Perpendicularity Proofs, Rectangle/Point-Construction Problems, and Gradient-Pythagoras Combinations

~55 min · WMA11 · 2.1

WMA11 · 2.1 · 55 min

A perpendicularity proof has two halves, and the exam only pays for both. The half everyone remembers is the calculation — multiply two gradients, or square two sides — and the half that quietly loses marks on real scripts is the sentence after it, the one that actually states the two lines are perpendicular. This lesson is built around that exact gap, confirmed on the richest verified example in the whole WMA11 research pass, plus the two traps that sit either side of it: an arithmetic slip in the length route that looks completely innocent on the page, and a 'find the fourth vertex' question where the obvious method is measurably the less reliable one.

Key terms in this lesson

Before you read on

Two or three questions on exactly what this lesson teaches. Being wrong here is fine — it's the fastest way to find out what to pay attention to next.

The equation of a straight line — three forms, and the one number that ties them together

Every straight line in this topic is described by exactly one number that matters more than any other: its gradient, $m$ , the rate at which $y$ changes as $x$ increases. Everything else — which of the equivalent forms you write it in, how you find it from two points, how you use it to test for parallel or perpendicular lines — is bookkeeping around that one number (spec 2.1).

Form one: $y - y_1 = m(x - x_1)$ , a line through the point $(x_1, y_1)$ with gradient $m$ . This is the form to reach for the moment you know a point and a gradient, because it needs no rearranging to write down — both pieces of information go straight in.

Form two: $ax + by + c = 0$ , the general form. It has one property the first form doesn't: it reads the gradient off without fully rearranging into $y = mx + c$ , because $m = -\frac{a}{b}$ directly — rearrange $ax + by + c = 0$ to $y = -\frac{a}{b}x - \frac{c}{b}$ and the coefficient of $x$ is exactly $-\frac{a}{b}$ . Recognising this saves a line of algebra every time a question hands you a line in this form and asks about its gradient, which is precisely what the third prequestion above tested.

Finding the equation of a line through two given points is form one, done in two steps rather than one: first compute the gradient between the two points, $m = \frac{y_2 - y_1}{x_2 - x_1}$ , then substitute that gradient and EITHER point into form one. Either point works — both lie on the same line, so both produce the same final equation once simplified, and checking that the OTHER point also satisfies your finished equation is a free, thirty-second verification that catches almost any slip in the working.

Finding the equation of a line parallel or perpendicular to a given line, through a given point (spec 2.1's own explicit third bullet) is the same two-step process with one extra step in front of it: find the given line's gradient first, apply the parallel or perpendicular condition to get the new gradient, and only then substitute into form one with the given point. Skipping straight to substituting the ORIGINAL line's gradient — forgetting to apply the condition at all — is the single most common way this three-step process collapses to two.

Parallel and perpendicular — the two gradient conditions, and the sentence the exam checks for

Spec 2.2 states two conditions, and they are genuinely the whole of it: two non-vertical lines are PARALLEL exactly when $m_1 = m_2$ , and PERPENDICULAR exactly when $m_1 m_2 = -1$ (equivalently, $m_2 = -\frac{1}{m_1}$ — reciprocate, then negate, in either order). Two lines with equal gradients never meet, or are the same line; two lines whose gradients multiply to $-1$ cross at a right angle. Both conditions fail for a genuinely vertical line, whose gradient is undefined rather than a number — a rare case in this topic, but worth naming rather than silently assuming every line has a well-defined $m$ .

The condition itself is one line of algebra, and on this paper it is very rarely the whole of the question. The richest single finding in this course's whole coordinate-geometry research is that a correct application of $m_1 m_2 = -1$ can still lose a mark — not because the number is wrong, but because nothing on the page then says what that number MEANS. A real examiner report records exactly this, on a real proof question: candidates who had 'completed the correct work' still 'failed to give a suitable explanation and/or conclusion... without any reference to the fact that it indicated the two lines were perpendicular or that the angle was a right angle' (Jan 2023, Q2). The rest of this lesson is built around that one finding and the two traps documented alongside it on the same question, plus a third, independent trap from a different series' hardest question.

