Can someone help me review the distance formula and how to calculate a midpoint?

- anonymous

Can someone help me review the distance formula and how to calculate a midpoint?

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- anonymous

ok, sure...do you have any points to work with?

- anonymous

Hold on and I'll get a few my teacher gave me to work with

- anonymous

sure

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## More answers

- anonymous

W(3, -12) M(2, -1)

- anonymous

M is the midpoint

- anonymous

ok, good start. so draw it on a graph first

- anonymous

Ok one second

- anonymous

you need to find the other point, let's call it P...

- anonymous

Ok i have the graph drawn

- anonymous

nice, so what is the midpoint formula?

- anonymous

(X1+x2)/2 ; (Y1+y2)/ 2

- anonymous

ok cool, so you have W:(3, -12), and you have the midpoint...so M=(2,-1)...right so use the formulas. let's look at (x1+x2)/2

- anonymous

we have that \[2 = (x1 + x2) / 2]\, and that x1 = the x-coordinate from W

- anonymous

x1 = 3, x2 = ? --> 2 = ( 3 + x2 )/2...solve for x2

- anonymous

what did you get?

- anonymous

this would be the x-coordinate for the point P we made up

- anonymous

I got -2 for X2

- anonymous

hey, you can refresh the page when it gets stuck ;)

- anonymous

ok.. I was freaking there for a sec

- anonymous

the x-coordinate for W is 3, so you say x2 = -2, let's put that back in the Midpoint equation for X:
=> so we have 3 from W and now we have -2 from x2 , this is (3 + -2) = 1, then 1/2
but the goal is 2

- anonymous

So this is how I got -2 for an answer
2=(3+X2)/2
-3 -3
(-1)=X2/2
x2 x2
-2= X2

- anonymous

so you should multiply both sides by 2
1. 2 = ( 3 + x2 ) /2
2. Multiply both sides by 2 (to get ride of the 2 on the right side)
2*2 = ( 3 + x2 ) /2 *2
= 4 = (3 +x2)
3. now we have [ 4 = (3 + x2) ], solve for x2 get x2 = 1

- anonymous

Oh... okay. I didn't know to do that.

- anonymous

you skipped ahead and subtracted 3 from both sides, but 3 is in the parantheses. it's like we can get the 3 because the whole thing ( 3 + x2 ) is being divided by 2, so the equation actually is 3/2 + x2/2

- anonymous

*can't get the 3...

- anonymous

you want to try the same process to find y?

- anonymous

Alright.. Thanks for telling me that, my math teacher thought it was okay not to tell us that. And if you want too.

- anonymous

yes, let's solve for the y-coordinate of P...we have P = ( x2, y2 ) --> P = ( 1, y2 )

- anonymous

so you can look at your graph, and find 1 on the x-axis...now you're half way there to finding where P actually is

- anonymous

Okay

- anonymous

ok, so you're stuck at x=1, now you can only move up or down, changing y, so you have M on the graph and W on the graph, so you can make a guess at where P is just by checking out the graph...

- anonymous

Is it possible to use the slope intercept to find Y? or would that just add confusion?

- anonymous

you could, but you have a simple midpoint equation to find y... (y1 + y2) /2 = -1, with y1=-12, so ( -12 + y2 ) /2 = -1...solve it!

- anonymous

I ended up getting -14

- anonymous

what are your steps? remember that you have to multiply by 2 here before dealing with -12

- anonymous

(-12 + Y2)/2=-1
*2 *2
(-12 + y2)= -2
+-12 +-12
Y2= -14

- anonymous

ah, if you have -12 to both sides, you actually have (-24 + y2) = -14...so you should add 12 to both sides, to get y2 = -2 + 12 = 10

- anonymous

the goal was to get rid of -12 on the left side, but you can't just get rid of it, you have to keep the relationship that you already have....so if you get rid of -12 on the left side by adding 12 (thus getting 0), you have to do the same thing to the right side or else you've changed the relationship...it's all about find ways to reduce the problem

- anonymous

right, so what are the coordinates of our point P?

- anonymous

Okay, so the -2 we divided by is still on the right side of the equal sign?

- anonymous

right, you multiplied each side by 2 to get rid of the /2 on (y1 + y2), that would make the right side (-1) * (2) = -2

- anonymous

sorry, :(, I have to go, but I will be back in 15-20minutes

- anonymous

once you get the midpoint stuff down, you can work on the the distance formula, which is: \[d = \sqrt{ (x2 - x1)^2 + (y2 - y1)^2 }\]

- anonymous

so, brb

- anonymous

Okay I'll be here

- anonymous

all right, are you finished with midpoint stuff?

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