Length, gradient and midpoint

AS · P1 · 14 min

Coordinate geometry turns pictures into algebra: a point becomes a pair of numbers, and questions about lengths, slopes and positions become calculations. Three formulas, for the length of a line segment, its midpoint and its gradient, sit underneath every coordinate geometry question on Paper 1, from equations of lines to circles and tangents. This note builds each one from a picture, then shows how examiners combine them with unknown coordinates to make longer problems.

Coordinates and line segments

A point in the plane is written $(x, y)$ : $x$ is the distance across from the $y$ -axis and $y$ the distance up from the $x$ -axis. Either can be negative.

A line segment $AB$ is the straight piece of line between the points $A$ and $B$ . Its length is written $AB$ or $|AB|$ . A line carries on for ever in both directions, so a line has a gradient but not a length.

Throughout this note, $A$ is the point $(x_1, y_1)$ and $B$ is the point $(x_2, y_2)$ . The two numbers that drive every formula are

\text{change in } x = x_2 - x_1, \qquad \text{change in } y = y_2 - y_1

Write these two differences down first in every question. Almost every error in this topic is a sign slip inside one of them.

The length of a line segment

Draw a horizontal line from $A$ and a vertical line from $B$ . They meet at a right angle, making a right-angled triangle with $AB$ as the hypotenuse. The horizontal side has length $x_2 - x_1$ and the vertical side $y_2 - y_1$ (up to sign), so Pythagoras gives the length.

(1, 1) -- (6, 5) (1, 1) -- (6, 1) (6, 1) -- (6, 5) (1, 1) (6, 5)

In the figure, $A(1, 1)$ and $B(6, 5)$ : the horizontal side is $5$ , the vertical side is $4$ , and $AB = \sqrt{5^2 + 4^2} = \sqrt{41}$ .

Key result

AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Because the differences are squared, it does not matter which point you call $A$ : $(x_2 - x_1)^2 = (x_1 - x_2)^2$ . It also does not matter if a difference is negative, since its square is positive.

Leave lengths as exact surds unless a decimal is asked for, and simplify them: $\sqrt{72} = \sqrt{36}\sqrt{2} = 6\sqrt{2}$ . Later parts of a question often need $AB^2$ , which is a whole number, so keep that too.

The midpoint of a line segment

The midpoint is halfway across and halfway up, so each coordinate is the mean of the two end coordinates.

Key result

\text{Midpoint of } AB = \left(\frac{x_1 + x_2}{2},\ \frac{y_1 + y_2}{2}\right)

The same idea run backwards finds a missing endpoint. If $M$ is the midpoint of $AB$ , then going from $A$ to $M$ is the same step as going from $M$ to $B$ . So

B = (2x_M - x_1,\ 2y_M - y_1)

You do not need to memorise this: just add the step from $A$ to $M$ on to $M$ again.

Dividing a segment in a given ratio

Sometimes a point $P$ lies on $AB$ with $AP : PB = 1 : 2$ , say. Then $P$ is one third of the way from $A$ to $B$ . Take the step from $A$ to $B$ , multiply it by the fraction, and add it to $A$ :

P = \left(x_1 + \tfrac{1}{3}(x_2 - x_1),\ y_1 + \tfrac{1}{3}(y_2 - y_1)\right)

For $AP : PB = m : n$ , the fraction is $\dfrac{m}{m + n}$ . This is not a formula on the syllabus, but it comes up inside longer questions, and the "fraction of the step" idea is all you need.

Gradient

The gradient of a line measures how steep it is: how far it rises for each unit it moves to the right.

Definition

The gradient $m$ of the line through $(x_1, y_1)$ and $(x_2, y_2)$ , with $x_1 \neq x_2$ , is

m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\text{change in } y}{\text{change in } x}

The order must match top and bottom: if you start the numerator from $B$ , start the denominator from $B$ too. Swapping both gives the same answer; swapping only one changes the sign.

Gradient	What the line does
$m > 0$	rises from left to right
$m < 0$	falls from left to right
$m = 0$	horizontal, equation $y = k$
undefined	vertical, equation $x = k$ (the change in $x$ is zero)

If a line makes an angle $\theta$ with the positive $x$ -axis, measured anticlockwise, then $m = \tan\theta$ . A line at $45^\circ$ has gradient $1$ ; a line at $135^\circ$ has gradient $-1$ .

Collinear points

Three points $A$ , $B$ , $C$ are collinear if they lie on one straight line. Since they share the point $B$ , it is enough to show that the gradients of $AB$ and $BC$ are equal. Equal gradients alone would only make the segments parallel; the shared point is what puts them on the same line, so mention it.

Parallel and perpendicular lines

Parallel lines have the same steepness, so they have equal gradients.

