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Mid-Point Of A Line

8.4 Mid-point of a line (EMA6H)

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:

a765960b01019d087415e28113620f7f.png
\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

81975ba476907c7931543631ce263b8f.png

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*}

Write the answer

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

05d2db513b2526122fc1101894389a73.png

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

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

Write the final answer

\(M\left(\frac{1}{2};\frac{1}{2}\right)\) 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.

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

650aabdd8141850aaf827098f7939c8b.png

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 \({y}_{2}\)

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

Write the final answer

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

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

eda2a41b247014831065da41e7b426ff.png

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 \(y\)

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

Write the final answer

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

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Exercise 8.5

You are given the following diagram:

0c0389c4d35c5102de7a0a7f586bdb08.png

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:

17bff7db7c557a5a8c3491b9f621280e.png

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:

52fd9154414afcefa8cbd88c55a0c002.png

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 \(S(4;-5)\).