Mechanism

Where $m_1 m_2 = -1$ actually comes from

Take a line through the origin with gradient $m_1 = \frac{p}{q}$ (positive $p$ , $q$ , for concreteness — the argument works identically for any sign). A point on this line, one 'gradient triangle' along from the origin, is $A = (q, p)$ : run $q$ , rise $p$ , gradient $\frac{p}{q}$ . Now rotate the segment $OA$ by $90°$ about $O$ . A $90°$ anticlockwise rotation about the origin sends the point $(x, y)$ to $(-y, x)$ — check it on the simplest case: $(1, 0)$ , one unit along the positive $x$ -direction, maps to $(0, 1)$ , one unit along the positive $y$ -direction, which is exactly a quarter-turn anticlockwise. Apply that same rule to $A = (q, p)$ : it lands on $A' = (-p, q)$ . The line $OA'$ is perpendicular to $OA$ — a $90°$ rotation turns every line through the centre of rotation into a line perpendicular to where it started, by definition of what a right angle is — and its gradient is $\frac{q}{-p} = -\frac{q}{p} = -\frac{1}{m_1}$ . So the perpendicular line's gradient is exactly the negative reciprocal of the original, and $m_1 m_2 = \frac{p}{q} \times \left(-\frac{q}{p}\right) = -1$ — not a rule to memorise, but the one number a $90°$ turn produces every time, whatever $p$ and $q$ actually are.

Diagram — Rectangle PQRS, and the one right angle a proof at Q actually needs

x-axis: x · y-axis: y

PQ: From P(1, 1) to Q(4, 7). Gradient = (7 − 1)/(4 − 1) = 2.
QR: From Q(4, 7) to R(−4, 11). Gradient = (11 − 7)/(−4 − 4) = −1/2 — the negative reciprocal of PQ's gradient, which is exactly the algebraic signature of a right angle at Q.
RS: From R(−4, 11) to S(−7, 5). Gradient = (5 − 11)/(−7 + 4) = 2 — parallel to PQ, which is what closes the shape into a rectangle rather than an open zigzag.
SP: From S(−7, 5) back to P(1, 1). Gradient = (1 − 5)/(1 + 7) = −1/2 — parallel to QR.

P (1, 1): One of the two vertices given directly in this lesson's worked question.
Q (4, 7): The vertex the perpendicularity proof is actually about — where PQ meets QR at 90°.
R (−4, 11): The third given vertex, used both for the proof at Q and for finding S.
S (−7, 5): The fourth vertex — not given, and the whole subject of this lesson's method-comparison block on how to find it.
Midpoint of both diagonals (−1.5, 6): PR and QS share this single midpoint, because the diagonals of a rectangle bisect each other. This ONE fact is the entire content of the faster route to S covered later in this lesson.

Common error: Computing gradient(PQ) × gradient(QR) = −1 and stopping there.

Correct: State the conclusion explicitly: since gradient(PQ) × gradient(QR) = −1, PQ is perpendicular to QR (the angle at Q is 90°). A mark scheme gates credit on that sentence appearing on the page, not on the arithmetic alone.

examiner-report · Jan 2023 · Q2(a)

Marked, line by line

$P(1, 1)$ , $Q(4, 7)$ and $R(-4, 11)$ are three vertices of a rectangle $PQRS$ . (a) Prove that $PQ$ is perpendicular to $QR$ . (3) — VERIDIAN-original question, built around the confirmed real proof-rigour finding on Jan 2023 Q2(a) (the coordinates are original; the real question's own points were not captured by the research pass).

3 marks available

01
Gradient of $PQ = \dfrac{7 - 1}{4 - 1} = \dfrac{6}{3} = 2$
Method mark for a correct attempt at a gradient, using the change-in-y-over-change-in-x pattern with the given coordinates in a consistent order.
M1
02
Gradient of $QR = \dfrac{11 - 7}{-4 - 4} = \dfrac{4}{-8} = -\dfrac{1}{2}$
Both gradients correct. The sign is the whole difficulty in this line: $-4 - 4 = -8$ , not $-4 + 4$ — a genuinely easy slip when the second point's x-coordinate is itself negative.
A1
03
$2 \times \left(-\dfrac{1}{2}\right) = -1$ , so $PQ$ is perpendicular to $QR$ (the angle at $Q$ is a right angle).
Dependent accuracy mark for the explicit conclusion — connecting the product $-1$ to the fact it actually proves. This is the mark a real examiner report records as commonly lost even when every calculation above it is correct: 'a small number of candidates simply left it at that without any reference to the fact that it indicated the two lines were perpendicular or that the angle was a right angle' (Jan 2023, Q2). The real mark scheme states this mark's dependency explicitly: 'A1: Depends on both previous marks and requires an explanation and a conclusion' (Jan 2023, Q2 general guidance for part (a)) — it is a DEPENDENT accuracy mark, not an independent B mark: earned only once the method mark and the gradient-accuracy mark above are both already secured, on top of which it additionally requires the explanation.
A1

In your own words

In one sentence: why does correctly calculating gradient(PQ) × gradient(QR) = $-1$ not, by itself, count as a complete proof that PQ is perpendicular to QR?