For perpendicular lines, rotate a line through $90^\circ$ . A step of $(1, m)$ along the line becomes a step of $(-m, 1)$ , which has gradient $\dfrac{1}{-m} = -\dfrac{1}{m}$ . So the gradients multiply to $-1$ .

Key result

For two lines with gradients $m_1$ and $m_2$ (neither vertical):

parallel $\iff m_1 = m_2$
perpendicular $\iff m_1 m_2 = -1$ , that is, $m_2 = -\dfrac{1}{m_1}$

To get a perpendicular gradient, flip the fraction and change its sign: $\tfrac{2}{3}$ becomes $-\tfrac{3}{2}$ ; $-4$ becomes $\tfrac{1}{4}$ .

y = (2/3)x + 1 y = (-3/2)x + 4 y = (2/3)x - 2

The two lines with gradient $\tfrac{2}{3}$ are parallel; the line with gradient $-\tfrac{3}{2}$ is perpendicular to both.

The rule fails for horizontal and vertical lines, because a vertical line has no gradient. Spot these by eye: a horizontal line ( $m = 0$ ) is perpendicular to any vertical line.

Using an unknown coordinate

Write the unknown point with a letter, such as $(k, 5)$ or $(t, 2t)$ if it lies on a known line.
Translate the condition into an equation: a length, a midpoint, equal gradients, or a gradient product of $-1$ .
Simplify. Distance conditions are easier squared: use $AB^2$ , not $AB$ .
Solve, and keep every solution unless the question rules one out (for example " $k > 0$ ").

Worked examples

Length and midpoint

The points $A$ and $B$ have coordinates $(-3, 7)$ and $(5, 1)$ . Find the length of $AB$ and the coordinates of its midpoint.

Solution

Change in $x$ : $5 - (-3) = 8$ . Change in $y$ : $1 - 7 = -6$ .

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

\text{Midpoint} = \left(\frac{-3 + 5}{2},\ \frac{7 + 1}{2}\right) = (1,\ 4)

A missing endpoint

The midpoint of $PQ$ is $M(2, -1)$ . The point $P$ is $(-4, 3)$ . Find the coordinates of $Q$ .

Solution

The step from $P$ to $M$ is $+6$ in $x$ and $-4$ in $y$ . The same step from $M$ reaches $Q$ :

Q = (2 + 6,\ -1 - 4) = (8,\ -5)

Check: the midpoint of $(-4, 3)$ and $(8, -5)$ is $\left(\tfrac{4}{2}, \tfrac{-2}{2}\right) = (2, -1)$ . Correct.

An unknown from a length

The distance between the points $(k, 3)$ and $(2, -1)$ is $5$ . Find the possible values of $k$ .

Solution

Square the distance to avoid the square root:

\begin{aligned} (k - 2)^2 + (3 - (-1))^2 &= 25 \\ (k - 2)^2 + 16 &= 25 \\ (k - 2)^2 &= 9 \\ k - 2 &= \pm 3 \end{aligned}

So $k = 5$ or $k = -1$ . Both are valid: the points $(5, 3)$ and $(-1, 3)$ are both $5$ units from $(2, -1)$ , one on each side.

Collinear points

(a) Show that $A(-2, -5)$ , $B(1, 1)$ and $C(4, 7)$ are collinear.

(b) The point $D(k, 13)$ lies on the same line. Find $k$ .

Solution

(a)

m_{AB} = \frac{1 - (-5)}{1 - (-2)} = \frac{6}{3} = 2, \qquad m_{BC} = \frac{7 - 1}{4 - 1} = \frac{6}{3} = 2

$AB$ and $BC$ have the same gradient and share the point $B$ , so $A$ , $B$ and $C$ are collinear.

(b) $BD$ must also have gradient $2$ :

\frac{13 - 1}{k - 1} = 2 \quad\Rightarrow\quad 12 = 2k - 2 \quad\Rightarrow\quad k = 7

A right angle with an unknown

The points $A(1, 2)$ , $B(5, k)$ and $C(7, 4)$ are such that angle $ABC$ is a right angle. Find the possible values of $k$ .

Solution

The right angle is at $B$ , so $AB \perp BC$ .

m_{AB} = \frac{k - 2}{5 - 1} = \frac{k - 2}{4}, \qquad m_{BC} = \frac{4 - k}{7 - 5} = \frac{4 - k}{2}

Perpendicular means the product is $-1$ :

\begin{aligned} \frac{k - 2}{4} \times \frac{4 - k}{2} &= -1 \\ (k - 2)(4 - k) &= -8 \\ -k^2 + 6k - 8 &= -8 \\ k^2 - 6k &= 0 \\ k(k - 6) &= 0 \end{aligned}

So $k = 0$ or $k = 6$ . The points $(5, 0)$ and $(5, 6)$ both lie on the circle with diameter $AC$ , which is why there are two answers (the angle in a semicircle is a right angle).

