Length and Midpoint

Extended only: what E3.4 asks you to do

The syllabus lists exactly two calculations here: calculate the length of a line segment and find the coordinates of the midpoint of a line segment. Both start from the same information - the coordinates of the two ends - and both are Extended content, so they do not appear on the Core paper.

There is no new formula to learn for the length. You already know Pythagoras' theorem, and the segment joining two points is the hypotenuse of a right-angled triangle whose two shorter sides you can read straight off the coordinates. The midpoint needs one short formula: average the two $x$-coordinates, then average the two $y$-coordinates.

Because a midpoint is an average, it always lies exactly halfway along the segment. That single fact is worth more than the formula, because it tells you immediately whether an answer is sensible.

• Length: Pythagoras applied to the horizontal and vertical distances between the two points
• Midpoint: the average of the two $x$-coordinates and the average of the two $y$-coordinates
• Both answers come from the same two pairs of numbers, so most questions give you the coordinates once and ask for both

The right-angled triangle hidden in a line segment

Take $A(1,2)$ and $B(4,6)$. To get from $A$ to $B$ you move across and then up: the horizontal distance is $4-1=3$ and the vertical distance is $6-2=4$. Those two moves are the legs of a right-angled triangle, and the segment $AB$ is its hypotenuse.

The horizontal distance is always $x_2-x_1$ and the vertical distance is always $y_2-y_1$, where $(x_1,y_1)$ and $(x_2,y_2)$ are the two ends. Starting from the left-hand point keeps both numbers positive, which makes the arithmetic easier, and the answer comes out the same either way round.

Sketching the triangle costs a few seconds and prevents almost every error in this subtopic. When the two points are in different quadrants the sketch shows you which way each distance goes, and it stops you adding a distance where you should subtract it.

The length of a line segment

Pythagoras' theorem says the square on the hypotenuse equals the sum of the squares on the other two sides. Applied to two points that gives $A{B}^2=(x_2-x_1{)}^2+(y_2-y_1{)}^2$, so the length is $AB=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$. This is the distance formula: it is Pythagoras with the coordinates already built in.

For $A(1,2)$ and $B(4,6)$: $AB=\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5$. Square each distance, add the two squares, then take the square root - in that order. The square root comes last, and the answer is the length itself, not its square.

The squares make the order of the two points irrelevant. $(1-4{)}^2=9$ and $(4-1{)}^2=9$, so swapping $A$ and $B$ gives the same length. That is a useful check when an answer looks wrong.

• $AB=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$
• Subtract, square, add, then take the square root
• A horizontal or vertical segment needs no formula at all: from $(2,3)$ to $(7,3)$ the length is $7-2=5$

Negative coordinates, and lengths that are not whole numbers

Negative coordinates cause no trouble in the formula, because every difference is squared. For $P(-4,3)$ and $Q(5,-9)$: the horizontal distance is $5-(-4)=9$ and the vertical distance is $-9-3=-12$. Then $PQ=\sqrt{9^2+(-12{)}^2}=\sqrt{81+144}=\sqrt{225}=15$.

Most lengths do not come out as whole numbers. For $A(1,2)$ and $B(4,4)$: $AB=\sqrt{3^2+2^2}=\sqrt{13}$. The exact answer is the surd $\sqrt{13}$, and as a decimal it is 3.6055..., which is 3.61 correct to 3 significant figures.

Read the question's last line before deciding what to write. If it asks for a decimal place or a number of significant figures, give the rounded decimal. If it says nothing, give the exact surd and put the decimal beside it - a correct rounded decimal is accepted in place of the surd.

• $\sqrt{13}=3.61$ correct to 3 significant figures
• 3.6055... rounds up to 3.61, because the next digit is 5 - check before you round
• $\sqrt{25}=5$ and $\sqrt{169}=13$, so a whole-number answer is possible whenever the two distances are the legs of a Pythagorean triple such as 3, 4, 5 or 5, 12, 13

The midpoint of a line segment

The midpoint is the point exactly halfway between the two ends, so its coordinates are the averages: $M=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$. Add the two $x$-coordinates and halve, then add the two $y$-coordinates and halve. Adding first matters - halving one end on its own gives a point a quarter of the way along, not halfway.

For $A(-3,8)$ and $B(9,-2)$: the $x$-coordinate is $\frac{-3+9}{2}=3$ and the $y$-coordinate is $\frac{8+(-2)}{2}=3$, so the midpoint is $(3,3)$. The signs travel with the numbers, and the two halves are worked out separately.

Midpoints are often not whole numbers. For $A(2,1)$ and $B(5,6)$ the midpoint is $(3.5,3.5)$, and $(\frac{7}{2},\frac{7}{2})$ is just as good. Do not round a midpoint to a whole number unless the question tells you to.

Working backwards from the midpoint, and using both results

If you are given the midpoint $M$ and one end $A$, the other end $B$ is the same distance past $M$ again: $B=2M-A$, worked out one coordinate at a time. With $M(4,7)$ and $A(1,3)$: the $x$-coordinate is $2\times 4-1=7$ and the $y$-coordinate is $2\times 7-3=11$, so $B$ is $(7,11)$.

The two calculations are often asked together in one question. For $A(2,3)$ and $B(8,11)$: the distances are 6 and 8, so the length is $\sqrt{36+64}=\sqrt{100}=10$, and the midpoint is $(\frac{2+8}{2},\frac{3+11}{2})=(5,7)$.

The length and the midpoint also combine into the perpendicular bisector of a segment: the line through the midpoint drawn at right angles to the segment. Finding its equation is Extended work on perpendicular gradients, and the midpoint is always the first thing you need for it.

• Given the midpoint $M$ and one end $A$, the other end is $B=2M-A$
• Every point on the perpendicular bisector is the same distance from $A$ as it is from $B$
• The length is never shorter than the larger of the two distances you measured