General

@LowWillpower: Oh I understand. We aren't really learning different types of formulas (like the distance one) yet, but she likes to give homework on the next thing we're going to learn in class so we're kind of prepared. Thank you so much for this reference, I know it will help me in class.

You seem to know a lot about this area. Are you a teacher or did you just learn it previously?

[quote=ImNoMerchant]Well, to be honest we didn't really go depth of Pythagorean theorem so I kinda just tested it out but I got the right answer.
Now I have another method of doing the problem. Thanks to you. [/quote]
All you need to know for pythagorean theorem is a^2 + b^2 = d^2. If you picture drawing a line between two points in an x-y system, it will make a triangle, the line will be the hypotenuse d, because x and y are defined at right angles. x1 and y1 become the bottom tip of the triangle, from there you can see you know two lengths, but just need to find the third (y2 - y1), which we can call b, so (x2 - x1) = a

Taking a^2 + b^2 = d^2 and rearranging:

b^2 = d^2 - a^2

and

b = sqrt( d^2 - a^2)

Substitute x2 - x1 for a and y2 - y1 for b, and you have all you need to land right back at

y2 = sqrt( d^2 - (x2 - x1)) + y1

The distance formula is just hypotenuse theorem.

[quote=LowWillpower]This question would almost make more sense to solve b first so you could just then plug it in for a.

If you re-arrange it's pretty easy btw, since you can easily re-arrange so y2 - y1 = √(d² - (x2 - x1)² ). It's just a triangle solved using Pythagoras Theorem, d is the hypotenuse and x is one of your directions, find the length of the opposite side so that you can find y2 relative to y1.

Also, you messed up the other guy because in your partial solution picture, you showed 13 = √((2 - 7)² + ( y2 [b]+[/b] 6)² but then factored it out as if it was (y2 - 6)² like it should be.

Also, you'll notice in any answer, there is a √ term. You may recall that √ gives a + and - answer. In this case, you can actually take either root, because in both cases it is about how much y changes while x is changing, and the direction isn't relevant.[/quote]

Well, to be honest we didn't really go depth of Pythagorean theorem so I kinda just tested it out but I got the right answer.
Now I have another method of doing the problem. Thanks to you.

[quote=sparkshooter]My Part B is pretty ugly.
I got:
y2 = y1 + sqrt(d^2 - (x2-x1)^2)
and
y2 = y1 - sqrt(d^2 - (x2-x1)^2)[/quote]
Not really very ugly. Pretend this is a triangle, d is the hypotenuse, x2 - x1 is the base, and y2 - y1 is the height. When y1 = 0 (triangle sitting on the ground) y2 becomes the opposite length of a right angle triangle, it's height. As you change how high x starts, it's like raising up the entire triangle. The +/- in the root corresponds to it pointing up or down, since directionality isn't relevant.

My Part B is pretty ugly.
I got:
y2 = y1 + sqrt(d^2 - (x2-x1)^2)
and
y2 = y1 - sqrt(d^2 - (x2-x1)^2)

[quote=ImNoMerchant]@TheDarkHero3: Wow, thanks so much man. You have been a huge help.

@sparkshooter: I understand now, thank you.

@LowWillpower: Thank you for the explanation man. I finally get this now. [/quote]
NP, just out of curiosity, did you actually find the Pythagorean part helpful? Just wondering cause I find it usually helps if you can connect a situation a few different ways.

@TheDarkHero3: Will do. Thanks again.

[quote=ImNoMerchant]@TheDarkHero3: Wow, thanks so much man. You have been a huge help.

@sparkshooter: I understand now, thank you. [/quote]
Anytime! Make sure to be careful when rearranging equations.

@TheDarkHero3: Wow, thanks so much man. You have been a huge help.

@sparkshooter: I understand now, thank you.

@LowWillpower: Thank you for the explanation man. I finally get this now.

[quote=ImNoMerchant]Yeah I kinda over thought the whole thing and went 2 steps too far.
Btw do you know what b) is asking? I tried asking my teacher for help for b) but she explained it really weird.[/quote]
Yeah Part A should be -6 or 18. Both answers work.
Part B is just asking for the equation where y2 is on one side and d, x1, and x2 are on the other side.

This question would almost make more sense to solve b first so you could just then plug it in for a.

If you re-arrange it's pretty easy btw, since you can easily re-arrange so y2 - y1 = √(d² - (x2 - x1)² ). It's just a triangle solved using Pythagoras Theorem, d is the hypotenuse and x is one of your directions, find the length of the opposite side so that you can find y2 relative to y1.

