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# 8.4 Mid-point of a line

## Finding the mid-point of a line

On graph paper, accurately plot the points $$P\left(2;1\right)$$ and $$Q\left(-2;2\right)$$ and draw the line $$PQ$$.

• Fold the piece of paper so that point $$P$$ is exactly on top of point $$Q$$.

• Where the folded line intersects with line $$PQ$$, label point $$S$$

• Count the blocks and find the exact position of $$S$$.

• Write down the coordinates of $$S$$.

To calculate the coordinates of the mid-point $$M\left(x;y\right)$$ of any line between the points $$A\left({x}_{1};{y}_{1}\right)$$ and $$B\left({x}_{2};{y}_{2}\right)$$, we use the following formulae:

\begin{align*} x & = \frac{{x}_{1} + {x}_{2}}{2} \\ y & = \frac{{y}_{1} + {y}_{2}}{2} \end{align*}

From this we obtain the mid-point of a line:

Mid-point $$M\left(x;y\right) = \left(\dfrac{{x}_{1}+{x}_{2}}{2} \; ; \; \dfrac{{y}_{1}+{y}_{2}}{2}\right)$$

This video shows some examples of finding the mid-point of a line.

Video: 2GDF

## Worked example 10: Calculating the mid-point

Calculate the coordinates of the mid-point $$F\left(x;y\right)$$ of the line between point $$E\left(2;1\right)$$ and point $$G\left(-2;-2\right)$$.

### Draw a sketch

From the sketch, we can estimate that $$F$$ will lie on the $$y$$-axis, with a negative $$y$$-coordinate.

### Assign values to $$\left({x}_{1};{y}_{1}\right)$$ and $$\left({x}_{2};{y}_{2}\right)$$

${x}_{1} = -2 \quad {y}_{1} = -2 \quad {x}_{1} = 2 \quad {y}_{2}=1$

### Write down the mid-point formula

$F\left(x;y\right) = \left(\frac{{x}_{1} + {x}_{2}}{2};\frac{{y}_{1} + {y}_{2}}{2}\right)$

### Substitute values into the mid-point formula

\begin{align*} x & = \frac{{x}_{1} + {x}_{2}}{2} \\ & = \frac{-2 + 2}{2} \\ & = 0 \\ y & = \frac{{y}_{1} + {y}_{2}}{2} \\ & = \frac{-2 + 1}{2} \\ & = -\frac{1}{2} \end{align*}

The mid-point is at $$F\left(0;-\frac{1}{2}\right)$$.

Looking at the sketch we see that this is what we expect for the coordinates of $$F$$.

For worked example 10 (calculating the mid-point) learners can check their answer using the distance formula.

Using the distance formula, we can confirm that the distances from the mid-point to each end point are equal:

\begin{align*} PS & = \sqrt{{\left({x}_{1} - {x}_{2}\right)}^{2} + {\left({y}_{1} - {y}_{2}\right)}^{2}} \\ & = \sqrt{{\left(0 - 2\right)}^{2} + {\left(-\text{0,5} - 1\right)}^{2}} \\ & = \sqrt{{\left(-2\right)}^{2} + {\left(-\text{1,5}\right)}^{2}} \\ & = \sqrt{4 + \text{2,25}} \\ & = \sqrt{\text{6,25}} \end{align*}

and

\begin{align*} QS & = \sqrt{{\left({x}_{1}-{x}_{2}\right)}^{2}+{\left({y}_{1}-{y}_{2}\right)}^{2}} \\ & = \sqrt{{\left(0-\left(-2\right)\right)}^{2} + {\left(-\text{0,5}-\left(-2\right)\right)}^{2}} \\ & = \sqrt{{\left(0 + 2\right)}^{2}{+\left(-\text{0,5} + 2\right)}^{2}} \\ & = \sqrt{{\left(2\right)}^{2}{+\left(-\text{1,5}\right)}^{2}} \\ & = \sqrt{4 + \text{2,25}} \\ & = \sqrt{\text{6,25}} \end{align*}

As expected, $$PS=QS$$, therefore $$F$$ is the mid-point.