Same question, every valid method

$P(1, 1)$ , $Q(4, 7)$ and $R(-4, 11)$ are three vertices of a rectangle. Prove that angle $PQR$ is a right angle. (Same points as the marked solution above — this is the same proof, by a genuinely different route.)

2 valid methods · every one reaches gradient(PQ) × gradient(QR) = −1, equivalently PQ² + QR² = PR² = 125, so angle PQR = 90° — either way, PQ is perpendicular to QR. · 3 marks available

01
Gradient of $PQ = 2$ , gradient of $QR = -\dfrac{1}{2}$ — see the full mark-by-mark working above.
Both gradients found and correct — the M1 and A1 already broken down line by line above.
M1 A1
02
$2 \times \left(-\dfrac{1}{2}\right) = -1$ , so $PQ$ is perpendicular to $QR$ .
The dependent concluding statement, earned only once the method mark and the gradient-accuracy mark above are both secured — see the commonWrongPath above for exactly what is lost without it.
A1

The default. Two short gradient calculations and a one-line conclusion — faster than the alternative below whenever the points are given as coordinates rather than as lengths.

Same question, every valid method

$P(1, 1)$ , $Q(4, 7)$ and $R(-4, 11)$ are three vertices of rectangle $PQRS$ , in order round the rectangle. Find the coordinates of $S$ .

2 valid methods · every one reaches $S(-7, 5)$ · 2 marks available

01
$RS$ is parallel to $PQ$ (gradient $2$ ), through $R(-4, 11)$ : $y - 11 = 2(x + 4)$ , i.e. $y = 2x + 19$ . $SP$ is parallel to $QR$ (gradient $-\frac{1}{2}$ ), through $P(1, 1)$ : $y - 1 = -\frac{1}{2}(x - 1)$ , i.e. $y = -\frac{1}{2}x + \frac{3}{2}$ . Setting them equal: $2x + 19 = -\dfrac{1}{2}x + \dfrac{3}{2} \Rightarrow \dfrac{5}{2}x = -\dfrac{35}{2} \Rightarrow x = -7$ , then $y = 2(-7) + 19 = 5$ .
Method mark for attempting the equations of BOTH unknown sides and solving them simultaneously to find $x$ or $y$ — the real mark scheme scores forming both equations AND solving them as a SINGLE method mark, not two: 'Attempts the equation of line PS, the equation of line RS and solves simultaneously to find x or y' (Jan 2023, Q2(b), Alt 1). Generous on what counts as the method, unforgiving on the arithmetic — nothing here is scored until $x$ or $y$ is actually reached.
M1
02
$S = (-7, 5)$ .
States the coordinates in full — the only other mark this method carries. On a real script this is also the line most often never reached: an examiner report on this exact question type records that candidates 'who attempted to find equations for RS and PS and solving simultaneously rarely produced enough work to score full marks' (Jan 2023, Q2) — two gradients and two full line equations before the simultaneous solve even starts is a lot of working to sustain without a slip, for a method worth no more than the 2 marks (M1 A1) below either way.
A1

The 'obvious' route — every step follows directly from facts already taught (parallel sides, equation of a line through a point) — but confirmed, on a real script, as the LESS reliable one in practice, for the same 2 marks (M1 A1) the shortcut below also earns. Four separate pieces of algebra (two gradients, two line equations, one simultaneous solve) is a lot of surface area for a single sign error to hide in before the final answer is checked against anything.

Points a fixed distance from a given point — where the right angle actually is

A different, independently-verified question type combines the perpendicular condition with Pythagoras in a different order: instead of PROVING a right angle exists between two given lines, you're asked to FIND a point that satisfies a distance condition, using a right angle the geometry guarantees but the question doesn't draw for you. This question type comes from a different series than the rectangle proof above — Oct 2020 Q6, described by its own examiner report as 'one of the most challenging questions on the paper', where 'the majority of candidates failed to score more than 2 of the 5 marks available' on the part built this way.

