Question

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?

Answer 1

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! :)

Answer 2

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! :)

Answer 3

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 \(x_{2}\).

Answer 4

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.

Answer 5

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?

Answer 6

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.\[2\cdot(-1)= \left( \dfrac{5 + x_2}{ 2 } \right)\cdot2\]

Answer 7

Then you will have a much simpler looking equation to solve for x.

Answer 8

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!

Answer 9

noooo.....

Answer 10

oh!i wouldnt have to divide by 2

Answer 11

\[2\cdot(-1)= \left( \dfrac{5 + x_2}{ \cancel2 } \right)\cdot \cancel2\]

What do you get?

Answer 12

so then you would be left with -2=5+x2

Answer 13

right! Now solve that for \(x_2\) and you get......?

Answer 14

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..

Answer 15

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. :)

Answer 16

if the formulas give you troubles ... then find a different way to organize the data :)

Answer 17

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!