Year 10 Maths - 10A Linear Relations

Length and Midpoint of a Line Segment

Q: What is the midpoint formula?

M equals the point ((x1 plus x2) over 2, (y1 plus y2) over 2). It is the average of the two x-coordinates and the average of the two y-coordinates.

Q: What is the distance formula?

The distance d between two points is the square root of (x2 minus x1) squared plus (y2 minus y1) squared. It is Pythagoras' theorem with the run and the rise as the shorter sides.

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Length and Midpoint of a Line Segment - Video - Distance Between Points Watch
Length and Midpoint of a Line Segment - Video - How to Find the Distance Between Two Points Watch
Length and Midpoint of a Line Segment - Video - How To Find The Midpoint Using The Midpoint Formula Watch

Theory

The midpoint of an interval is the point halfway between its ends: average the \(x\)-coordinates and average the \(y\)-coordinates. The length of an interval comes from Pythagoras' theorem, with the run and the rise as the two shorter sides. This Year 10A maths page revises both formulas from Years 7 to 9: negative coordinates, fractions, surds in the coordinates, simplest surd form, exact surds and rounding, finding a missing end point or coordinate, and lengths in shapes and on scaled maps.

An interval (or line segment) is the part of a line between two end points.

The midpoint \(M\) is the point halfway between the two ends. It is an average: its \(x\)-coordinate is the mean of the two \(x\)-coordinates, and its \(y\)-coordinate is the mean of the two \(y\)-coordinates. Midpoints often land on a half, such as \(\left(\dfrac{5}{2}, 3\right)\); that is a correct answer, so do not round it.

The length of an interval is the straight-line distance between its ends. Go across from one point and up (or down) to the other. The distance across is the run, the difference of the \(x\)-coordinates; the distance up is the rise, the difference of the \(y\)-coordinates.

The run and the rise are the two shorter sides of a right-angled triangle, and the interval is the hypotenuse, so Pythagoras' theorem gives its length. A horizontal interval's length is just the run; a vertical interval's length is just the rise.

A length is never negative, and it does not matter which point you start from.

\(M\) sits halfway along \(AB\). Its \(x\)-coordinate \(5\) is the average of \(2\) and \(8\); its \(y\)-coordinate \(4\) is the average of \(1\) and \(7\).

Go across, then up. The run and the rise are the two shorter sides, and the interval is the hypotenuse: \(d^2 = 4^2 + 2^2 = 20\), so \(d = \sqrt{20} \approx 4.5\) units.

The midpoint of the interval joining \((x_1, y_1)\) and \((x_2, y_2)\) is:

\[M = \left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right)\]

M = (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})

Working backwards, from one end \(A\) and the midpoint \(M\) to the other end \(B\):

\[x_B = 2x_M - x_A \qquad y_B = 2y_M - y_A\]

x_{B} = 2 x_{M} - x_{A}

For the points \((x_1, y_1)\) and \((x_2, y_2)\):

\[\text{run} = x_2 - x_1 \qquad \text{rise} = y_2 - y_1\]

run = x_{2} - x_{1}

\[d^2 = \text{run}^2 + \text{rise}^2\]

\[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]

d = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}

One missing coordinate. If only one value is unknown, write the average as an equation and solve. For \(A(2, p)\), \(B(6, 10)\) and midpoint \((4, 7)\): \(\dfrac{p + 10}{2} = 7\), so \(p + 10 = 14\) and \(p = 4\).

Quarter points. To split an interval into four equal parts, find its midpoint \(M\), then the midpoint of each half. For \(A(-1, 2)\) and \(B(7, 10)\): \(M = (3, 6)\), the midpoint of \(A\) and \(M\) is \((1, 4)\) (one quarter of the way along), and the midpoint of \(M\) and \(B\) is \((5, 8)\) (three quarters of the way).

Surd or decimal? Leave the answer as a surd such as \(\sqrt{41}\) unless the question asks for decimal places. Then round only at the very end.

Grids with a scale. If \(1\) unit stands for a real distance, such as \(50\) m, find the length in grid units first, then multiply by the scale at the end.

Shapes on the plane. Find each side the same way and add them for the perimeter. To compare two sides, compare their squares. In a parallelogram the two diagonals have the same midpoint, and the centre of a circle is the midpoint of any diameter.

A missing coordinate from a length. The points \(A(2,\ 1)\) and \(B(k,\ 13)\) are \(13\) units apart, and \(k \gt 2\). Square both sides of the distance formula: \((k - 2)^2 + 12^2 = 13^2\), so \((k - 2)^2 = 25\) and \(k - 2 = 5\) or \(-5\). That gives \(k = 7\) or \(k = -3\); only \(k = 7\) is greater than \(2\). Expect two answers until the question rules one out.

Simplest surd form. Look for the largest square factor: \(\sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}\) and \(\sqrt{45} = \sqrt{9 \times 5} = 3\sqrt{5}\). Surds in the coordinates are simplified first, so that like surds can be added: \(\sqrt{12} = 2\sqrt{3}\).

Fractions in the coordinates. Add the fractions, then halve. For \(\left(-\dfrac{5}{2},\ \dfrac{1}{4}\right)\) and \(\left(\dfrac{7}{2},\ \dfrac{3}{4}\right)\): \(x_M = \dfrac{1}{2}\left(-\dfrac{5}{2} + \dfrac{7}{2}\right) = \dfrac{1}{2}\) and \(y_M = \dfrac{1}{2}\left(\dfrac{1}{4} + \dfrac{3}{4}\right) = \dfrac{1}{2}\).

