**Introduction**

If we want to find the midpoint of segment ** CD** in

*Figure 1*, where the coordinates of points

**and**

*C***are and , it is clear that the length of**

*D***is units. To determine the midpoint of**

*CD***, we want to get the coordinates of the point which is units away from both points**

*CD***and**

*C***. 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**

*D**y-*coordinate of the midpoint would be or . This means that the midpoint of

*is .*

**CD**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

**is .**

*AB*

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

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

**, we draw vertical segment**

*AB***coinciding with the**

*PQ**y*-axis with endpoints having

*y*-coordinates the same as those of the

*y*-coordinates of segment

**. We also draw a horizontal line**

*AB*(see Figure 3)**coinciding the**

*RS**x*-axis with endpoints having

*x*-coordinates the same as those of the

*x*-coordinates of segment

**. Looking at Figure 3, it is clear that the coordinates of the midpoint of the**

*AB***is .**

*PQ*If we draw a horizontal line from towards segment ** AB** (see yellow dashed segment), and draw a vertical line from the intersection

**to segment**

*M***, 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:**

*RS*“If the midpoint of ** PQ **is and the midpoint of

**is , is the midpoint of**

*RS**, ?*

**AB**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

**is the midpoint of**

*M***.**

*AB*To show that ** M** is the midpoint of

**, it is sufficient to show that**

*AB***is congruent to**

*AM***. This leads us to Figure 5, where we label the right angles of the two triangles**

*MB***and**

*T***. We will now show that triangle**

*U***is congruent to triangle**

*AUM***. If so, then we can show that**

*MTB***is congruent to**

*AM***since they are the corresponding sides of the given triangles.**

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

Since ** PB **and

**are horizontal segments, we can consider**

*MT***as a transversal of parallel segments**

*AB***and the segment containing**

*PB***. It follows that angle**

*MT***and angle**

*TBM***are congruent because they are corresponding angles. It is also clear that**

*UMA***is congruent to**

*BT***and**

*MU*,**angle**

**is congruent to angle**

*T***since they are both right angles. Therefore, by the**

*U***ASA**congruence theorem,

**is congruent to**

*AUM***. (For an explanation of parallel lines and transversals, click here).**

*MTB*Since corresponding parts of congruent triangles are congruent, ** AM** is congruent to

**. Hence,**

*MB***is the midpoint of**

*M***.**

*AB*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

**are , the coordinates of**

*B**is equal to .*

**M**** **

** **

** **

**Delving Deeper**

There are also other ways to show that the midpoint of * AB *is

**. and lies on the midpoint formula. The details of the solutions are left to the reader as an exercise.**

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