The recurring difficulty is not the Pythagoras step itself — it's identifying WHICH point in the diagram is actually guaranteed to form a right angle. When a point $C$ lies on the perpendicular bisector of a segment $AB$ , the geometry guarantees a right angle at exactly one place: $M$ , the midpoint of $AB$ , where the perpendicular bisector meets $AB$ . No other labelled point in the same diagram — not $A$ , not $B$ , not the origin, not any other point that happens to sit nearby — carries that guarantee, however tempting it looks on a quick sketch. A confirmed real trap on this exact question type: 'other candidates who did not score full marks... assumed that AB was 12, they applied the Pythagorean triple 3,4,5 from the origin, rather than from the midpoint of AB' (Oct 2020, Q6) — building the right triangle from a point the geometry never promised a right angle at.

Worked, in full

$A(0, 3)$ and $B(6, 3)$ . $C$ lies on the perpendicular bisector of $AB$ with $AC = 5$ . Find both possible positions of $C$ .

01
Identify the one point guaranteed to give a right angle. $C$ lies on the perpendicular bisector of $AB$ , and that line meets $AB$ at exactly one point: $M$ , the midpoint of $AB$ — the only place in this whole figure where a right angle is actually promised by the geometry, not assumed from the look of a sketch.
Earns: M1 — identifies M, the midpoint of AB, as the vertex of the right angle needed to set up Pythagoras. Confirmed, on a genuinely difficult real series, as exactly the step that goes missing on real scripts: candidates who 'applied the Pythagorean triple 3,4,5 from the origin, rather than from the midpoint of AB' (Oct 2020, Q6) built their right triangle from the wrong point in the diagram, not from a wrong calculation once the (wrong) point was already chosen.
02
Find $M$ and $AM$ . $M$ = midpoint of $A(0, 3)$ and $B(6, 3)$ = $(3, 3)$ . $AM$ , HALF of $AB$ , is $3$ — not the full length $AB$ ( $=6$ ): the right triangle here is $A$ – $M$ – $C$ , and $M$ is the vertex that makes it right-angled, so $AM$ , not $AB$ , is the leg that belongs in it.
Earns: A1 — correct M and correct AM. Using the full AB instead of half of it, or measuring from any point besides M, produces a triangle the geometry never actually guarantees is right-angled.
03
Apply Pythagoras to the right-angled triangle $AMC$ (right angle at $M$ , since $MC$ lies along the perpendicular bisector and $AM$ lies along $AB$ ): $MC^2 = AC^2 - AM^2 = 5^2 - 3^2 = 25 - 9 = 16$ , so $MC = 4$ .
Earns: M1 — correct use of Pythagoras' theorem from the right vertex, M, with the given AC as the hypotenuse.
04
$AB$ is horizontal, so its perpendicular bisector is vertical: moving from $M(3, 3)$ a distance of $4$ straight up or down stays on that line. $C = (3, 3 + 4) = (3, 7)$ , or $C = (3, 3 - 4) = (3, -1)$ .
Earns: A1 — both positions of C. A fixed distance from a point, applied to a point constrained to a line, almost always has exactly two solutions — one on each side of the foot of the perpendicular — and stopping at one of them is the same 'half an answer' failure this lesson has already seen twice: one rectangle vertex found where four were needed, one squared coordinate difference computed where the whole distance formula was needed.

Source — Examiner report, Oct 2020

"other candidates who did not score full marks... assumed that AB was 12, they applied the Pythagorean triple 3,4,5 from the origin, rather than from the midpoint of AB"

Complete it yourself

Complete the chain — $A(2, 0)$ and $B(2, 8)$ . $D$ lies on the perpendicular bisector of $AB$ with $AD = 5$ . Find both possible positions of $D$ .

01
$AD = 5$ , but the right angle this problem is built on is not at $A$ , and it is not at the origin — it is at $M$ , the midpoint of $AB$ , because $M$ is the one point where the perpendicular bisector genuinely meets $AB$ at $90°$ .
02
Find $M$ and $AM$ . $M$ = midpoint of $A(2, 0)$ and $B(2, 8)$ = $(2, 4)$ . $AM$ , half of $AB$ , is $4$ .

