# Midpoint Formula

• MHB
53. A line segment has (x_1, y_1) as one endpoint and (x_m, y_m) as its midpoint. Find the other endpoint (x_2, y_2) of the line segment in
terms of x_1, y_1, x_m, and y_m.

54. Use the result of Exercise 53 to find the endpoint (x_2, y_2) of each line segment with the given endpoint (x_1, y_1) and midpoint (x_m, y_m).

(a) (x_1, y_1) = (1, −2)
(x_m, y_m) = (4, −1)

(b) (x_1, y_1) = (−5, 11)
(x_m, y_m) = (2, 4)

I need help with 53 and 54.

jonah1
Beer soaked query follows.
53. A line segment has (x_1, y_1) as one endpoint and (x_m, y_m) as its midpoint. Find the other endpoint (x_2, y_2) of the line segment in
terms of x_1, y_1, x_m, and y_m.

54. Use the result of Exercise 53 to find the endpoint (x_2, y_2) of each line segment with the given endpoint (x_1, y_1) and midpoint (x_m, y_m).

(a) (x_1, y_1) = (1, −2)
(x_m, y_m) = (4, −1)

(b) (x_1, y_1) = (−5, 11)
(x_m, y_m) = (2, 4)

I need help with 53 and 54.
What have you done so far?

HOI
53. A line segment has (x_1, y_1) as one endpoint and (x_m, y_m) as its midpoint. Find the other endpoint (x_2, y_2) of the line segment in
terms of x_1, y_1, x_m, and y_m.
Call the other endpoint (x_2, y_2). Then x_m= (x_1+ x_2)/2 so 2x_m= x_1+ x_2 and x_2= 2x_m- x_1. Similarly y2= 2y_m- y_1.

54. Use the result of Exercise 53 to find the endpoint (x_2, y_2) of each line segment with the given endpoint (x_1, y_1) and midpoint (x_m, y_m).

(a) (x_1, y_1) = (1, −2)
(x_m, y_m) = (4, −1)
x_2= 2(4)- 1= 8- 1= 7
y_2= 2(-1)- (-2)= -2+ 2= 0.
(x_2, y_2)= (7, -0). Check: ((7+1)/2, (-2- 0)/2)= (8/2, -2/2)= (4, -1).

(b) (x_1, y_1) = (−5, 11)
(x_m, y_m) = (2, 4)

I need help with 53 and 54.
Why don't you try the last one?

Call the other endpoint (x_2, y_2). Then x_m= (x_1+ x_2)/2 so 2x_m= x_1+ x_2 and x_2= 2x_m- x_1. Similarly y2= 2y_m- y_1.

x_2= 2(4)- 1= 8- 1= 7
y_2= 2(-1)- (-2)= -2+ 2= 0.
(x_2, y_2)= (7, -0). Check: ((7+1)/2, (-2- 0)/2)= (8/2, -2/2)= (4, -1).

Why don't you try the last one?

You said:

"Call the other endpoint (x_2, y_2). Then x_m= (x_1+ x_2)/2 so 2x_m= x_1+ x_2 and x_2= 2x_m- x_1. Similarly y2= 2y_m- y_1."

HOI
m++
You said:

"Call the other endpoint (x_2, y_2). Then x_m= (x_1+ x_2)/2 so 2x_m= x_1+ x_2 and x_2= 2x_m- x_1. Similarly y2= 2y_m- y_1."

There is NO "2x" in what I wrote. If you mean "2x_m", multiply both sides of x_m= (x_1+ x_2)/2 by 2.

jonah1
Beer soaked ramblings follow.
You said:

"Call the other endpoint (x_2, y_2). Then x_m= (x_1+ x_2)/2 so 2x_m= x_1+ x_2 and x_2= 2x_m- x_1. Similarly y2= 2y_m- y_1."

Fourteen years of Precalculus review and you're still stumped by the midpoint formula?
I can see that your quest for Calculus understanding will be a very rough and even longer ride.
For someone who once pretended he's on his way to mastering Calculus as https://mathhelpboards.com/members/harpazo.8631/, it is disappointing to see you struggling once again with basic stuff.

Maybe you can see this more clearly if Countryboy decided to give his excellent answer in LaTex.
Too hammered right now to do it, so see a somewhat similar application of your present dilemma at https://mathforums.com/threads/symmetrical-line.356080/

But then again, I might already be in your prestigious Ignore List and you might not see this at all.

HOI
I didn't use LaTeX because the original post did not! Here:

https://mathhelpboards.com/goto/post?id=124850
53. A line segment has $(x_1, y_1)$ as one endpoint and $(x_m, y_m)$ as its midpoint. Find the other endpoint $(x_2, y_2)$ of the line segment in
terms of $x_1$, $y_1$, $x_m$, and $y_m$.
Call the other endpoint $(x_2, y_2)$. Then $x_m= (x_1+ x_2)/2$ so $2x_m= x_1+ x_2$and $x_2= 2x_m- x_1$. Similarly $y2= 2y_m- y_1$.

54. Use the result of Exercise 53 to find the endpoint $(x_2, y_2)$ of each line segment with the given endpoint $(x_1, y_1)$ and midpoint $(x_m, y_m)$.

(a) $(x_1, y_1) = (1, −2)$
$(x_m, y_m) = (4, −1)$
$x_2= 2(4)- 1= 8- 1= 7$
$y_2= 2(-1)- (-2)= -2+ 2= 0$.
(x_2, y_2)= (7, -0). Check: ((7+1)/2, (-2- 0)/2)= (8/2, -2/2)= (4, -1).

Last edited:
I didn't use LaTeX because the original post did not! Here:

Call the other endpoint $(x_2, y_2)$. Then $x_m= (x_1+ x_2)/2$ so $2x_m= x_1+ x_2$and $x_2= 2x_m- x_1$. Similarly $y2= 2y_m- y_1$.

$x_2= 2(4)- 1= 8- 1= 7$
$y_2= 2(-1)- (-2)= -2+ 2= 0$.
(x_2, y_2)= (7, -0). Check: ((7+1)/2, (-2- 0)/2)= (8/2, -2/2)= (4, -1).

Nicely done. I thank you.

jonah1
Beer soaked ramblings follow.
Nicely done. I thank you.
Translation: I love cajoling members into doing my work without showing any kind of effort whatsoever on my part.

HOI
You're such a sweetheart!