Also, you messed up the other guy because in your partial solution picture, you showed 13 = √((2 - 7)² + ( y2 [b]+[/b] 6)² but then factored it out as if it was (y2 - 6)² like it should be.

Also, you'll notice in any answer, there is a √ term. You may recall that √ gives a + and - answer. In this case, you can actually take either root, because in both cases it is about how much y changes while x is changing, and the direction isn't relevant.

[quote=ImNoMerchant]Yeah I kinda over thought the whole thing and went 2 steps too far.
Btw do you know what b) is asking? I tried asking my teacher for help for b) but she explained it really weird.[/quote]
I tried posting b) earlier, but my internet crashed ><
Rearrange the formula for y2
d=√(x2 - x1)² + (y2 - y1)²
d²=(x2-x1)²+(y2-y1)²
d²-(x2-x1)²=(y2-y1)²
√(d²-(x2-x1)²)=(y2-y1)
√(d²-(x2-x1)²)-y1=y2
We can check this by using our previous results. If x1=7 x2=2 y1=6 and d=13, then we should get -6 and 18 for the value of y.
√(d²-(x2-x1)²)-y1=y2
√((13)²-(2-7)²)-6=y2
√(144)-6=y2
root 144 is +-12, so we make separate cases for each.
12-6=y2
y2=6
OR
-12-6=y2
y2=-18
Therefore,
√(d²-(x2-x1)²)-y1=y2 is how to get y2 if you have the other variables.

[quote=sparkshooter]This is simplification, you just have to avoid math errors.[/quote]

Yeah I kinda over thought the whole thing and went 2 steps too far.
Btw do you know what b) is asking? I tried asking my teacher for help for b) but she explained it really weird.

This is simplification, you just have to avoid math errors.

[quote=TheDarkHero3]I don't think you would change it to a positive. I am quite sure its a negative because you subtract y1, and since y1 is 6, subtracting 6 would be -6. You can even check your answer after you get the solutions for y.
0=y^2-12y-108
y equals -6 and 18.
Plug those in to the original question.
d=√(x2 - x1)² + (y2 - y1)²
13=√(25)+(-6-6)^2
13=√(25+144)
13=13
Same thing happens if you plug in 18 for y2.
If you were to change the 12y to a +, then y would equal to -18 and 6.
Plug those in
13=√(25)+(6-6)^2
13=5 which isn't true
Therefore 12y should be negative.[/quote]
Thanks a lot man.
I really appreciate it.

I think I messed up the other guy by not putting a square root sign in front of the original formula.

[quote=ImNoMerchant]Sorry to question you, but can you please explain why you would change it to a positive?

Edit: Oh and you were right, it is the square root of that. i just forgot it when writing it here but I used it when i was solving my work.[/quote]
I don't think you would change it to a positive. I am quite sure its a negative because you subtract y1, and since y1 is 6, subtracting 6 would be -6. You can even check your answer after you get the solutions for y.
0=y^2-12y-108
y equals -6 and 18.
Plug those in to the original question.
d=√(x2 - x1)² + (y2 - y1)²
13=√(25)+(-6-6)^2
13=√(25+144)
13=13
Same thing happens if you plug in 18 for y2.
If you were to change the 12y to a +, then y would equal to -18 and 6.
Plug those in
13=√(25)+(6-6)^2
13=5 which isn't true
Therefore 12y should be negative.

[quote=DiGiT]Make sure to change the -12y to +12y. You should be able to get the answer[/quote]

Sorry to question you, but can you please explain why you would change it to a positive?

Edit: Oh and you were right, it is the square root of that. i just forgot it when writing it here but I used it when i was solving my work.

Make sure to change the -12y to +12y. You should be able to get the answer

[quote=DiGiT]Well you actually got kinda far. What you have to do once you get 108 = y^2 - 12y is make it y^2 - 12y - 108 = 0 and then solve for y by factoring or quadratic formula.[/quote]
Ohh, wow thanks. (I feel so stupid.)
Lemme try it out.

Well you actually got kinda far. What you have to do once you get 108 = y^2 - 12y is make it y^2 - 12y - 108 = 0 and then solve for y by factoring or quadratic formula.
Edit: Btw, I'm pretty sure I was right when I said that distance formula is supposed to be square root. You seemed to have done it in your work when you squared both sides to get 169 = 25 + y^2 - 12y + 36. Also, I just realized I think it should be +12y not -.

[quote=DiGiT]I thought distance formula was the square root of that...[/quote]
I'm not entirely sure, my teacher gave this to me. (She might have made it up.)

I thought distance formula was the square root of that...