Named traps

correct-calculation-no-concluding-statement: Confirmed on the richest verified example in this course's whole coordinate-geometry research: "Some candidates did not appreciate the rigour required in the proof in question 2 part (a) in that, having completed the correct work, failed to give a suitable explanation and/or conclusion... a small number of candidates simply left it at that without any reference to the fact that it indicated the two lines were perpendicular or that the angle was a right angle" (Jan 2023, Q2). The calculation — gradients found, product taken, equal to $-1$ — is not the answer to "prove PQ is perpendicular to QR"; it is the EVIDENCE for that answer, and the mark scheme requires the answer itself to be written down as a sentence. This is the single most valuable exam-technique habit this lesson teaches: finish every "prove that..." or "show that..." question with a sentence that restates, in words, exactly what the question asked you to show.
squares-differenced-instead-of-difference-squared: Confirmed verbatim on the same question, in the length/Pythagoras route: "Some did not use the difference of the coordinates squared but the difference of the squares of the coordinates e.g. 11² − 7² [instead of (11−7)²]" (Jan 2023, Q2). The distance formula needs $(x_2 - x_1)^2$ — subtract first, THEN square — not $x_2^2 - x_1^2$ , squaring each coordinate separately and then subtracting. The two calculations are wildly different in general ( $11^2 - 7^2 = 72$ against $(11-7)^2 = 16$ here) and the wrong one gives no warning sign on the page — it is a clean, plausible-looking number, which is exactly what makes it dangerous.
fourth-vertex-via-simultaneous-equations-often-incomplete: Confirmed on the same question's part (b): "Candidates who attempted to find equations for RS and PS and solving simultaneously rarely produced enough work to score full marks" — while using the midpoint of the rectangle's diagonals was "a more unusual but often successful approach" (Jan 2023, Q2). The simultaneous-equations method is not wrong — the mark scheme credits it exactly as fully as the alternative — but real scripts attempting it consistently under-deliver, most likely because it needs four separate correct pieces of algebra (two gradients, two full line equations) before the answer even appears, against one geometric fact and one midpoint calculation the other way. When three vertices of a rectangle (or any parallelogram) are given and a fourth is needed, try the diagonal-midpoint route first.
pythagoras-built-from-the-wrong-reference-point: From a different series' hardest question, by its own examiner report's account "one of the most challenging questions on the paper", where "the majority of candidates failed to score more than 2 of the 5 marks available": "other candidates who did not score full marks... assumed that AB was 12, they applied the Pythagorean triple 3,4,5 from the origin, rather than from the midpoint of AB" (Oct 2020, Q6). When a point is constrained to lie on the perpendicular bisector of a segment, the ONLY point in the figure guaranteed to form a right angle with it is the midpoint of that segment — not the origin, not either endpoint, not any other point that happens to be labelled nearby. Before applying Pythagoras to any right-angle-shaped question, name out loud which vertex the right angle is actually AT, and why the geometry (not the picture) guarantees it there.

Beyond the spec

Spec 2.1 and 2.2 ask only for a SPECIFIC line's equation, or a SPECIFIC point satisfying a distance condition — not the general shape traced out by every point that could satisfy it. Circle geometry is explicitly P2/P3 content, not examined at P1. But the worked chain above — fix a point A, ask which points sit a given distance from it — is, without any extra machinery, exactly how a , so recognising the connection now costs nothing and pays back the moment the formal version arrives.

Let $A = (a, b)$ be a fixed point and let $C = (x, y)$ be ANY point with $AC = r$ , a fixed positive number. The distance formula gives $AC = \sqrt{(x-a)^2 + (y-b)^2}$ , and squaring both sides — which changes nothing about which points satisfy the equation, since both sides are non-negative — gives $(x - a)^2 + (y - b)^2 = r^2$ . That is exactly the calculation the worked chain above ran, done ONCE, symbolically, for every point that could be the answer, instead of separately for two specific ones. It is also exactly why the shape traced out by 'a fixed distance from a fixed point' is called a circle: every point on it is, by definition, the same distance $r$ from the same centre $A$ , and $(x-a)^2 + (y-b)^2 = r^2$ is simply that definition written in coordinates. The worked chain's own $C = (3, 7)$ and $C = (3, -1)$ are two of the infinitely many points on the circle $x^2 + (y-3)^2 = 25$ — the two, specifically, that also happen to lie on one particular vertical line.

