Coordinate Geometry Revision Notes — 9709 Paper 1

Everything two points give you

Syllabus requirement · §1.3.2

“

Including calculations of distances, gradients, midpoints, points of intersection and use of the relationship between the gradients of parallel and perpendicular lines.

”

Why coordinates exist at all

Putting a grid on the page turns a point into a pair of numbers and a line into an equation, so geometry questions become arithmetic. That is the idea this whole topic runs on. When a question feels geometric and you cannot see the answer, write the coordinates down and compute.

Give a question two points and there are exactly three things it can ask for straight away: the gradient of the line joining them, the midpoint of the segment, and its length. All three come from one picture — the right-angled triangle with the segment as its hypotenuse — so it is worth learning them as one idea rather than three formulae.

Gradient is the run-and-rise triangle; length is Pythagoras on the same triangle; midpoint is the average of the coordinates. One picture, three formulae.

Building the three formulae

Take two points and call them $A(x_1, y_1)$ and $B(x_2, y_2)$ . The subscripts are just labels: point 1 and point 2. Drop a horizontal line from $A$ and a vertical line from $B$ , and they meet at a corner making a right-angled triangle.

The run — how far you move across — is $x_2 - x_1$ .
The rise — how far you move up — is $y_2 - y_1$ .

Everything below is those two numbers used three different ways.

① Gradient. Gradient means steepness: how much $y$ goes up for each $1$ that $x$ goes across. That is rise divided by run:

m = \frac{\text{rise}}{\text{run}} = \frac{y_2 - y_1}{x_2 - x_1}

② Length. The rise and the run are the two short sides of a right-angled triangle whose hypotenuse is $AB$ . Pythagoras gives

AB^2 = (\text{run})^2 + (\text{rise})^2 = (x_2-x_1)^2 + (y_2-y_1)^2

and then $AB$ is the square root of that.

③ Midpoint. To get to the middle of $AB$ you start at $A$ and go half the run across and half the rise up:

x_M = x_1 + \tfrac12(x_2 - x_1) = \tfrac12 x_1 + \tfrac12 x_2 = \frac{x_1 + x_2}{2}

and the same for $y$ . So the midpoint is just the average of the two $x$ 's and the average of the two $y$ 's — which is why it is the one formula in this topic nobody ever needs to look up.

For A(x₁, y₁) and B(x₂, y₂)

m = \frac{y_2 - y_1}{x_2 - x_1}

Gradient — rise over run

\left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right)

Midpoint — average each coordinate

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

Length — Pythagoras on the triangle

AB^2 = (x_2-x_1)^2 + (y_2-y_1)^2

Length squared — usually the more useful one

Doing all three on clean numbers

Take $A(1, 2)$ and $B(7, 6)$ — the points in the figure above.

Run: $x_2 - x_1 = 7 - 1 = 6$ .
Rise: $y_2 - y_1 = 6 - 2 = 4$ .

m = \frac{4}{6} = \frac{2}{3}

M = \left(\frac{1+7}{2},\ \frac{2+6}{2}\right) = \left(\frac{8}{2},\ \frac{8}{2}\right) = (4, 4)

AB^2 = 6^2 + 4^2 = 36 + 16 = 52 \quad\Rightarrow\quad AB = \sqrt{52} = \sqrt{4 \times 13} = 2\sqrt{13}

Three checks worth building into your habits:

The gradient is positive and less than $1$ , and indeed the line in the picture climbs gently. A gradient of, say, $-6$ would be visibly wrong.
The midpoint $(4,4)$ sits between $1$ and $7$ , and between $2$ and $6$ . If your midpoint is not between the two points, you have subtracted instead of added.
$\sqrt{52}$ simplifies. Pull out the largest square factor ( $4$ ) to get $2\sqrt{13}$ ; mark schemes ask for surd form more often than you expect.

Now with negatives, because that is where marks go

Take $A(-3, 5)$ and $B(2, -7)$ . Nothing changes except that you must write the subtraction out before evaluating it.

