anonymous
  • anonymous
please help!! I am usually really good with midpoint but this has stumped me and i dont know why! if you can explain all your steps that would be great! The midpoint of CD is E(–1, 0). One endpoint is C (5, 2). What are the coordinates of the other endpoint?
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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DebbieG
  • DebbieG
Just use the midpoint formulas to solve each for the coordinates. The x=coordinate of the midpoint is: \[M_{x}=\dfrac{ x_1 + x_2}{ 2 }\] So plug in your midpoint's x, and the known point's x, and solve for the other x-coordinate. do the same for y, and that will give you the 2 coordinates for the endpoint! :)
anonymous
  • anonymous
so it would look something like m= -1+5= 4/2=2 so 2 is the x coordinate? then for the y it would be m=0+2=2/2=1 so 1 =y and my endpoint is (2,1) ? Thanks for your help too btw! :)
DebbieG
  • DebbieG
Not quite... here is the equation for the x-coordinate of the midpoints (which I'm calling \(M_{x})\)\[M_{x}=\dfrac{ x_1 + x_2}{ 2 }\] Youre given the midpoint, so you know that x-coordinate, \(M_{x}=-1\). You also know the x of one point is 5. Now plug those both in: \[-1=\dfrac{5 + x_2}{ 2 }\] And solve for \(x_{2}\).

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DebbieG
  • DebbieG
That's the x-coordinate of the missing point. Do similarly for y to find the y coordinate of the missing point. And there are the coordinates for the point.
anonymous
  • anonymous
so how would you solve that? Multiply both sides by the 5 or something? thats the part that confuses me the most! after you have that -1+ 5+x2 divided by 2, what do you do to solve it?
DebbieG
  • DebbieG
Start by "clearing the fraction". That fraction will go away like magic *poof* if you simply multiply both sides of the equation by the denominator.\[2\cdot(-1)= \left( \dfrac{5 + x_2}{ 2 } \right)\cdot2\]
DebbieG
  • DebbieG
Then you will have a much simpler looking equation to solve for x.
anonymous
  • anonymous
ohh okay okay! So then i would have -2=5/2? and then i divide 5 by 2 and get 2.5...hmm that doesnt seem right..i dont know why this is confusing me so much!! its so simple!
DebbieG
  • DebbieG
noooo.....
anonymous
  • anonymous
oh!i wouldnt have to divide by 2
DebbieG
  • DebbieG
\[2\cdot(-1)= \left( \dfrac{5 + x_2}{ \cancel2 } \right)\cdot \cancel2\] What do you get?
anonymous
  • anonymous
so then you would be left with -2=5+x2
DebbieG
  • DebbieG
right! Now solve that for \(x_2\) and you get......?
anonymous
  • anonymous
so you would subtract 5 and get -7=x2 wow i dont know why i was having so much trouble with that!! thank you so much! so then for y..
DebbieG
  • DebbieG
That's right! You've got the idea now. And no problem, happy to help. Do the same for y. If you want, let me know what you get and I'll tell you if it's correct. :)
amistre64
  • amistre64
if the formulas give you troubles ... then find a different way to organize the data :)
anonymous
  • anonymous
The formula's are the ones i had learned to do before too, they were just fonsuing me for some reason. okay im gonna try it for y!
amistre64
  • amistre64
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amistre64
  • amistre64
5 to -1 is -6; -1 -6 = -7 2 to 0 is -2; 0 - 2 = -2
DebbieG
  • DebbieG
Nice, @amistre64 :) I can't disagree with that approach!! Although I would also say, it is also good to PRACTICE with the formulas so that you gain more comfort and proficiency with them. :)
anonymous
  • anonymous
wow that makes it look simple! thanks Amistre64! but if i were to do it the other way, i would have 0= 2+y2/2 then multiply by 2 so i would have 0=2+y2 then subtract 2 and have -2= y2?
anonymous
  • anonymous
its nice to know both ways too! the other way seems good to check my work since its so much simpler!
amistre64
  • amistre64
:) good luck
DebbieG
  • DebbieG
Yes, exactly! And notice that both approaches result in the same answer! (always how it should be!!) :)
anonymous
  • anonymous
thank you guys very much!! very helpful :)

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