Retrieval — with feedback on every choice

Question 1

2 marks

Line $l$ passes through $A(1, 4)$ and $B(5, -4)$ . What is the gradient of a line perpendicular to $l$ ?

Question 2

3 marks

Line $l$ has equation $2x - 5y + 10 = 0$ . Which is the equation of the line through the origin, perpendicular to $l$ ?

Question 3

3 marks

A mark scheme awards 3 marks for 'prove angle $ABC = 90°$ ': M1 for a correct gradient method, A1 for both gradients correct, and a second A1 — dependent on both previous marks — for the concluding statement. A student writes: 'gradient $AB = -3$ , gradient $BC = 3$ . $-3 \times 3 = -9$ .' What is the correct assessment?

Question 4

1 mark

To find $PQ^2$ where $P(7, 3)$ and $Q(11, 15)$ , which is the correct calculation?

Question 5

2 marks

$P$ , $Q$ , $R$ are three vertices of a rectangle $PQRS$ , given in order round the rectangle. Which method is confirmed, on a real script, to be the more reliable way to find $S$ ?

Question 6

2 marks

$C$ lies on the perpendicular bisector of $AB$ , and you're asked to find $C$ given a fixed distance $AC$ . Which point must be used as the right-angle vertex when setting up Pythagoras?

Reference — not a study method, a lookup

Gradient: m = (y₂ − y₁)/(x₂ − x₁). Line ax + by + c = 0 has gradient −a/b — read it off without full rearranging.
Parallel: m₁ = m₂. Perpendicular: m₁m₂ = −1, i.e. m₂ = −1/m₁ — reciprocate, then negate.
A 'prove/show perpendicular' calculation is not the answer — the answer is the sentence stating it IS perpendicular.
Distance formula: square the DIFFERENCE of each coordinate pair, then add. Not the difference of the squares.
3 vertices of a rectangle given, 4th one wanted: try the diagonals' shared midpoint before simultaneous equations.
Point on a line, fixed distance from A: the right angle is at the FOOT of the perpendicular from A, not at A, and not at the origin.

Not affiliated with or endorsed by Pearson Edexcel. This lesson's WMA11 material draws on exactly two verified series — January 2023 Q2 and October 2020 Q6 — which is comparable in depth to two of this course's other single-topic WMA11 lessons (discriminant-with-a-parameter and trig-graphs-period-and-domain also draw on exactly two series each) but thinner than others with three or four independently-verified series available (sector-area-and-arc-length, graph-transformations-in-words, integrate-twice-two-constants). Every quotation attributed to an examiner report was independently verified against the primary Pearson document. Neither Jan 2023 Q2's nor Oct 2020 Q6's own coordinates or full question text were captured by the research pass this lesson was written from — only the examiner reports' prose commentary on them — so every worked scenario here (the rectangle PQRS, the points-on-a-perpendicular-bisector questions, every prequestion and MCQ) is VERIDIAN-original, never a reproduction of either real question. Where a real examiner report quotes a specific number from a script (Jan 2023 Q2's '11² − 7²'), this lesson's own coordinates were deliberately chosen so that number falls out of them, to show the real quoted arithmetic worked through against a fully consistent example — not because it reproduces the real question's own points, which this research pass did not capture. Because the questions are original, the per-line mark allocations attached to them are modelled on verified mark-scheme conventions (what M, A and B marks mean, when a mark is independent versus dependent) rather than transcribed from a real mark scheme, which for an original question does not exist.

Question 12 marks

Line $l$ passes through $A(1, 4)$ and $B(5, -4)$ . What is the gradient of a line perpendicular to $l$ ?

$\frac{1}{2}$
Correct. Gradient of $l = \dfrac{-4 - 4}{5 - 1} = \dfrac{-8}{4} = -2$ , so the perpendicular gradient is $-\dfrac{1}{-2} = \dfrac{1}{2}$ .
B $-2$
This is $l$ 's own gradient — the condition needed here is perpendicular, not parallel, so this answers a different question.
C $-\frac{1}{2}$
Reciprocated, but not negated: this is the plain reciprocal of $-2$ , which satisfies $m_1 m_2 = 1$ , not $m_1 m_2 = -1$ .
D $2$
Negated, but not reciprocated: $2 \times (-2) = -4$ , not $-1$ .

Traps tested: Parallel condition used for perpendicular · Reciprocated but not negated · Negated but not reciprocated

Question 23 marks

Line $l$ has equation $2x - 5y + 10 = 0$ . Which is the equation of the line through the origin, perpendicular to $l$ ?