\text{run} = 2 - (-3) = 2 + 3 = 5

\text{rise} = -7 - 5 = -12

m = \frac{-12}{5} = -\frac{12}{5}

Negative gradient — correct, since the line falls from left to right.

M = \left(\frac{-3+2}{2},\ \frac{5+(-7)}{2}\right) = \left(-\frac12,\ -1\right)

AB^2 = 5^2 + (-12)^2 = 25 + 144 = 169 \quad\Rightarrow\quad AB = 13

Notice that the rise was negative but $(-12)^2$ is positive, so length never cares about direction. That is the practical difference between the gradient formula (signs matter enormously) and the length formula (signs cannot survive the squaring).

Keep the order consistent

In the gradient formula, whichever point you call "2" for the $y$ 's must be the same one for the $x$ 's. Doing $\dfrac{y_2 - y_1}{x_1 - x_2}$ gives the right size and the wrong sign — and a wrong-signed gradient produces a perfectly plausible-looking line equation that scores zero.

It genuinely does not matter which point you call $1$ : $\dfrac{6-2}{7-1} = \dfrac46$ and $\dfrac{2-6}{1-7} = \dfrac{-4}{-6} = \dfrac46$ agree. Just pick an order and keep it.

Gradient	What the line does	Typical give-away in a question
$m > 0$	rises left to right	—
$m < 0$	falls left to right	—
$m = 0$	horizontal, equation $y = c$	"parallel to the $x$ -axis"
$m$ undefined	vertical, equation $x = k$ — the run is $0$ and you cannot divide by it	two points with the same $x$ -coordinate

A vertical line has no gradient at all — not gradient zero. Saying m = 0 for a vertical line is a guaranteed lost mark.

Parallel and perpendicular

Two lines are parallel exactly when they have the same steepness, so

m_1 = m_2.

Perpendicular is the interesting one. Take a line of gradient $m_1 = \dfrac{a}{b}$ , so its gradient triangle goes $b$ across and $a$ up. Now rotate that triangle through $90^\circ$ : what was " $b$ across, $a$ up" becomes " $a$ across the other way, $b$ up" — the rise and run swap places, and one of them changes sign. The rotated gradient is therefore

m_2 = \frac{b}{-a} = -\frac{b}{a} = -\frac{1}{m_1}.

Multiply the two together and the fractions cancel:

m_1 m_2 = \frac{a}{b} \times \left(-\frac{b}{a}\right) = -1.

So: flip the fraction upside down, and change the sign. Both operations, every time. Doing only one of them is the single most common error in this topic.

m_1 m_2 = -1 \qquad\Longleftrightarrow\qquad m_2 = -\frac{1}{m_1}

Perpendicular gradients

Flip the fraction, change the sign. Gradient ⅔ becomes −3⁄2.

$m_1$	$m_2 = -1/m_1$	Check
$\tfrac23$	$-\tfrac32$	$\tfrac23 \times -\tfrac32 = -1$ ✓
$-\tfrac35$	$\tfrac53$	two sign changes cancel — the answer is positive
$4$	$-\tfrac14$	write $4$ as $\tfrac41$ first if flipping confuses you
$-2$	$\tfrac12$	$-\tfrac21 \to -\tfrac12 \to$ change sign $\to \tfrac12$
$1$	$-1$	the two diagonals of a square
$0$ (horizontal)	undefined (vertical)	the one case where $m_1m_2 = -1$ fails — treat it separately

Practise these until they are instant. Almost every circle question needs one.

The two conditions that generate equations

Most "find the unknown coordinate" questions are one of these two sentences in disguise:

" $PQ$ is perpendicular to $QR$ " → write both gradients, set the product to $-1$ , clear the fractions. Because both gradients contain the unknown, the product is a quadratic, so expect two answers and look for the sentence that rules one out.

" $AC$ and $BC$ are equal in length" → write $AC^2 = BC^2$ (never with square roots — square both sides immediately). The $p^2$ terms cancel and you are left with a linear equation.

