# Calculating the midpoint of a line segment

**Introduction**

In geometry, the midpoint is the centre point of a straight line. In a straight line, the midpoint is equidistance (equal distance) from both ends of the straight line. The midpoint of a straight line is the centroid of the line segment, which gives the exact centre point in a line segment. A midpoint bisects (cuts in half) a line. Let $$AB$$ be a straight line, and $$M$$ be the midpoint of $$AB$$. Then, $$AM=BM$$.

So, point $$M$$ bisects the straight line $AB$$ into two halves.

**E3.3: Calculate the coordinates of the midpoint of the straight line from the coordinates of its endpoints.**

Here, $$AB$$ is a straight line, and $$M$$ is the midpoint of the straight line. Let the coordinates of endpoint $$A$$ of the straight line be $$\left (x_{1}, y_{1} \right)$$, and the coordinates of endpoint $$B$$ of the straight line be $$\left(x_{2},y_{2} \right)$$.

Then, the formula to find the midpoint of the straight line is $$\left (\frac{x_{1}+x_{2}}{2} ,\frac{y_{1}+y_{2}}{2}\right)$$, where $$(x_{1},x_{2})$$ are the coordinates of $$x-\text{axis}$$, and $$ y_{1},y_{2}$$ are the coordinates of $$y-\text{axis}$$.

The concept of the midpoint is used in many real-life situations. For example, if we have a stick and we want to cut it in two equal parts, but we do not have any measuring device, in that situation, we use the concept of the midpoint.

Suppose $$\left ( 13,2 \right )$$ and $$\left ( 7,10 \right )$$ are two endpoints of a straight line. How do we find the coordinates of the midpoint of the straight line?

We use the formula of the midpoint of a straight line.

Let's give this a go.

$$ \left ( x_{1},y_{1} \right )= \left ( 13,2 \right )$$.

$$ \left ( x_{2},y_{2} \right )= \left ( 7,10 \right )$$.

Now, let us put the value of $$ x_{1}$$, $$y_{1}$$, $$x_{2}$$, and $$y_{2}$$ in

$$\left (\frac{x_{1}+x_{2}}{2} ,\frac{y_{1}+y_{2}}{2}\right )$$ to get the midpoint of the line.

$$\left (\frac{13+7}{2} ,\frac{2+10}{2}\right )$$,

Solve further.

$$\left (\frac{20}{2} ,\frac{12}{2}\right ) =\left (10,6\right)$$.

$$\left (10,6\right) $$

Therefore, the midpoint of the straight line is $$\left (10,6\right) $$.

**Worked examples of calculating the midpoint of a line segment**

**Example 1**: If $$\left (4,3 \right)$$ are the coordinates of the midpoint of a straight line, and the coordinates of endpoint $$A$$ are \left (5,4\right ), find the coordinates of the endpoint $$B$$ of the straight line.

**Step 1: Apply the formula of the midpoint of a straight line**

$$\left ( \frac{x_{1}+x_{2}}{2} ,\frac{y_{1}+y_{2}}{2}\right )$$

Put the values from the question.

$$\left (\frac{x_{1}+x_{2}}{2}\right )=4$$ and $$\left (\frac{y_{1}+y_{2}}{2}\right )=3$$

**Step 2: Put the value of $$(x_{1},x_{2})$$ in the above equation.**

$$\left (\frac{5+x_{2}}{2} \right )=4$$ and $$\left ( \frac{4+y_{2}}{2} \right )=3$$

**Step 3: Solve the above equation further.**

$$5+x=8$$ and $$4+y=6$$

**Step 4: Answer in preferred notation.**

$$5+x=8$$ gives $$x=3$$

$$4+y=6$$ gives $$y=2$$

Therefore, the coordinates of the endpoint $$B$$ are $$\left (3,2 \right) $$.

**Example 2**: Find the coordinates of the centre of a circle whose endpoints of the diameter are $$ \left (6,8\right)$$ and $$ \left (4,10\right)$$.

**Step 1: Let assume some points.**

$$\left (x_{1},y_{1}\right)=\left (6,8\right)$$

$$\left (x_{2},y_{12\right) =\left (4,10\right)$$

From the properties of a circle, it is clear the midpoint of the diameter is the centre of a circle. To find the centre of the circle, we have to find the midpoint of the diameter.

**Step 2: Apply the formula of the midpoint of a straight line.**

$$\left (\frac{x_{1}+x_{2}}{2} ,\frac{y_{1}+y_{2}}{2}\right )$$.

**Step 3: Substitute the values of $$x_{1}$$, $$x_{2}$$, $$y_{1}$$, and $$y_{2}$$ in the above equation.**

$$\left (\frac{6+4}{2},\frac{8+10}{2} \right )$$

**Step 4: Solve further.**

$$\left (\frac{10}{2},\frac{18}{2} \right )= \left (5,9 \right)$$

**Step 5: Answer in preferred notation.**

Therefore, the coordinates of the centre of the circle are $$\left (5,9\right)$$.