
    2x + 5y = 5 is:

    2x/√13 + 5/√13 = 5/√13

    in general: the normal form of a general form equation ax + by + c = 0 looks like this
    xcosØ + ysinØ = p , where p = c/√(a^2 + b^2)

    where Ø is the angle that the normal makes with the x-axis

    e.g. for our example,
    cosØ = 2/√13 and sinØ = 5/√13
    Ø = appr 56.3°

    I really don't recall ever using this form, or ever teaching it in my 35 years of teaching.
    Last time I saw this was in university back in 1961,
    and I actually had to look it up

  2. isaiah

    thank you so have really helped me today...i have a test tomorrow and i think ill do much better than i wouldve without your help

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