Both are worked in full below.

A gradient condition fixing one unknown coordinate

9709/15 M/J 2025 Q7(a)2 marks

In the parallelogram $ABCD$ , the coordinates of $A$ are $(3, 7)$ , the coordinates of $B$ are $(6, p)$ and the coordinates of $D$ are $(1, p)$ . It is given that the gradient of $AB$ is $-\tfrac23$ .

Find the value of $p$ .

Show full working

1
Only $A$ and $B$ matter here, so ignore $D$ completely for now. Write the gradient of $AB$ using the formula, taking $A(3,7)$ as point 1 and $B(6,p)$ as point 2: $m_{AB} = \frac{p - 7}{6 - 3}$
Substitute the letter p exactly as if it were a number. The formula does not care that one coordinate is unknown.
2
Simplify the denominator: $m_{AB} = \frac{p-7}{3}$
3
The question says this gradient equals $-\tfrac23$ , so set them equal: $\frac{p-7}{3} = -\frac23$
This is the whole trick of the question — a given gradient is an equation, not a piece of description.
4
Multiply both sides by $3$ to clear the left-hand denominator: $p - 7 = 3 \times \left(-\frac23\right) = -\frac{6}{3} = -2$
5
Add $7$ to both sides: $p = -2 + 7 = 5$
6
Check. With $p = 5$ , $B$ is $(6,5)$ and $m_{AB} = \dfrac{5-7}{6-3} = \dfrac{-2}{3} = -\dfrac23$ . ✓
Substituting back takes ten seconds and catches every sign slip.

Answer

$p = 5$

Whenever a question hands you a gradient, a length or a midpoint as information, it is giving you an equation. Write the formula with the unknown left in as a letter, set it equal to what you are told, and solve.

A right angle → a quadratic → two possible points

9709/13 M/J 2025 Q9(a)4 marks

Three points $P$ , $Q$ and $R$ have coordinates $P(-13, 5)$ , $Q(5, 1)$ and $R(2, k)$ , where $k$ is a constant. It is given that the angle $PRQ$ is a right angle.

Show that one of the possible values of $k$ is $10$ , and find the other possible value.

Show full working

1
"Angle $PRQ$ is a right angle" means the angle at $R$ is $90^\circ$ . The two lines meeting at $R$ are $RP$ and $RQ$ , so those are the two gradients we need.
Read the middle letter. Angle PRQ is at R, not at P — getting this wrong pairs up the wrong two lines and every subsequent mark is lost.
2
Gradient of $PR$ , from $P(-13,5)$ to $R(2,k)$ : $m_{PR} = \frac{k - 5}{2 - (-13)} = \frac{k-5}{15}$
2 − (−13) = 2 + 13 = 15. Write the double negative out; guessing it is the classic slip.
3
Gradient of $RQ$ , from $R(2,k)$ to $Q(5,1)$ : $m_{RQ} = \frac{1 - k}{5 - 2} = \frac{1-k}{3}$
4
Perpendicular means the product of the gradients is $-1$ : $\frac{k-5}{15} \times \frac{1-k}{3} = -1$
5
Multiply the two fractions — numerators together, denominators together: $\frac{(k-5)(1-k)}{45} = -1$
6
Multiply both sides by $45$ to clear the fraction: $(k-5)(1-k) = -45$
Clear the denominator before expanding. Expanding first leaves you carrying a /45 through every line.
7
Expand the left-hand side. $(k-5)(1-k) = k - k^2 - 5 + 5k = -k^2 + 6k - 5$ : $-k^2 + 6k - 5 = -45$
8
Multiply every term by $-1$ so the $k^2$ term is positive — much easier to factorise: $k^2 - 6k + 5 = 45$
9
Subtract $45$ from both sides to get the quadratic equal to zero: $k^2 - 6k - 40 = 0$
10
Factorise: we need two numbers multiplying to $-40$ and adding to $-6$ . Those are $-10$ and $+4$ : $(k - 10)(k + 4) = 0$
Expand back to check: k² + 4k − 10k − 40 = k² − 6k − 40 ✓.
11
So $k = 10$ or $k = -4$ . The first is the value the question told us to show, so we have shown it; the other possible value is $k = -4$ .
In a 'show that … and find the other', you must produce both roots from the algebra. Verifying k = 10 works and then guessing the second value earns almost nothing.