How to find a midpoint

Add the two \(x\)-coordinates and halve. Then add the two \(y\)-coordinates and halve.
Keep any halves as fractions or decimals.
Going backwards? Double the midpoint, then subtract the known end.

How to find the length of an interval

Find the run (\(x_2 - x_1\)) and the rise (\(y_2 - y_1\)). Take care when subtracting a negative.
Square the run and the rise separately, then add them to get \(d^2\).
Take the square root to get \(d\). Leave it as a surd, or round at the end if the question asks.
Check on a sketch: \(d\) must be longer than both the run and the rise, and the midpoint must sit halfway along.

Example 1 — Surds in the coordinates

Find the midpoint of the interval joining \(P(\sqrt{18},\ 1)\) and \(Q(-\sqrt{2},\ -4)\).

Solution

Simplify the surds first, so they are like surds: \(\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}\).

\(x_M\)	\(=\)	\(\dfrac{3\sqrt{2} + (-\sqrt{2})}{2}\)
\(\)	\(=\)	\(\dfrac{2\sqrt{2}}{2} = \sqrt{2}\)
\(y_M\)	\(=\)	\(\dfrac{1 + (-4)}{2} = -\dfrac{3}{2}\)

M = (\sqrt{2}, - \frac{3}{2})

The midpoint is \(\boldsymbol{\left(\sqrt{2},\ -\dfrac{3}{2}\right)}\). Leave the surd and the fraction exact.

Example 2 — Exact surd, then rounded

Find the distance between \(C(-4,2)\) and \(D(1,-3)\)
(i) exactly, in simplest surd form
(ii) correct to \(1\) decimal place.

Solution

\(\text{run}\)	\(=\)	\(1 - (-4) = 5\)
\(\text{rise}\)	\(=\)	\(-3 - 2 = -5\)
\(d^2\)	\(=\)	\(5^2 + (-5)^2\)
\(\)	\(=\)	\(25 + 25 = 50\)
\(d\)	\(=\)	\(\sqrt{50} = \sqrt{25 \times 2}\)
\(\)	\(=\)	\(\mathbf{5\sqrt{2}} \text{ units}\)
\(\)	\(=\)	\(7.071\ldots\)
\(\)	\(\approx\)	\(\mathbf{7.1} \text{ units}\)

d = \sqrt{50} \approx 7.1

The negative rise does no harm, because squaring makes it positive. Part (i) wants the exact value, so simplify \(\sqrt{50}\) to \(5\sqrt{2}\); round only for part (ii).

Example 3 — Working backwards

The midpoint of the interval joining \(A(4,-3)\) and \(B\) is \(M(1,2)\). Find the coordinates of \(B\).

Solution

Double the midpoint, then subtract the end you already know.

\(x_B\)	\(=\)	\(2 \times 1 - 4 = -2\)
\(y_B\)	\(=\)	\(2 \times 2 - (-3) = 7\)

B = (- 2, 7)

So \(B = \mathbf{(-2, 7)}\). Check: the average of \(4\) and \(-2\) is \(1\), and the average of \(-3\) and \(7\) is \(2\).

Example 4 — Is the triangle isosceles?

A triangle has vertices \(A(0,0)\), \(B(6,2)\) and \(C(2,6)\). Show that it is isosceles.

Solution

Work out the square of each side. There is no need to take square roots to compare them.

\(AB^2\)	\(=\)	\(6^2 + 2^2 = 40\)
\(AC^2\)	\(=\)	\(2^2 + 6^2 = 40\)
\(BC^2\)	\(=\)	\((2 - 6)^2 + (6 - 2)^2\)
\(\)	\(=\)	\(16 + 16 = 32\)

{AB}^{2} = {AC}^{2} = 40

\(AB^2 = AC^2\), so \(AB = AC\) and the triangle is isosceles. Since \(32 + 40 \ne 40\), it is not right-angled.

Common pitfalls

Add the coordinates for a midpoint. Subtracting gives the run and the rise. Those are for length and gradient, not the midpoint.

Keep the half. A midpoint of \(\left(\dfrac{5}{2}, 3\right)\) is correct. Rounding it to \((3, 3)\) is not.

Square first, then add. \((4 + 3)^2\) is not \(4^2 + 3^2\). Square the run and the rise separately, and take the square root at the end.

Watch the double negative. A run from \(x = -4\) to \(x = 1\) is \(1 - (-4) = 5\), not \(-3\).

Frequently asked questions

What is the midpoint formula?

\(M = \left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right)\). It is the average of the two \(x\)-coordinates and the average of the two \(y\)-coordinates.

What is the distance formula?

\(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\). It is Pythagoras' theorem with the run and the rise as the shorter sides.

How do you find an endpoint when you know the midpoint?

Double each coordinate of the midpoint, then subtract the matching coordinate of the end you know. With \(A(4, -3)\) and midpoint \((1, 2)\), the other end is \((2 - 4, 4 + 3) = (-2, 7)\).

Can a midpoint have fractions?

Yes. If two coordinates add to an odd number, the midpoint lands halfway between grid lines, such as \(\dfrac{3}{2}\). Leave it as a fraction or a decimal.

Can a distance be negative?

No. A distance is a length, so it is always positive or zero. A negative run or rise becomes positive when it is squared.

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Perpendicular and Parallel Lines

Length and Midpoint of a Line Segment

📖 Theory

How to find a midpoint

How to find the length of an interval

Common pitfalls

Frequently asked questions

Theory