**Introduction**

If we want to find the midpoint of segment *CD* in *Figure 1*, where the coordinates of points *C* and *D* are and , it is clear that the length of *CD* is units. To determine the midpoint of *CD*, we want to get the coordinates of the point which is units away from both points *C* and *D*. Hence, we have to divide by , and add the result to or subtract the result from . Summarizing, the expression that would describe the value of the *y-*coordinate of the midpoint would be or . This means that the midpoint of **CD** is .

If we want to get the midpoint of *AB*, using the same reasoning above, the expression that would describe the *x*-coordinate of the midpoint would be or . This means that the midpoint of *AB *is .

Figure 1 – Horizontal line AB and vertical line CD in the coordinate plane.

Generalizing our observation above, if we have a vertical segment with their endpoints having coordinates and (see Figure 2), we can get its midpoint using the following formula or

Figure 2 – Generalized coordinates of a vertical and a horizontal line.

Simplifying, both the expressions above result to . Similarly, for a horizontal segment with endpoints having coordinates and can be computed by the expression .

**Midpoint of a Slanting Segment**

The preceding derivations are only valid for vertical and horizontal segments. In Figure 3, segment *AB* is neither horizontal nor vertical. To investigate the midpoint of *AB*, we draw vertical segment *PQ* coinciding with the *y*-axis with endpoints having *y*-coordinates the same as those of the *y*-coordinates of segment *AB *(see Figure 3). We also draw a horizontal line *RS* coinciding the *x*-axis with endpoints having *x*-coordinates the same as those of the *x*-coordinates of segment *AB*. Looking at Figure 3, it is clear that the coordinates of the midpoint of the *PQ* is .

If we draw a horizontal line from towards segment *AB* (see yellow dashed segment), and draw a vertical line from the intersection *M* to segment *RS*, it seems that the intersection of the yellow dashed line and segment *RS* is which is the midpoint of *RS*. From here, it is tempting to ask the following question:

“If the midpoint of *PQ *is and the midpoint of *RS* is , is the midpoint of **AB**, ?

Figure 3 – A non-horizontal segment AB with midpoint M.

Generalizing the questions above, we might want to ask “If the midpoint the point with coordinates *A * and B equal to ?”

If we extend the* QA* and the horizontal yellow dashed line to the right, we can from two right triangles as shown in *Figure 4*. From the statements above, we want to show that point *M* is the midpoint of *AB*.

To show that *M* is the midpoint of *AB*, it is sufficient to show that *AM* is congruent to *MB*. This leads us to Figure 5, where we label the right angles of the two triangles *T* and *U*. We will now show that triangle *AUM* is congruent to triangle *MTB*. If so, then we can show that *AM* is congruent to *MB* since they are the corresponding sides of the given triangles.

**Proof that ***AUM* is congruent to *MTB*

Since *PB *and *MT* are horizontal segments, we can consider *AB* as a transversal of parallel segments *PB* and the segment containing *MT*. It follows that angle *TBM* and angle *UMA* are congruent because they are corresponding angles. It is also clear that *BT* is congruent to *MU*, and** **angle* T* is congruent to angle *U* since they are both right angles. Therefore, by the **ASA** congruence theorem, *AUM* is congruent to *MTB*. (For an explanation of parallel lines and transversals, click here).

Since corresponding parts of congruent triangles are congruent, *AM* is congruent to *MB*. Hence, *M* is the midpoint of *AB*.

Figure 5 – The triangle produced by extending the horizontal lines passing through the three points.

Note that our proof did not talk about coordinates, but the general case. That is, if the coordinates of *A* are and the coordinates of *B* are , the coordinates of **M** is equal to .

** **

** **

** **

**Delving Deeper**

There are also other ways to show that the midpoint of **AB **is *M*. and lies on the midpoint formula. The details of the solutions are left to the reader as an exercise.

- From Figure 3, draw two right triangles with hypotenuse
*AM* and hypotenuse *AB* and show that *AM* is half of *AB*.
- Using the distance formula, show that the distance between point
*A *and point *M *is the same as the distance between point *M* and point *B*.
- Show that
*AM* and *MB* has the same slope.
- Get the equation of the line containing
*AB,* and substitute the coordinates of *M* to the equation of line *AB*.