Answer

$k = 10$ (as required) or $k = -4$

A perpendicularity condition on an unknown coordinate always produces a quadratic, because the unknown appears in both gradients. Two answers is the expected outcome, not a sign that you have gone wrong. Only discard one if the question restricts it (here it did not — both are genuine).

An equal-length condition → a linear equation

9709/13 O/N 2022 Q11(a)3 marks

The coordinates of points $A$ , $B$ and $C$ are $A(5, -2)$ , $B(10, 3)$ and $C(2p, p)$ , where $p$ is a constant.

Given that $AC$ and $BC$ are equal in length, find the value of the fraction $p$ .

Show full working

1
Write both lengths squared. Never write $\sqrt{\ }$ signs here — you would only square them again a line later. $AC^2 = (2p - 5)^2 + (p - (-2))^2 = (2p-5)^2 + (p+2)^2$
If AC = BC then AC² = BC², and vice versa, because both lengths are positive. Squaring first removes the surds from the entire question.
2
$BC^2 = (2p - 10)^2 + (p - 3)^2$
3
Set them equal: $(2p-5)^2 + (p+2)^2 = (2p-10)^2 + (p-3)^2$
4
Expand the left-hand side one bracket at a time. $(2p-5)^2 = 4p^2 - 20p + 25$ and $(p+2)^2 = p^2 + 4p + 4$ , so $\text{LHS} = 4p^2 - 20p + 25 + p^2 + 4p + 4 = 5p^2 - 16p + 29$
Expand each bracket separately and only then add. Trying to do both at once is where sign errors breed.
5
Now the right-hand side. $(2p-10)^2 = 4p^2 - 40p + 100$ and $(p-3)^2 = p^2 - 6p + 9$ , so $\text{RHS} = 4p^2 - 40p + 100 + p^2 - 6p + 9 = 5p^2 - 46p + 109$
6
So $5p^2 - 16p + 29 = 5p^2 - 46p + 109$
7
Subtract $5p^2$ from both sides. The quadratic terms always cancel here — that is what makes equal-distance conditions easy: $-16p + 29 = -46p + 109$
If your p² terms do not cancel, you have made an arithmetic error. Go back and re-expand rather than pressing on into a quadratic.
8
Add $46p$ to both sides: $30p + 29 = 109$
9
Subtract $29$ : $30p = 80$
10
Divide by $30$ and cancel the common factor of $10$ : $p = \frac{80}{30} = \frac{8}{3}$

Answer

$p = \dfrac{8}{3}$

"Equidistant from $A$ and $B$ " is exactly the statement " $C$ lies on the perpendicular bisector of $AB$ " — which is why the $p^2$ terms cancel and the equation is linear. You will meet the same idea again as the standard way to locate a circle's centre, in the circle geometry section.

Common mistakes

$m = \dfrac{y_2 - y_1}{x_1 - x_2}$
$m = \dfrac{y_2 - y_1}{x_2 - x_1}$
Mixing the order between numerator and denominator flips the sign. The size looks right, so nothing warns you.
Perpendicular gradient of $-\tfrac35$ is $\tfrac35$
It is $\tfrac53$
Negative reciprocal means flip AND change sign. Doing only one of the two is the classic error.
Writing $AC = BC$ and then squaring each side separately, term by term
$AC^2 = BC^2$ written out in full, then expanded
√(a) + √(b) is not √(a+b). Square the whole length, not its pieces.
A vertical line has gradient $0$
A vertical line has no gradient; a horizontal line has gradient $0$
Vertical means the run is zero, and you cannot divide by zero. Its equation is x = k.
Leaving a length as $\sqrt{52}$ when the question says "in surd form"
$\sqrt{52} = \sqrt{4 \times 13} = 2\sqrt{13}$
Surd form means fully simplified. Look for the largest square factor every time.