$y = -\frac{5}{2}x$
Correct. Rearranging $l$ : $5y = 2x + 10$ , so $y = \frac{2}{5}x + 2$ — $l$ 's gradient is $\frac{2}{5}$ . The perpendicular gradient is $-\frac{5}{2}$ , and through the origin the equation is simply $y = -\frac{5}{2}x$ (no constant term needed, since $(0,0)$ already satisfies it).
B $y = \frac{5}{2}x$
Reciprocated correctly, but not negated: $\frac{2}{5} \times \frac{5}{2} = 1$ , not $-1$ .
C $y = \frac{2}{5}x$
This is $l$ 's OWN gradient, used unchanged — the equation of a line PARALLEL to $l$ through the origin, not perpendicular to it.
D $y = -\frac{2}{5}x$
Negated, but not reciprocated: $\left(-\frac{2}{5}\right) \times \frac{2}{5} = -\frac{4}{25}$ , not $-1$ .

Traps tested: Reciprocated but not negated · Parallel condition used for perpendicular · Negated but not reciprocated

Question 33 marks

M1 A1 are potentially earnable for the gradients if they're correct, but the second, dependent A1 is unavailable regardless — and the conclusion drawn would be wrong even if stated, since $-9 \neq -1$
Correct on both counts. This student's gradients don't actually satisfy the perpendicular condition at all ( $-3 \times 3 = -9$ , not $-1$ ) — so even a perfectly written concluding sentence ('therefore perpendicular') would be FALSE, and the dependent A1 could never be earned here regardless of wording. Always check the product actually equals $-1$ before reaching for the concluding sentence — the sentence records what the arithmetic proved, it doesn't fix arithmetic that proved the wrong thing.
B3/3 — the gradients multiply to a negative number, which is enough to show perpendicularity
A negative product is necessary but nowhere near sufficient. The perpendicular condition is a SPECIFIC number, $-1$ , not merely 'negative' — $-3 \times 3 = -9$ is negative and still fails the test completely.
C2/3 — same gap as the concluding-sentence trap taught earlier in this lesson: correct working, missing conclusion
This assumes the working itself is correct, which it isn't checked here — and in this case it happens to be internally consistent but conclusion-wise WRONG, not just incomplete: $-9 \neq -1$ means these two lines are not perpendicular at all, whatever gradients $-3$ and $3$ actually belong to.
D0/3 — nothing here can be trusted
Too harsh on the method mark, which is independent of whether the FINAL numbers happen to be right — M1 credits a correct approach to finding a gradient, which this student may well have used even if the specific values turn out to be wrong for the actual points in the question (not given here).

Traps tested: Any negative product assumed sufficient · Assumes working correct without checking · Abandons partial credit reasoning

Question 41 mark

To find $PQ^2$ where $P(7, 3)$ and $Q(11, 15)$ , which is the correct calculation?

$(11 - 7)^2 + (15 - 3)^2$
Correct — square each coordinate DIFFERENCE, then add: $4^2 + 12^2 = 16 + 144 = 160$ .
B $11^2 - 7^2 + 15^2 - 3^2$
This is the documented real trap, generalised: squaring the raw coordinates first and subtracting them, instead of subtracting first and squaring the result. It gives a completely different, and wrong, number: $121 - 49 + 225 - 9 = 288$ , not $160$ .
C $11^2 + 7^2 + 15^2 + 3^2$
This adds every coordinate's square with no differencing at all — it doesn't correspond to any distance, and would give the same (wrong) answer regardless of where P and Q actually were relative to each other.
D $(11 + 7)^2 + (15 + 3)^2$
Coordinates added instead of subtracted. The distance formula measures the GAP between two points, which is a difference, not a sum — adding coordinates instead answers a question about a different, unrelated pair of numbers.

Traps tested: Squares differenced instead of difference squared · Coordinates combined with no differencing · Coordinates added instead of subtracted

Question 52 marks

$P$ , $Q$ , $R$ are three vertices of a rectangle $PQRS$ , given in order round the rectangle. Which method is confirmed, on a real script, to be the more reliable way to find $S$ ?