## Worked example 11: Calculating the mid-point

Find the mid-point of line $$AB$$, given $$A\left(6;2\right)$$ and $$B\left(-5;-1\right)$$.

### Draw a sketch

From the sketch, we can estimate that $$M$$ will lie in quadrant I, with positive $$x$$- and $$y$$-coordinates.

### Assign values to $$\left({x}_{1};{y}_{1}\right)$$ and $$\left({x}_{2};{y}_{2}\right)$$

Let the mid-point be $$M\left(x;y\right)$$

${x}_{1} = 6 \quad {y}_{1} = 2 \quad {x}_{2} = -5 \quad {y}_{2} = -1$

### Write down the mid-point formula

$M\left(x;y\right) = \left(\frac{{x}_{1} + {x}_{2}}{2};\frac{{y}_{1} + {y}_{2}}{2}\right)$

### Substitute values and simplify

$M\left(x;y\right) = \left(\frac{6 - 5}{2};\frac{2 - 1}{2}\right) = \left(\frac{1}{2};\frac{1}{2}\right)$

$$M\left(\frac{1}{2};\frac{1}{2}\right)$$ is the mid-point of line $$AB$$.

We expected $$M$$ to have a positive $$x$$- and $$y$$-coordinate and this is indeed what we have found by calculation.

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## Worked example 12: Using the mid-point formula

The line joining $$C\left(-2;4\right)$$ and $$D\left(x;y\right)$$ has the mid-point $$M\left(1;-3\right)$$. Find point $$D$$.

### Draw a sketch

From the sketch, we can estimate that $$D$$ will lie in Quadrant IV, with a positive $$x$$- and negative $$y$$-coordinate.

### Assign values to $$\left({x}_{1};{y}_{1}\right)$$ and $$\left({x}_{2};{y}_{2}\right)$$

Let the coordinates of $$C$$ be $$\left({x}_{1};{y}_{1}\right)$$ and the coordinates of $$D$$ be $$\left({x}_{2};{y}_{2}\right)$$.

${x}_{1} = -2 \quad {y}_{1} = 4 \quad {x}_{2} = x \quad {y}_{2} = y$

### Write down the mid-point formula

$M\left(x;y\right) = \left(\frac{{x}_{1} + {x}_{2}}{2};\frac{{y}_{1} + {y}_{2}}{2}\right)$

### Substitute values and solve for $${x}_{2}$$ and $${y}_{2}$$

$\begin{array}{rlcrl} 1 & = \dfrac{-2 + {x}_{2}}{2} & \qquad\qquad & -3& = \dfrac{4 + {y}_{2}}{2} \\ 1\times 2& = -2 + {x}_{2} & \qquad\qquad & -3 \times 2 & = 4 + {y}_{2} \\ 2 & = -2 + {x}_{2} & \qquad\qquad & -6 & = 4 + {y}_{2} \\ {x}_{2} & = 2 + 2 & \qquad\qquad & {y}_{2} & = -6 - 4 \\ {x}_{2} & =4 & \qquad\qquad & {y}_{2} & = -10 \end{array}$

The coordinates of point $$D$$ are $$\left(4;-10\right)$$.

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## Worked example 13: Using the mid-point formula

Points $$E\left(-1;0\right)$$, $$F\left(0;3\right)$$, $$G\left(8;11\right)$$ and $$H\left(x;y\right)$$ are points on the Cartesian plane. Find $$H\left(x;y\right)$$ if $$EFGH$$ is a parallelogram.

### Draw a sketch

Method: the diagonals of a parallelogram bisect each other, therefore the mid-point of $$EG$$ will be the same as the mid-point of $$FH$$. We must first find the mid-point of $$EG$$. We can then use it to determine the coordinates of point $$H$$.

