Coordinate Geometry

• The one-line summary: get a gradient, use a point, simplify, then check the point still fits. That single routine answers 3.5 and 3.6 on both routes, and 3.7 as well on the Extended route — only the source of the gradient changes.
• Vertical lines are the exception to almost everything, and parallel means copy the gradient. Those two sentences defuse the Core danger zones entirely. On the Extended route add a third: perpendicular means flip and negate — and a bisector has to pass through the midpoint as well.
• The intercept shortcut. For any equation in the form \(ax+by=c\), setting \(x=0\) gives the \(y\)-intercept and setting \(y=0\) gives the \(x\)-intercept. Two substitutions, two points, one ruled line — often faster than building a table. The form \(ax+by=c\) is Extended E3.5, so this shortcut belongs to the Extended route.
• Count the triangle for a gradient. Parallel means copy the gradient. Vertical lines have no gradient at all and are written \(x=k\).
• Perpendicular means flip it and negate it, and check the product is \(-1\). A bisector must pass through the midpoint as well.
• Core: ten minutes today, twelve on day 3, ten on day 10, ten on day 30 — forty-two minutes in total. Extended: fifteen, twenty-five, twenty-five and ten — seventy-five minutes. Either way it is about one revision session, spread so that it survives to the examination.

• Core C3.1 Plot and read a point as an ordered pair \((x,\,y)\), moving horizontally first.
• Core C3.1 State which quadrant a point lies in from the signs of its coordinates, and recognise that a point on the \(x\)-axis has \(y=0\) and a point on the \(y\)-axis has \(x=0\).
• Core C3.2 Draw the straight-line graph of an equation given in the form \(y=mx+c\), by tabulating values across the stated domain or by using two points, then ruling and labelling the line.
• Core C3.2 Draw a straight-line graph from a table of values you have been given.
• Core C3.2 Draw and recognise \(x=k\) as a vertical line with undefined gradient and \(y=k\) as a horizontal line with gradient \(0\).
• Core C3.3 Find the gradient of a straight line from a grid, by drawing a gradient triangle between two points where the line crosses grid intersections, counting the two changes and simplifying the fraction.
• Core C3.3 Interpret the sign and size of a gradient, and say when a gradient is zero and when it is undefined.
• Core C3.5 Obtain the equation of a straight line in the form \(y=mx+c\) when its graph is given, by reading the \(y\)-intercept and counting the gradient off the grid.
• Core C3.5 Interpret an equation already in the form \(y=mx+c\) — state the gradient and the \(y\)-intercept of, for example, \(y=6x+3\) — and write down the equation of a vertical line as \(x=k\).
• Core C3.5 Give every equation in a fully simplified form.
• Core C3.6 Find the gradient of a line parallel to a given line, and the equation of the parallel line through a given point.
• Extended E3.2 Draw the straight-line graph of an equation given in any linear form — \(y=7-4x\) or \(3x+2y=5\), for example — rearranging into \(y=\ldots\) before tabulating.
• Extended E3.3 Calculate the gradient of a straight line from the coordinates of two points on it, using \(m=\dfrac{y_2-y_1}{x_2-x_1}\) with a consistent subtraction order.
• Extended E3.4 Calculate the length of a line segment using \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\), leaving a surd exact and fully simplified when an exact answer is required.
• Extended E3.4 Find the coordinates of the midpoint of a line segment as the average of the two \(x\)-coordinates and the average of the two \(y\)-coordinates.
• Extended E3.5 Obtain the equation of a straight line in different forms, including \(ax+by=c\) with integer coefficients, and convert freely between \(y=mx+c\), \(y-y_1=m(x-x_1)\) and \(ax+by=c\).
• Extended E3.5 Read the gradient and the \(y\)-intercept out of an equation that is not solved for \(y\) — for example \(5x+4y=8\) — by rearranging first.
• Extended E3.5 Obtain the equation of a line from two given points.
• Extended E3.6 Find the gradient and equation of a line parallel to a line given in any form, and show that two lines with equal gradients are distinct rather than coincident.
• Extended E3.7 Find the gradient of a line perpendicular to a given line by taking the negative reciprocal, using \(m_1m_2=-1\), and handle the horizontal–vertical exception.
• Extended E3.7 Find the equation of the line through a given point perpendicular to a given line.
• Extended E3.7 Find the equation of the perpendicular bisector of the line joining two given points, checking both of its conditions.

Where this shows up. Every digital map, every screen, every CAD drawing and every plotted data set is a coordinate plane. A phone locating a shop is comparing coordinate pairs; a graph of cost against time is reading a gradient as a rate. The algebra you learn here is the reason a computer can decide whether two drawn walls meet at a right angle without ever using a protractor.

