Slope of Perpendicular Line to x - 3y = 9

AI Thread Summary
To find the slope of a line perpendicular to the equation x - 3y = 9, the first step is to rewrite the equation in slope-intercept form (y = mx + b). After rearranging, the slope of the original line is determined to be 1/3. The slope of a line perpendicular to this is the negative reciprocal, which is -3. The discussion highlights the importance of understanding slopes and suggests seeking tutoring for better comprehension, especially when juggling multiple subjects. Ultimately, the correct slope for the perpendicular line is confirmed to be -3.
AngelShare
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I don't know if I'm having a brain fart or what but I'm drawing a blank...:smile:

What is the slope of a line perpendicular to x - 3y = 9?

Notes:
Perpendicular lines have slopes that are negative reciprocals of each other.

Example: So, if the original line has a slope of -3/2, the line perpendicular to the original line has a slope of 2/3.

So, if I work that problem out, I get...

-3y = 9 - x
y = 9 - x/-3
y =

That is where I'm locked up...what do I do with the x? It's been a while since I've done this (The last time I turned any homework in in this class was eight days ago. I've been juggling catching this class up and keeping up in Chemistry so...:bugeye: ) so I'm, more than likely, just having a brain fart.:-p
 
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y=mx + b

where m is equal to the slopeNow, what is your slope for the equation x - 3y = 9
when you solve for y?
Essentially, what is the coefficient multiplying x?

For example, in y=4x + 2
the slope is 4

Now to get the negative reciprocal, you know how to do...I'm going to be painfully honest with you, you should have known this. If you are juggling catching this class up and keeping up in chemistry, I suggest you go into tutoring to let all the math just really kick in. If you start understanding it, you will have to put in far less effort afterwards. Chemistry, try to do some group studying!

Good luck!
 
Tutoring and group studying sounds like a "painfully honest" and simple solution but it isn't. I'm not a public school student, I'm a cyber student. I don't even have books let alone teachers. I had trouble with Trig so my Dad tried to find a tutor several ways. He checked with a college near here (Juniata), my public high school, and my cyber school- nothing. The college didn't get back to us for a while, my high school sent out an email to the math department only to get no replies, and my cyber school suggested the college...when we had already checked. And as for group studying, nothing like that existed at my school and it certainly doesn't in this one. If it did, I doubt as many kids would be failing when the grading system was changed so that anything beneath a 65% is failing when it used to be 70%. As my "mentor" put it, "We aren't teachers, we're 'mentors" because, technically, we aren't teaching you; you're teaching yourself."

And, to explain why I'm working on catching up in Math just in case I seem like a slacker:-p , as I stated already, I had trouble with Trig. I solved that problem by dropping that class and replacing it with my current one. It took me two weeks to do so because my teacher argued with me saying that I could do it even though I clearly could not. That means, I had to catch up...I was behind by 35 assignments two weeks ago. Currently, I'm all caught up.:wink:

Now, onto the actual question.:biggrin:

I know I'm to be solving for y as I began to do above. However, I got to a point where I couldn't figure out what to do. Is this right?

y = 1/3x - 3

So, the slope in that would be 1/3 making my answer...-3?
 
AngelShare said:
I know I'm to be solving for y as I began to do above. However, I got to a point where I couldn't figure out what to do. Is this right?
y = 1/3x - 3
So, the slope in that would be 1/3 making my answer...-3?
That's correct :smile:
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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