Year 11 Methods (Unit 1 & 2) Coordinate Geometry And Linear Relations

Distance And Midpoints

Q: What is the distance formula?

The distance between two points is the square root of the quantity: the x gap squared plus the y gap squared. It is Pythagoras theorem applied to the horizontal and vertical gaps between the points.

Q: What is the midpoint formula?

The midpoint is the average of the x values and the average of the y values. Add the two x coordinates and halve, then add the two y coordinates and halve.

Q: Does it matter which point I call point one?

No. For distance each gap is squared so the order does not change the answer, and for the midpoint the addition is symmetric, so either point can be first.

Q: When should I leave the distance as a surd?

Leave it as an exact simplified surd, such as two root thirteen, unless the question asks for a decimal. If a decimal is asked for, evaluate the surd and round as instructed.

Q: How do I find an endpoint when I know the midpoint?

Set each midpoint coordinate equal to the average of the endpoints and solve. For the x coordinate, x equals two times the midpoint x minus the known endpoint x.

20 practice questions 1 video lesson Theory + worked examples

The distance and midpoint formulas are a foundational skill assumed for Queensland Year 11 Mathematical Methods (QCAA). The distance between two points is the length of the interval joining them, and the midpoint is the point halfway along it.

You will find a length using Pythagoras' theorem on the horizontal and vertical gaps, average coordinates to locate a midpoint, and solve for an unknown coordinate — core skills in coordinate geometry.

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Theory

In Year 11 Mathematical Methods (QCAA), the distance between two points and the midpoint of the interval joining them are the first tools of coordinate geometry. The distance comes from Pythagoras' theorem, \(d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\), and the midpoint is the average of the endpoints, \(M=\left(\dfrac{x_1+x_2}{2},\,\dfrac{y_1+y_2}{2}\right)\). This page shows how to compute both, work in exact surd form, and solve reverse problems.

The distance between \(A(x_1,y_1)\) and \(B(x_2,y_2)\) is the length of the interval \(AB\). Drawing the horizontal and vertical gaps forms a right-angled triangle whose hypotenuse is \(AB\), so Pythagoras' theorem gives the length. The gaps are \(x_2-x_1\) and \(y_2-y_1\); their order does not matter because each is squared.

The midpoint \(M\) of the interval \(AB\) is the point exactly halfway between the two endpoints. Each coordinate of \(M\) is the average of the matching coordinates of \(A\) and \(B\).

Distance squares, midpoint averages. For a length, square the two gaps, add, then take the square root. For a midpoint, add the two \(x\)-values and halve, then do the same for the \(y\)-values.

The distance \(AB\) is the hypotenuse of a right triangle with legs \(x_2-x_1\) and \(y_2-y_1\).

The midpoint \(M\) is halfway along \(AB\); each coordinate is an average.

For points \(A(x_1,y_1)\) and \(B(x_2,y_2)\), the distance is

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

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

and the midpoint 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})

Exact vs decimal: leave the distance as a simplified surd such as \(2\sqrt{13}\) for an exact answer, or evaluate it (\(\approx 7.21\)) when a decimal is asked for.

How to find a distance or a midpoint

Label the points as \((x_1,y_1)\) and \((x_2,y_2)\) so every substitution is clear.
For distance, find the gaps \(x_2-x_1\) and \(y_2-y_1\), square each, add them, then take the square root and simplify the surd.
For a midpoint, average the \(x\)-values and average the \(y\)-values to get \(M\).

Example 1 — Midpoint of an interval

Find the midpoint of the interval joining \(A(-3,\,2)\) and \(B(5,\,8)\).

Solution

Average the \(x\)-coordinates:

\(\dfrac{x_1+x_2}{2}\)	\(=\)	\(\dfrac{-3+5}{2}\)
	\(=\)	\(\dfrac{2}{2}\)
	\(=\)	\(1\)

Average the \(y\)-coordinates:

\(\dfrac{y_1+y_2}{2}\)	\(=\)	\(\dfrac{2+8}{2}\)
	\(=\)	\(\dfrac{10}{2}\)
	\(=\)	\(5\)

Midpoint: \(M=(1,\,5)\).

M = (1, 5)

Example 2 — Distance between two points

Find the distance between \(A(1,\,2)\) and \(B(7,\,10)\).

Solution

Find the two gaps:

\(x_2-x_1\)	\(=\)	\(7-1=6\)
\(y_2-y_1\)	\(=\)	\(10-2=8\)

Substitute into the distance formula:

\(d\)	\(=\)	\(\sqrt{6^2+8^2}\)
	\(=\)	\(\sqrt{36+64}\)
	\(=\)	\(\sqrt{100}\)
	\(=\)	\(10\)

Distance: \(AB=10\) units.

d = 10

Example 3 — Exact distance (surd form)

Find the exact distance between \(A(-2,\,1)\) and \(B(4,\,5)\).

Solution

Find the two gaps:

\(x_2-x_1\)	\(=\)	\(4-(-2)=6\)
\(y_2-y_1\)	\(=\)	\(5-1=4\)

Substitute, then simplify the surd:

\(d\)	\(=\)	\(\sqrt{6^2+4^2}\)
	\(=\)	\(\sqrt{36+16}\)
	\(=\)	\(\sqrt{52}\)
	\(=\)	\(\sqrt{4\times 13}\)
	\(=\)	\(2\sqrt{13}\)

Exact distance: \(AB=2\sqrt{13}\) units.

d = 2 \sqrt{13}

Example 4 — Finding an endpoint from the midpoint

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

Solution

The \(x\)-average equals \(3\):

\(\dfrac{-2+x}{2}\)	\(=\)	\(3\)
\(-2+x\)	\(=\)	\(6\)
\(x\)	\(=\)	\(8\)

The \(y\)-average equals \(-1\):

\(\dfrac{4+y}{2}\)	\(=\)	\(-1\)
\(4+y\)	\(=\)	\(-2\)
\(y\)	\(=\)	\(-6\)

Endpoint: \(B=(8,\,-6)\).

B = (8, - 6)

Common pitfalls

Subtracting midpoint coordinates. The midpoint uses a plus sign: you add the coordinates and halve, \(\dfrac{x_1+x_2}{2}\). Subtracting is for the distance gaps only.

Forgetting to square both gaps. The distance is \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\); you must square each gap before adding, not add then square.

Leaving a surd unsimplified. \(\sqrt{52}\) is not fully simplified; pull out the largest square factor to get \(2\sqrt{13}\).

Frequently asked questions

What is the distance formula?

For \(A(x_1,y_1)\) and \(B(x_2,y_2)\), \(d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\). It is Pythagoras' theorem applied to the horizontal and vertical gaps.

What is the midpoint formula?

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

Does it matter which point I call point one?

For distance, no — each gap is squared, so the order does not change the answer. For a midpoint the addition is symmetric, so order does not matter either.

When should I leave the distance as a surd?

Leave it as an exact simplified surd (for example \(2\sqrt{13}\)) unless the question asks for a decimal, in which case evaluate it and round as instructed.

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

Set each midpoint coordinate equal to its average and solve. For example, \(\dfrac{x_1+x}{2}=M_x\) gives \(x=2M_x-x_1\).

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