• “Every straight line can be written as \(y=mx+c\).” Not true A vertical line cannot. \(x=4\) has no \(m\) and no \(c\), because its gradient is undefined. Any argument that starts “let the line be \(y=mx+c\)” silently assumes the line is not vertical.
• “A vertical line has gradient zero.” Reversed A horizontal line \(y=k\) has gradient \(0\). A vertical line \(x=k\) has gradient undefined, because the change in \(x\) is \(0\) and you cannot divide by \(0\).
• “Parallel lines have gradients that are related somehow.” Say it precisely Parallel lines have equal gradients. Nothing is done to the number at all: you copy it. If you find yourself doing arithmetic on a gradient while answering a parallel question, stop.
• “It doesn't matter which point I call the first one.” Extended E3.3 Half true It genuinely does not matter — provided you use the same order in the numerator and the denominator. \(\dfrac{y_2-y_1}{x_2-x_1}\) and \(\dfrac{y_1-y_2}{x_1-x_2}\) are equal. Mixing them gives you the right size with the wrong sign.
• “For a perpendicular, just change the sign of the gradient.” Extended E3.7 Incomplete You must take the negative reciprocal: turn the fraction upside down and change the sign. From \(\dfrac{3}{4}\) you get \(-\dfrac{4}{3}\), not \(-\dfrac{3}{4}\). Check with \(m_1m_2=-1\) every time. Confusing this with the parallel rule is worth guarding against deliberately, because it produces a tidy-looking answer that is nevertheless a completely different line.
• “A perpendicular bisector just has to be perpendicular.” Extended E3.7 Half a line It must satisfy both conditions: perpendicular to the segment and passing through its midpoint. An answer that meets only one of the two is not the perpendicular bisector, however correct the gradient looks.

• Two positions have no Core row at all. The syllabus prints “Extended content only” against 3.4 and 3.7, so a Core candidate is never asked for the length of a segment, the midpoint of a segment, a perpendicular gradient or a perpendicular bisector. If you are on the Core route and a practice question asks for one of those, it was written for the other route.
• Working matters as much as the answer. The syllabus is explicit: if candidates are asked to show their working, they cannot gain full marks without clearly communicating their method, even when the final answer is correct. On the Core route that means writing the gradient triangle you counted; on the Extended route it means writing \(m=\dfrac{y_2-y_1}{x_2-x_1}\) with the numbers in it, then the arithmetic, then the equation.
• The two Extended-only positions chain together. Sections 3.4 and 3.7 have no Core counterpart at all, and because a question may assess more than one part of the subject content, an Extended paper can build a whole multi-part question out of them — midpoint, length, gradient and perpendicular bisector, all from one pair of coordinates. That is why they are worth knowing as a single connected method rather than as four separate formulas.
• Accuracy conventions that apply on all four papers. A plotted point must be within half of the smallest grid square of its true position, and should be marked with a small cross. A linear graph must be ruled, and drawn across the whole domain the question gives. An equation should be fully simplified unless the question says otherwise: \(y=2x-5\), not \(y=\dfrac{4x-10}{2}\). When a question specifies a form, the answer has to be in that form. A correct line written in a form the question did not ask for has not answered the question. Read the form before you start. Answers are expected in their simplest form unless the question says otherwise, and where a question asks for an exact value, a fraction or a surd is what it wants. How accurately you may read a value off a graph is set by the scale of that graph: within half of the smallest square, and no finer. Algebraic and graphical calculators are not permitted on any paper.
• What the calculator will not do for you. It will not write your method, and the method is what you have to communicate in the examination. Use it the way you would use an answer key: calculate first, compare second, and when the two disagree, find which line of your working went wrong rather than copying the result.
• Fully simplified, every time. The syllabus expects equations of a line to be given in a fully simplified form. \(y=\dfrac{8x-4}{2}\) is not an answer; \(y=4x-2\) is. Cancel before you write the final line, and put fractions in their lowest terms.
• Match the requested form exactly. “In the form \(ax+by=c\) where \(a\), \(b\) and \(c\) are integers” means no fractions and no decimals anywhere. If you reach \(y=\tfrac{2}{3}x-\tfrac{1}{6}\), multiply through by \(6\) to get \(6y=4x-1\), then rearrange to \(4x-6y=1\). A correct line written in a form the question did not ask for has not yet answered the question, and converting takes one line.
• Tabulate three points, not two. Two points determine a line, so two is technically sufficient — but a single arithmetic slip in one of them produces a confidently drawn wrong line with nothing to reveal it. A third point is a free check: if all three are collinear the line is almost certainly right, and if one is off you know immediately which one to recompute.
• The habit these three share. In every one of them, a later part reuses a number from an earlier part instead of recomputing it. That is not just faster — it is the structure the question was written with, and following it keeps your working consistent. If part (c) makes you recalculate something you already found in part (a), read the question again.
• Marking yourself honestly. An equation that is algebraically equivalent to the printed answer is correct: \(3x+2y=14\) and \(y=-\tfrac{3}{2}x+7\) are the same line. An equation that merely looks similar is not. The test is not whether it resembles the answer, but whether the given point satisfies it and the gradient matches — apply that test rather than comparing shapes.
• After marking. Sort your errors into two piles: arithmetic slips, and method errors. Slips are fixed by writing more slowly and checking. Method errors — using the wrong rule, substituting the wrong point, misreading a form — are fixed by going back to the section that owns them. The two need completely different responses, and treating a method error as “a silly mistake” guarantees you make it again.
• Interleave rather than block. Once Topic 3 is solid, stop practising it on its own. Mix coordinate-geometry questions in with algebra and mensuration questions, so that you have to recognise which method a question wants before you can apply it. Recognition is the skill an examination actually tests, and a page of questions all from the same topic never trains it.