_{1}+ x

_{2})/2, (y

_{1}+ y

_{2})/2

(6, 10), (-8, 5)

Tutors, sign in to answer this question.

The midpoint of (6, 10) & (-8, 5) = (x_{1} + x_{2})/2, (y_{1} + y_{2})/2

x = (6 + -8)/2 ; (-2/2) ; = -1

y = (10 + 5)/2 ; 7.5

The midpoint of this line is (-1, 7.5)

Pretend the segment joins the bottom left with the top right of this figure. The entire x distance (left to right) can be split into two parts--the horizontal lines of the top triangle and the bottom triangle. The total change in y is split over the two vertical lines in the figure below. The hypotenuses of these triangles are the same length because their sides are the same length. So the top right hypotenuse is equal to the bottom left hypotenuse making the point where the triangles touch the midpoint of the segment.

x

, l

, l <- change in y / 2

, l

, l

x --------------

, l ^ change in x / 2

, l

, l <- change in y / 2

, l

x--------------

^ change in x / 2

Vertical change:

The total change in Y is 10-5 = 5. The change in y / 2 = 5/2 = 2.5

Added to the lowest value (or sutracted from the high value is 10-2.5 = 5+2.5 = 7.5. This is the new Y value of the midpoint.

Horizontal Change:

The total change in x is 6 - -8 = 14. The total change in x / 2 = 14/2 = 7.

When added to the lower value or subtracted from the higher value we get our new x value at the midpoint of -8 + 7 = 6 - 7 = -1.

Our segment has a midpoint of (-1, 7.5).

It's important to note that this process is similar to taking the average of x values and y values to find the midpoint--but will work for finding a point anywhere along the line segment. To find the point one third of the way from one end to the other (divide by 3 instead of by 2), or one fourth of the way (divide by 4) etc. Taking the average is the shortcut that helps you find only the midpoint.

**Clarification:

To clarify the above description, what I meant to say is that

the average works only to find the midpoint. If you want to find points other than the midpoint you need to use the change in x / 2 added to the lower or subtracted from the higher value of x but instead of taking the halfway point (dividing by 2) you would need to use a different fraction: 1/3 or dividing by 3 for a point one third of the way... Apparently there were points of confusion, sorry.

Already have an account? Log in

By signing up, I agree to Wyzant’s terms of use and privacy policy.

Or

To present the tutors that are the best fit for you, we’ll need your ZIP code.

Your Facebook email address is associated with a Wyzant tutor account. Please use a different email address to create a new student account.

Good news! It looks like you already have an account registered with the email address **you provided**.

It looks like this is your first time here. Welcome!

To present the tutors that are the best fit for you, we’ll need your ZIP code.

Please try again, our system had a problem processing your request.

## Comments

x1 + 1/3(x2-x1) =

x1 - 1/3 x1 + 1/3 x2 =

2/3 x1 + 1/3 x2

which is NOT EQUAL to (x1 + x2)/3

Trying a general case:

x1 + (a/b)(x2-x1) =? (a/b)(x1 + x2)

x1 - (a/b) x1 + (a/b) x2 =? (a/b) x1 + (a/b) x2

(1 - (a/b)) x1 + (a/b) x2 ≠ (a/b) x1 + (a/b) x2

For 1 - a/b = a/b

1 = 2 (a/b)

a/b = 1/2

So ONLY 1/2 WORKS!