In the exam

433 marks in 2021–2025 — rank 5 of 8 Paper 1 topics

Coordinate geometry usually comes as one long multi-part question near the end of the paper. No single step is hard, but the steps are chained, so an early slip spoils everything after it.

The steps candidates actually need most are the ones in this section: a midpoint, a distance, a gradient, and a negative reciprocal. Make them automatic and the long questions become a list of short ones.

Your turn

The first two are pure speed drills — do them without writing much. The last two are the two condition-types that generate equations.

1
For $A(-3, 5)$ and $B(9, -1)$ , find
(a) the gradient of $AB$ ,
(b) the midpoint of $AB$ ,
(c) the exact length $AB$ , in simplified surd form.
Show solution
1. 1
  (a) Run $= 9 - (-3) = 12$ ; rise $= -1 - 5 = -6$ . $m = \frac{-6}{12} = -\frac12$
  Write the two subtractions separately before dividing. Negative signs inside a single fraction are where marks go.
2. 2
  (b) Average each coordinate: $M = \left(\frac{-3+9}{2},\ \frac{5+(-1)}{2}\right) = \left(\frac{6}{2},\ \frac{4}{2}\right) = (3, 2)$
3. 3
  (c) $AB^2 = 12^2 + (-6)^2 = 144 + 36 = 180$
4. 4
  $AB = \sqrt{180}$
  Now simplify. The largest square factor of $180$ is $36$ , since $180 = 36 \times 5$ : $AB = \sqrt{36}\sqrt{5} = 6\sqrt5$
  Hunt for the largest square factor, not the first one you spot — √180 = √4·√45 = 2√45 is not finished.
Answer
(a) $-\tfrac12$ (b) $(3, 2)$ (c) $6\sqrt5$
2
Write down the gradient of a line perpendicular to a line of gradient
(a) $\tfrac34$ (b) $-5$ (c) $-\tfrac27$ (d) $0$ .
Stuck? Show hint
Flip, then change the sign. For (d), think about what a line perpendicular to a horizontal line looks like.
Show solution
1. 1
  (a) Flip $\tfrac34 \to \tfrac43$ ; change the sign $\to -\tfrac43$ .
2. 2
  (b) Write $-5$ as $-\tfrac51$ . Flip $\to -\tfrac15$ ; change the sign $\to \tfrac15$ .
  Writing an integer as a fraction over 1 removes all the guesswork from flipping it.
3. 3
  (c) Flip $-\tfrac27 \to -\tfrac72$ ; change the sign $\to \tfrac72$ .
4. 4
  (d) Gradient $0$ is a horizontal line. The perpendicular to it is vertical, and a vertical line has no gradient — its equation has the form $x = k$ .
  This is the one case where m₁m₂ = −1 breaks down, and examiners do occasionally use it.
Answer
(a) $-\tfrac43$ (b) $\tfrac15$ (c) $\tfrac72$ (d) undefined — the perpendicular is vertical, $x = k$
3
The line joining $A(2, -1)$ and $B(k, 7)$ is perpendicular to the line joining $C(0, 3)$ and $D(4, 1)$ .

