Youtube Math Question

bio_man

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3 years ago

Youtube Math Question

I have a problem in math but I’m really lost and struggling on what to do. The question is “The slope of GH if GH (is perpendicular to) AB. A(5,4) and B(-1,-6). I’m also have difficulty in reciprocals and negative reciprocals and when I must use it. For example, for my slope, I got -10/-6 but I’m not sure if it’s correct or how to express it into the final answer. Do I just leave it and my final answer for slope is written as -10/-6? Thanks!

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bio_man Author

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Educator

3 years ago

Assuming that the coordinates of link GH are A(5,4) and B(-1,-6), we can find the slope like this:

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theepicseal

wrote...

3 years ago

Thanks! So this means that slope of AB should be
3
- —
5

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bio_man Author

wrote...

Educator

3 years ago

The slope of GH is 5 over 3, but if you want the slope of the line that's perpendicular to it, it'd be -3/5

You weren't very clear in your request, so I didn't know what to find.

Now, what's next?

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theepicseal

wrote...

3 years ago

Oh my apologies! The fraction in my last reply is the slope of GH if GH is perpendicular to AB?

Here is what I am dealing with

A=(5,4)
B=(-1,-6)

a) calculate the distance from A to B. round your answer to 1 decimal place.
b) find the midpoint between A and B
c) calculate the slope of AB
d) calculate the slope of GH if GH is perpendicular to AB
e) find the equation of the circle with centre (0,0) that passes through point A

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bio_man Author

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Educator

3 years ago Edited: 3 years ago, bio_man

That's much better!

Those are all the answers, let me know if you have questions

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theepicseal

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3 years ago

Wow thank you so much! I don’t have any questions. It’s making sense to me now, and got me on the right track to make sure I’m doing the correct steps. There’s also a last question regarding the A & B coordinates I gave in the previous reply.

It says:

f) determine the equation of the line through A and B

I’m kind of confused on what they’re asking and how to proceed with this

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bio_man Author

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Educator

3 years ago

Ok, so start with y=mx+b

substitute the slope we found earlier in for m

y=(5/3)x+b

now substitute a point in for x and y, and solve for b; let's choose the coordinates of A:

4=(5/3)(5)+b
4 - (25/3) = b
-13/3 = b

therefore, the equation is:

y=(5/3)x-(13/3)

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theepicseal

wrote...

3 years ago

Thanks!! I’m wondering though, what happened through the bottom two lines.

4=(5/3)(5)+b
4 - (25/3) = b (how did it go from + b to = b, the symbols in front of 4 and b just swapped? also, i am wondering how 5/3 turned into 25/3, are we only supposed to multiply the numerator by the value of “x” and the denominator remains?
-13/3 = b (what happened to the numbers to reach this as our answer? what has happened to our equation from the line above this one, to the current line -13/3 = b

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bio_man Author

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Educator

3 years ago

So I multiplied (5/3)(5) together to get 25/3. Then I moved it over to the other side where it became minus 25/3. My ultimate goal there is to solve for b by isolating it on one side of the equation.

Does that make sense?

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theepicseal

wrote...

#10

3 years ago

Yes I think it makes sense now! The only thing I think I need clarification is why and how we went from

4 - (25/3) = b

to

-13/3 = b

Slight Smile

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bio_man Author

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#11

Educator

3 years ago

Quote from: theepicseal (3 years ago)

Yes I think it makes sense now! The only thing I think I need clarification is why and how we went from

4 - (25/3) = b

to

-13/3 = b

Slight Smile

Think of 2 fractions being subtracted:

4/1 - 25/3

finder a common denomintor being 3

12/3 - 25/3

now subtract the numerators 12 minus 25 and retain the denominator

-13/3

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theepicseal

wrote...

#12

3 years ago

Oh! I gotcha! Thank you so much! I’m clear of questions as of now! I have a lot of math work tomorrow regarding this topic, solving linear equations with elimination and substitution so I’ll definitely come back here if I’m facing any struggles! Thank you so much! I’m so glad I commented on YouTube and discovered a whole new platform that I can use to reach out for help with such prompt replies! Slight Smile

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bio_man Author

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#13

Educator

3 years ago

You're welcome, I log in every few hours, so if you don't get a response from me right away, don't be concerned

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