### Assign values to $$\left({x}_{1};{y}_{1}\right)$$ and $$\left({x}_{2};{y}_{2}\right)$$

Let the mid-point of $$EG$$ be $$M\left(x;y\right)$$

${x}_{1} = -1 \quad {y}_{1} = 0 \quad {x}_{2} = 8 \quad {y}_{2} = 11$

### Write down the mid-point formula

$M\left(x;y\right) = \left(\frac{{x}_{1} + {x}_{2}}{2};\frac{{y}_{1} + {y}_{2}}{2}\right)$

### Substitute values calculate the coordinates of $$M$$

$M\left(x;y\right) = \left(\frac{-1 + 8}{2};\frac{0 + 11}{2}\right) = \left(\frac{7}{2};\frac{11}{2}\right)$

### Use the coordinates of $$M$$ to determine $$H$$

$$M$$ is also the mid-point of $$FH$$ so we use $$M\left(\dfrac{7}{2};\dfrac{11}{2}\right)$$ and $$F\left(0;3\right)$$ to solve for $$H\left(x;y\right)$$.

### Substitute values and solve for $$x$$ and $$y$$

$\begin{array}{rlcrl} \dfrac{7}{2} & = \dfrac{0 + x}{2} & \qquad \qquad & \dfrac{11}{2} & = \dfrac{3+y}{2} \\ 7 & = x + 0 & \qquad \qquad & 11 & = 3 + y \\ x & = 7 & \qquad \qquad & y & = 8 \end{array}$

The coordinates of $$H$$ are $$\left(7;8\right)$$.

Textbook Exercise 8.5

You are given the following diagram:

Calculate the coordinates of the mid-point ($$M$$) between point $$A (-\text{1};\text{3})$$ and point $$B (\text{3};-\text{3})$$.

Let the coordinates of $$A$$ be $$\left({x}_{1};{y}_{1}\right)$$ and the coordinates of $$B$$ be $$\left({x}_{2};{y}_{2}\right)$$.

${x}_{1} = -1 \quad {y}_{1} = 3 \quad {x}_{2} = 3 \quad {y}_{2} = -3$

Substitute values into the mid-point formula:

\begin{align*} M\left(x;y\right) & = \left(\frac{{x}_{1} + {x}_{2}}{2};\frac{{y}_{1} + {y}_{2}}{2}\right) \\ x & = \frac{{x}_{1} + {x}_{2}}{2} \\ & = \frac{-1 + 3}{2} \\ & = 1 \\ y & = \frac{{y}_{1} + {y}_{2}}{2} \\ & = \frac{3 + (-3)}{2} \\ & = 0 \end{align*}

The mid-point is at $$M\left(1;0\right)$$.

You are given the following diagram:

Calculate the coordinates of the mid-point ($$M$$) between point $$A (-\text{2};\text{1})$$ and point $$B (\text{1};-\text{3,5})$$.

Let the coordinates of $$A$$ be $$\left({x}_{1};{y}_{1}\right)$$ and the coordinates of $$B$$ be $$\left({x}_{2};{y}_{2}\right)$$.

${x}_{1} = -2 \quad {y}_{1} = 1 \quad {x}_{2} = 1 \quad {y}_{2} = -\text{3,5}$

Substitute values into the mid-point formula:

\begin{align*} M\left(x;y\right) & = \left(\frac{{x}_{1} + {x}_{2}}{2};\frac{{y}_{1} + {y}_{2}}{2}\right) \\ x & = \frac{{x}_{1} + {x}_{2}}{2} \\ & = \frac{-2 + 1}{2} \\ & = -\text{0,5} \\ y & = \frac{{y}_{1} + {y}_{2}}{2} \\ & = \frac{1 + (-\text{3,5})}{2} \\ & = -\text{1,25} \end{align*}

The mid-point is at $$M\left(-\text{0,5};-\text{1,25}\right)$$.

