If you have the coordinates (x, y) of the endpoints of a line segment, finding the **midpoint** of the segment is very simple. Just take the average of the two x-coordinates of the endpoints (add them and divide by two) to get the x-coordinate of the midpoint. Then do the same with the y-coordinates.

The midpoint "formula" is given below. It implies that the x coordinate of the midpoint is half way between the x coordinates of the endpoints, and likewise for the y-coordinates. This is proven using triangle congruence below the formula.

The coordinates of the midpoint of a line segment on a plane (2-dimensional) are:

$$\text{midpoint} = \left( \frac{x_2 + x_1}{2}, \; \frac{y_2 + y_1}{2} \right)$$

If we take line segment **AE** (right →) with midpoint **C**, we can draw all of the dashed lines parallel to the coordinate axes. All vertical lines are || and all horizontal lines are ||. Now because parallel lines yield congruent pairs of **corresponding angles** (**∠ 1 ≅ ∠ 3, ∠ 2 ≅ ∠ 4**), we have that **ΔABC ≅ ΔCDE** by the ASA (angle-side-angle) theorem.

The congruence of those triangles means that segments **AB** and **CD** are congruent, thus **B** bisects **AF**, and **BC ≅ DE**, thus **D** bisects **EF**.

So the coordinates of **C** are **x** = the midpoint of **AF** (the average of the x-coordinates of **A** and **E**) and **y** = midpoint of **ED** (the average of the y-coordinates of **A** and **B**.

X
### ASA

"Angle-Side-Angle" theorem

If two angles and the connecting side of one triangle are congruent to the corresponding two angles and connecting side of another, then the two triangles are identical

Calculate the midpoint of the segments with the following endpoints:

1. |
## Solution$$ \begin{align} \text{mp} &= \left( \frac{-1 + 2}{2}, \frac{-2 + 2}{2} \right) \\ \\ &= \left( \frac{1}{2}, 0 \right) \end{align}$$ |

2. |
## Solution$$ \begin{align} \text{mp} &= \left( \frac{-4+9}{2}, \frac{2-3}{2} \right) \\ \\ &= \left( \frac{5}{2}, \frac{-1}{2} \right) \end{align}$$ |

3. |
## Solution$$ \begin{align} \text{mp} &= \left( \frac{-4+3}{2}, \frac{-4-4}{2} \right) \\ \\ &= \left( \frac{-1}{2}, -4 \right) \end{align}$$ |

4. |
## Solution$$ \begin{align} \text{mp} &= \left( \frac{6-4}{2}, \frac{2+8}{2} \right) \\ \\ &= (1, 5) \end{align}$$ |

5. |
## Solution$$ \begin{align} \text{mp} &= \left( \frac{-1-1}{2}, \frac{-5+2}{2} \right) \\ \\ &= \left( -1, \frac{-3}{2} \right) \end{align}$$ |

6. |
## Solution$$ \begin{align} \text{mp} &= \left( \frac{6+6}{2}, \frac{20-4}{2} \right) \\ \\ &= \left( \frac{12}{2}, \frac{16}{2} \right) = (6, 8) \end{align}$$ |

7. Find the other endpoint of the segment with endpoint __midpoint__

The midpoint is

$$\left( \frac{-3 -x}{2}, \frac{-4 -y}{2} \right) = (2, 1)$$

This gives us an equation for the x-coordinate of the missing point, and one for the y. They are:

$$ \begin{align} \frac{-3 +x}{2} &= 2 \\ \\ -3 +x &= 4 \\ x &= 7 \end{align}$$

$$ \begin{align} \frac{-4 +y}{2} &= 1 \\ \\ -4 +y &= 2 \\ y &= 6 \end{align}$$

The point is (7, 6)

The midpoint idea can easily be extended to lines in three dimensions, where each point has a three-dimesional coordinate, like **(x, y, z)** or **(-1, 2, 7)**. It looks like this:

$$\text{mp} = \left( \frac{x_1 + x_2}{2}, \; \frac{y_1 + y_2}{2}, \; \frac{z_1 + z_2}{2} \right)$$

And we can extend that to lines in many dimensions. Even though the look of such line segments might be difficult for you to imagine, they do exist, and each has a midpoint given by:

$$\text{mp} = \left( \frac{x_1 + x_2}{2}, \; \frac{y_1 + y_2}{2}, \; \frac{z_1 + z_2}{2}, \; \dots \right)$$

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