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?

asked by isaiah
  1. 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

    posted by Reiny
  2. 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

    posted by isaiah

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