Using the fact that the diagonals of a rectangle share a midpoint
Correct — confirmed as 'a more unusual but often successful approach' (Jan 2023, Q2), needing one geometric fact and one midpoint calculation, with no gradients or line equations required at all.
BFinding the equations of the two unknown sides and solving them simultaneously
This IS a valid method, credited exactly as fully by the mark scheme — but a real examiner report records that candidates attempting it 'rarely produced enough work to score full marks' (Jan 2023, Q2), because it needs roughly twice the algebra to reach the same point.
CBoth are exactly equally reliable in practice
Both reach the same correct answer and both are creditable in full — but 'equally reliable' is specifically the claim the real examiner-report finding contradicts: one route measurably under-delivers on real scripts, even though nothing about the mark scheme favours the other.
DNeither works without first finding the equation of the whole rectangle
There is no such thing as 'the equation of a rectangle' to find — a rectangle isn't a single curve with one equation, and both methods above find S directly from the three given vertices, with no such intermediate step.

Traps tested: Obvious method assumed more reliable · Assumes equal reliability · Invents unnecessary intermediate step

Question 62 marks

$C$ lies on the perpendicular bisector of $AB$ , and you're asked to find $C$ given a fixed distance $AC$ . Which point must be used as the right-angle vertex when setting up Pythagoras?

The midpoint of $AB$
Correct. The perpendicular bisector meets $AB$ at its midpoint, and that meeting is where a right angle is actually guaranteed by the geometry — nowhere else in the figure has that guarantee.
BThe origin
The origin has no special relationship to $AB$ or its perpendicular bisector unless the question happens to place it there — using it as the right-angle vertex is exactly the confirmed real trap: 'assumed that AB was 12, they applied the Pythagorean triple 3,4,5 from the origin, rather than from the midpoint of AB' (Oct 2020, Q6).
CPoint $A$ itself
There is no guaranteed right angle at $A$ in this construction — $A$ is one endpoint of the segment being bisected, not the point where the perpendicular bisector crosses it.
DPoint $B$ itself
Same issue as using $A$ : $B$ is the other endpoint of the segment, with no guaranteed right angle there either. The right angle belongs to the midpoint, and only the midpoint.

Traps tested: Pythagoras built from the wrong reference point

Practice this for real

This site teaches the mechanism; the exam is sat on Pearson's own real questions. Go find and attempt these yourself — nothing here substitutes for actually sitting a timed paper.

Examiner report: Jan 2023 · Q2(a) — cited directly in this lesson
Examiner report: Oct 2020 · Q6(c) — cited directly in this lesson

Pearson's official past-papers portal

Select International Advanced Level → Mathematics → any series, then look for WMA11.

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Differentiating Negative and Fractional Indices, Then Finding an Exact Gradient

Two independently-verified WMA11 series test exactly this shape — rewrite a negative or fractional index, differentiate it, then evaluate the gradient at a point — and both times a chunk of marks disappeared not on the calculus itself but on what happened around it. Candidates who correctly differentiate then tack on a '+ c' anyway; candidates asked for an exact gradient who hand back a rounded decimal instead; candidates who mishandle the sign or the power on a negative-index term even while getting a neighbouring positive-power term right. None of those three is a calculus mistake, and this lesson is built directly around stopping all three.

40 min

Perpendicularity Proofs, Rectangle/Point-Construction Problems, and Gradient-Pythagoras Combinations

The equation of a straight line — three forms, and the one number that ties them together

Parallel and perpendicular — the two gradient conditions, and the sentence the exam checks for

Where m1m2=−1m_1 m_2 = -1m1​m2​=−1 actually comes from

Points a fixed distance from a given point — where the right angle actually is

A(0,3)A(0, 3)A(0,3) and B(6,3)B(6, 3)B(6,3). CCC lies on the perpendicular bisector of ABABAB with AC=5AC = 5AC=5. Find both possible positions of CCC.

Complete the chain — A(2,0)A(2, 0)A(2,0) and B(2,8)B(2, 8)B(2,8). DDD lies on the perpendicular bisector of ABABAB with AD=5AD = 5AD=5. Find both possible positions of DDD.

Deep dive

Practice this for real

Where $m_1 m_2 = -1$ actually comes from

$A(0, 3)$ and $B(6, 3)$ . $C$ lies on the perpendicular bisector of $AB$ with $AC = 5$ . Find both possible positions of $C$ .

Complete the chain — $A(2, 0)$ and $B(2, 8)$ . $D$ lies on the perpendicular bisector of $AB$ with $AD = 5$ . Find both possible positions of $D$ .