A point dividing a segment

The points $A(-1, 4)$ and $B(8, -2)$ are joined. The point $P$ lies on $AB$ with $AP : PB = 2 : 1$ . Find the coordinates of $P$ .

Solution

$P$ is $\tfrac{2}{3}$ of the way from $A$ to $B$ . The step from $A$ to $B$ is $(9, -6)$ , and $\tfrac{2}{3}$ of it is $(6, -4)$ :

P = (-1 + 6,\ 4 - 4) = (5,\ 0)

Check: $AP = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13}$ and $PB = \sqrt{9 + 4} = \sqrt{13}$ , in the ratio $2 : 1$ .

An equidistant point and an area

The points $A$ and $B$ have coordinates $(2, 1)$ and $(6, 5)$ . The point $C$ lies on the $y$ -axis and is equidistant from $A$ and $B$ .

(a) Find the coordinates of $C$ .

(b) Find the area of triangle $ABC$ .

Solution

(a) $C$ is on the $y$ -axis, so write $C = (0, y)$ . Equidistant means $AC^2 = BC^2$ :

\begin{aligned} (0 - 2)^2 + (y - 1)^2 &= (0 - 6)^2 + (y - 5)^2 \\ 4 + y^2 - 2y + 1 &= 36 + y^2 - 10y + 25 \\ 8y &= 56 \\ y &= 7 \end{aligned}

So $C = (0, 7)$ .

(b) The triangle is isosceles with $CA = CB$ , so the line from $C$ to the midpoint $M$ of $AB$ is perpendicular to $AB$ and is the height.

$M = (4, 3)$ . Then

AB = \sqrt{4^2 + 4^2} = 4\sqrt{2}, \qquad CM = \sqrt{4^2 + (-4)^2} = 4\sqrt{2}

Check the right angle: $m_{AB} = \tfrac{4}{4} = 1$ and $m_{CM} = \tfrac{3 - 7}{4 - 0} = -1$ , product $-1$ .

\text{Area} = \tfrac{1}{2} \times 4\sqrt{2} \times 4\sqrt{2} = 16

(2, 1) -- (6, 5) (6, 5) -- (0, 7) (0, 7) -- (2, 1) (0, 7) -- (4, 3) (4, 3)

Watch out

Mixing the order in the gradient. $\dfrac{y_2 - y_1}{x_1 - x_2}$ gives the wrong sign. Keep the same point first on top and bottom.

Subtracting a negative. $5 - (-3) = 8$ , not $2$ . Write the brackets in.

Perpendicular gradient with only one change. The perpendicular to $m = \tfrac{2}{3}$ is $-\tfrac{3}{2}$ . Writing $\tfrac{3}{2}$ (flip only) or $-\tfrac{2}{3}$ (sign only) is a common lost mark.

Square-rooting too early. In " $AC = BC$ " problems, square both distances first; the $y^2$ terms then cancel and the equation is linear.

Using $-\tfrac{1}{m}$ with a vertical or horizontal line. A line with gradient $0$ is perpendicular to a vertical line, which has no gradient at all. Recognise these from the coordinates.

Exam tip

Exact values. "Find the length" with no accuracy stated: give an exact surd in simplest form, or a decimal to 3 significant figures. A simplified surd is always safe.
Show your differences. Write $\dfrac{1 - (-5)}{1 - (-2)}$ before simplifying. If you slip, the method mark is still earned.
State the reason. For perpendicular lines, write " $m_1 m_2 = -1$ , so perpendicular". For collinear points, mention the shared point.
Multiple answers. Squaring creates two solutions. Give both unless the question restricts them, and say which one you reject and why.
Sketch. A rough sketch with the points plotted catches impossible answers, such as a midpoint that is not between the two points.

Summary

Length: $AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ . Keep it exact and simplified.
Midpoint: average the coordinates. To find a missing endpoint, repeat the step from the known end to the midpoint.
A point a fraction $\lambda$ of the way from $A$ to $B$ is $A + \lambda(B - A)$ , coordinate by coordinate.
Gradient: $m = \dfrac{y_2 - y_1}{x_2 - x_1}$ ; horizontal lines have $m = 0$ , vertical lines have no gradient.
Parallel: equal gradients. Perpendicular: $m_1 m_2 = -1$ .
Collinear: equal gradients and a shared point.
Conditions with unknown coordinates become equations; square distance conditions before solving.

Once you can find gradients and points, the next step is writing the equation of a line through them: see Equations of straight lines. Using these facts to identify shapes is covered in Proving properties of shapes.

