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