Find the mid-points of the following lines:

$$A(2;5)$$, $$B(-4;7)$$

\begin{align*} M_{AB} &= \left(\frac{x_1 + x_2}{2};\frac{y_1 + y_2}{2}\right) \\ &= \left(\frac{2 - 4}{2};\frac{5 + 7}{2}\right) \\ &= \left(\frac{-2}{2};\frac{12}{2}\right) \\ &= (-1;6) \end{align*}

$$C(5;9)$$, $$D(23;55)$$

\begin{align*} M_{CD} &= \left(\frac{x_1 + x_2}{2};\frac{y_1 + y_2}{2}\right) \\ &= \left(\frac{5 + 23}{2};\frac{9 + 55}{2}\right) \\ &= \left(\frac{28}{2};\frac{64}{2}\right) \\ &= (14;32) \end{align*}

$$E(x + 2;y - 1)$$, $$F(x - 5;y - 4)$$

\begin{align*} M_{EF} &= \left(\frac{x_1 + x_2}{2};\frac{y_1 + y_2}{2}\right) \\ &= \left(\frac{x + 2 + x - 5}{2};\frac{y - 1 + y - 4}{2}\right) \\ &= \left(\frac{2x - 3}{2};\frac{2y - 5}{2}\right) \end{align*}

The mid-point $$M$$ of $$PQ$$ is $$(3;9)$$. Find $$P$$ if $$Q$$ is $$(-2;5)$$.

The mid-point formula is:

$M\left(x;y\right) = \left(\frac{{x}_{1} + {x}_{2}}{2};\frac{{y}_{1} + {y}_{2}}{2}\right)$

Substituting values and solving for $${x}_{2}$$ and $${y}_{2}$$ gives:

$\begin{array}{rlcrl} 3 & = \dfrac{-2 + {x}_{2}}{2} & \qquad\qquad & 9 & = \dfrac{5 + {y}_{2}}{2} \\ 6 & = -2 + {x}_{2} & \qquad\qquad & 18 & = 5 + {y}_{2} \\ {x}_{2} & = 6 + 2 & \qquad\qquad & {y}_{2} & = 18 - 5 \\ {x}_{2} & = 8 & \qquad\qquad & {y}_{2} & = 13 \end{array}$

The coordinates of point $$P$$ are $$\left(8;13\right)$$.

$$PQRS$$ is a parallelogram with the points $$P(5;3)$$, $$Q(2;1)$$ and $$R(7;-3)$$. Find point $$S$$.

Draw a sketch:

The diagonals of a parallelogram bisect each other, therefore the mid-point of $$QR$$ will be the same as the mid-point of $$PS$$. We must first find the mid-point of $$QR$$. We can then use it to determine the coordinates of point $$H$$.

\begin{align*} M_{QR} &= \left(\frac{x_1 + x_2}{2};\frac{y_1 + y_2}{2}\right)\\ &= \left(\frac{2 + 7}{2};\frac{1 - 3}{2}\right) \\ &= \left(\frac{9}{2};\frac{-2}{2}\right) \\ &= \left(\frac{9}{2};-1\right) \end{align*}

Use mid-point $$M$$ to find the coordinates of $$S$$:

\begin{align*} M_{QR} &= (\frac{x_1 + x_2}{2};\frac{y_1 + y_2}{2}) \\ \left(\frac{9}{2};-1\right) &= \left(\frac{x + 5}{2};\frac{y + 3}{2}\right) \end{align*}

Solve for $$x$$:

\begin{align*} \frac{9}{2} &= \frac{x + 5}{2} \\ 9 &= x + 5 \\ x &= 4 \end{align*}

Solve for $$y$$:

\begin{align*} -1 &= \frac{y + 3}{2} \\ -2 &= y + 3 \\ y &= -5 \end{align*}

Therefore $$S(4;-5)$$.