Question

I need to know the steps of how to do this.
The midpoint of line CD is E(-1,0). One endpoint is C(5,2). What are the coordinates of the other endpoint?

Answer 1

The midpoint is half way along the line from C to D.

Answer 2

You can look at it and see how much each coordinate changes.. from C to E, and this will be the same change from E to D...

From C to E,
x changes -6
y changes -2

The same changes happen from E to D, take off 6 on the x and 2 on the y

D(x,y) = (-1-6 , 0 - 2) = (-7 , -2)

Answer 3

Or if you like the formulas, the midpoint M(x,y) between points (x1,y1) and (x2,y2) is

\[\large M(x,y) = (\frac{ x _{1}+x _{2} }{ 2 } , \frac{ y _{1}+y _{2} }{ 2 })\]

Answer 4

How would I do it using the formula?

Answer 5

They give you one endpoint and the midpoint.
C(x,y) = (5,2)
M(x,y) = (-1,0)


\[(-1,0) = (\frac{ 5 + x _{2} }{ 2 }, \frac{ 2 + y _{2} }{ 2 })\]

Answer 6

The other endpoint (x2,y2) can be found from that...

\[\large -1 = \frac{ 5+x _{2} }{ 2 } ~~~~~and~~~~~0=\frac{ 2+y _{2} }{ 2 }\]

Answer 7

That gives you the other endpoint D
\[x _{2}=-7~~~~~and~~~~~y _{2}=-2\]

D(-7,-2)

Answer 8

Can you explain how you got -7 from \[\frac{ 5+x _{2} }{ 2 }\]

Answer 9

Well that's complicated.

Answer 10

@DanJS ?

Answer 11

The midpoint formula is
\[\large M(x,y) = (\frac{ x _{1}+x _{2} }{ 2 } , \frac{ y _{1}+y _{2} }{ 2 })\]
that means the x-coordinate of midpoint M is half the sum of the x coordinates x1 and x2 of the two endpoints. The same for the y coordinates.

Answer 12

The midpoint is M(-1,0)
The given end point is C(5,2)
Putting that into the midpoint formula
\[\large M(-1,0) = (\frac{ 5+x _{2} }{ 2 } , \frac{ 2+y _{2} }{ 2 })\]

Answer 13

So you get matching the x and y coordinates the two equations.

\[-1=\frac{ 5+x _{2} }{ 2 }~~~~~and~~~~~0=\frac{ 2+y _{2} }{ 2 }\]

Answer 14

yes, but how did you get from that to (-7,-2)

Answer 15

That is the point (x2,y2) you are finding...
Solve the two equations
\[-1=\frac{ 5+x _{2} }{ 2 }\]

\[-2=5+x _{2}\]
\[x _{2}=-7\]

Answer 16

Ok, thanks for your patience.