Practice questions

Question

Find the length of the line segment joining $(-4, -1)$ and $(2, 7)$ , and the coordinates of its midpoint.
The point $(3, -2)$ is the midpoint of $PQ$ , where $P$ is $(7, 5)$ . Find $Q$ .
Find the gradient of the line through $(2, -3)$ and $(6, 5)$ , and the gradient of a line perpendicular to it.
The distance between the points $(a, 3)$ and $(1, a)$ is $\sqrt{10}$ . Find the possible values of $a$ .
The points $(1, -1)$ , $(3, 3)$ and $(k, 11)$ are collinear. Find $k$ .
$A$ is $(2, -3)$ and $B$ is $(14, 6)$ . The point $P$ lies on $AB$ with $AP : PB = 1 : 2$ . Find the coordinates of $P$ .
The points $A(-2, 1)$ , $B(2, 3)$ and $C(k, -5)$ are such that angle $ABC = 90^\circ$ . Find $k$ .
Find the point on the $x$ -axis that is equidistant from $A(1, 4)$ and $B(7, 2)$ .
The points $A(1, 3)$ , $B(7, 5)$ and $C(k, 9)$ form a triangle that is right-angled at $A$ . Find $k$ , show that the triangle is isosceles, and find its area.
The point $P(t, 2t)$ is equidistant from $A(-1, 3)$ and $B(5, 1)$ . (a) Find $t$ . (b) $M$ is the midpoint of $AB$ . Show that $PM$ is perpendicular to $AB$ , and find the exact length of $PA$ .

Answers

Differences $6$ and $8$ , so the length is $\sqrt{36 + 64} = 10$ . Midpoint $\left(\tfrac{-4 + 2}{2}, \tfrac{-1 + 7}{2}\right) = (-1, 3)$ .
The step from $P$ to $M$ is $(-4, -7)$ . Repeating it from $M$ : $Q = (3 - 4, -2 - 7) = (-1, -9)$ .
$m = \dfrac{5 - (-3)}{6 - 2} = \dfrac{8}{4} = 2$ . Perpendicular gradient $-\tfrac{1}{2}$ .
$(1 - a)^2 + (a - 3)^2 = 10$ , so $a^2 - 2a + 1 + a^2 - 6a + 9 = 10$ , giving $2a^2 - 8a = 0$ , $2a(a - 4) = 0$ . So $a = 0$ or $a = 4$ .
The gradient of the first two points is $\dfrac{3 - (-1)}{3 - 1} = 2$ . Then $\dfrac{11 - 3}{k - 3} = 2$ , so $8 = 2k - 6$ and $k = 7$ .
$P$ is $\tfrac{1}{3}$ of the way from $A$ to $B$ . The step is $(12, 9)$ ; a third of it is $(4, 3)$ . So $P = (6, 0)$ .
$m_{AB} = \dfrac{3 - 1}{2 - (-2)} = \dfrac{1}{2}$ , so $m_{BC} = -2$ : $\dfrac{-5 - 3}{k - 2} = -2$ , so $-8 = -2k + 4$ and $k = 6$ .
Let the point be $(x, 0)$ . $(x - 1)^2 + 16 = (x - 7)^2 + 4$ , so $x^2 - 2x + 17 = x^2 - 14x + 53$ , giving $12x = 36$ and $x = 3$ . The point is $(3, 0)$ . Check: both distances squared are $20$ .
$m_{AB} = \tfrac{2}{6} = \tfrac{1}{3}$ , so $m_{AC} = -3$ : $\dfrac{9 - 3}{k - 1} = -3$ , so $6 = -3k + 3$ and $k = -1$ . Then $AB^2 = 36 + 4 = 40$ and $AC^2 = 4 + 36 = 40$ , so $AB = AC = \sqrt{40}$ and the triangle is isosceles. Area $= \tfrac{1}{2} \times AB \times AC = \tfrac{1}{2} \times 40 = 20$ .
(a) $(t + 1)^2 + (2t - 3)^2 = (t - 5)^2 + (2t - 1)^2$ . Expanding: $5t^2 - 10t + 10 = 5t^2 - 14t + 26$ , so $4t = 16$ and $t = 4$ . $P = (4, 8)$ . (b) $M = (2, 2)$ . $m_{PM} = \dfrac{8 - 2}{4 - 2} = 3$ and $m_{AB} = \dfrac{1 - 3}{5 - (-1)} = -\dfrac{1}{3}$ . The product is $-1$ , so $PM \perp AB$ . $PA = \sqrt{(4 + 1)^2 + (8 - 3)^2} = \sqrt{50} = 5\sqrt{2}$ .

Coordinates and line segments

The length of a line segment

The midpoint of a line segment

Dividing a segment in a given ratio

Gradient

Collinear points

Parallel and perpendicular lines

Worked examples

Practice questions

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