Find the value of $k$ .
Stuck? Show hint
Find the gradient of $CD$ first — it is fully numerical. Then you know what $m_{AB}$ must be.
Show solution
1. 1
  $CD$ has no unknowns, so start there: $m_{CD} = \frac{1 - 3}{4 - 0} = \frac{-2}{4} = -\frac12$
  Always compute the fully-known gradient first. It turns the perpendicularity condition into a single equation instead of two unknowns.
2. 2
  For $AB$ to be perpendicular to $CD$ , its gradient must be the negative reciprocal of $-\tfrac12$ : flip to $-\tfrac21$ , change the sign, giving $m_{AB} = 2$
3. 3
  Now write $m_{AB}$ from the coordinates: $m_{AB} = \frac{7 - (-1)}{k - 2} = \frac{8}{k-2}$
4. 4
  Set the two expressions equal: $\frac{8}{k-2} = 2$
5. 5
  Multiply both sides by $(k-2)$ : $8 = 2(k-2)$
6. 6
  Expand and solve: $8 = 2k - 4$ , so $2k = 12$ and $k = 6$
7. 7
  Check. With $k=6$ , $m_{AB} = \dfrac{8}{4} = 2$ , and $2 \times \left(-\tfrac12\right) = -1$ . ✓
Answer
$k = 6$
44 marks
Points $A(1, 2)$ and $B(9, 6)$ are fixed, and $C$ has coordinates $(k, 0)$ .

Given that angle $ACB$ is a right angle, find the two possible values of $k$ .
Stuck? Show hint
The right angle is at $C$ , so the two lines you need are $CA$ and $CB$ .
Show solution
1. 1
  The angle is at $C$ , so the perpendicular pair is $CA$ and $CB$ . $m_{CA} = \frac{2 - 0}{1 - k} = \frac{2}{1-k}$
2. 2
  $m_{CB} = \frac{6 - 0}{9 - k} = \frac{6}{9-k}$
3. 3
  Perpendicular, so the product is $-1$ : $\frac{2}{1-k} \times \frac{6}{9-k} = -1$
4. 4
  Multiply the fractions: $\frac{12}{(1-k)(9-k)} = -1$
5. 5
  Multiply both sides by $(1-k)(9-k)$ : $12 = -(1-k)(9-k)$
6. 6
  Expand the bracket product: $(1-k)(9-k) = 9 - k - 9k + k^2 = k^2 - 10k + 9$ . So $12 = -(k^2 - 10k + 9) = -k^2 + 10k - 9$
7. 7
  Bring everything to the left so the $k^2$ term is positive: $k^2 - 10k + 9 + 12 = 0 \quad\Rightarrow\quad k^2 - 10k + 21 = 0$
  Arranging so the squared term is positive makes factorising far more reliable.
8. 8
  Factorise: two numbers multiplying to $21$ and adding to $-10$ are $-3$ and $-7$ : $(k-3)(k-7) = 0$
9. 9
  So $k = 3$ or $k = 7$ . Both are genuine — geometrically, there are two points on the $x$ -axis from which $AB$ subtends a right angle.
  Those two points are where the circle with AB as diameter crosses the x-axis. You will meet this 'angle in a semicircle' fact again in the circle geometry section.
Answer
$k = 3$ or $k = 7$

Practise gradients, midpoints and lengthsReal past-paper questions · Equation of a straight line

Checking your access…

Can you do all of these?

Find the gradient, midpoint and length of $AB$ for $A(5,2)$ , $B(10,-1)$ in under a minute
Write the equation of the line through $(-2, 5)$ with gradient $\tfrac34$ in the form $ax+by+c=0$
Construct the perpendicular bisector of a given segment
Convert $x^2 + y^2 - 6x + 10y - 27 = 0$ into centre–radius form
Write the equation of the circle with $AB$ as diameter, given $A$ and $B$
Find the gradient of the tangent to a circle at a given point, without calculus
Find both intersection points of a line and a circle, as coordinate pairs
Decide whether a given line cuts, touches or misses a given circle
Use the discriminant to find the value of $k$ for which a line is a tangent to a curve or circle
Find the perpendicular distance from a point to a line, and compare it with the radius
Find the equation of the tangent and of the normal to a circle at a stated point
Recover the centre and radius of a circle from three conditions, or from a diameter's endpoints
Find where two graphs meet by solving their equations simultaneously, and read the discriminant to say how many times
Find the chord length and the distance from the centre to a chord
Find a circle's centre from three points on it, or from two points and the radius
Find the tangents to a circle with a given gradient
Find the least and greatest distance between two circles, or from a circle to an axis