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

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

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

sure

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

M is the midpoint

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

Ok one second

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

Ok i have the graph drawn

nice, so what is the midpoint formula?

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

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

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

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

what did you get?

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

I got -2 for X2

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

ok.. I was freaking there for a sec

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

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

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

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

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

*can't get the 3...

you want to try the same process to find y?

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

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

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

Okay

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...

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

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!

I ended up getting -14

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

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

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

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

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

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

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

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

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

so, brb

Okay I'll be here

all right, are you finished with midpoint stuff?

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