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.

Parallel Lines

Two distinct straight lines in the same plane that never meet, however far they are extended. For non-vertical lines this happens exactly when their gradients are equal, so a line parallel to y = mx + c has the same m but a different c, and finding a parallel line means copying the gradient and then using the given point to find the new intercept. Two vertical lines x = a and x = b are also parallel whenever a and b differ. Equal gradients together with equal intercepts do not describe two parallel lines: they describe the same line written twice. This requirement is worded identically for Core C3.6 and Extended E3.6.

Coordinates

An ordered pair (x, y) fixing the position of a point in a plane, measured against two perpendicular number lines: the x-coordinate gives the displacement from the origin parallel to the horizontal x-axis and is always written first, and the y-coordinate gives the displacement parallel to the vertical y-axis and is always written second. The point (0, 0) where the axes cross is the origin.

Length and Midpoint

The length of a line segment joining two points is the square root of the sum of the squares of the horizontal and vertical changes between them, which is Pythagoras applied to the right-angled triangle whose hypotenuse is the segment; it is always non-negative. The midpoint of the segment is the point whose coordinates are the mean of the two x-coordinates and the mean of the two y-coordinates, and it lies exactly halfway along the segment.

Perpendicular Lines

Two straight lines that cross at an angle of 90 degrees. When both have a defined gradient, the product of their gradients is minus one, so each gradient is the negative reciprocal of the other: the fraction is inverted and its sign is changed. The one perpendicular pair this rule cannot describe is a horizontal line, of gradient zero, meeting a vertical line, whose gradient is undefined. The perpendicular bisector of a line segment is the line perpendicular to it that also passes through its midpoint.

Gradient of Linear Graphs

A single number measuring the steepness and direction of a straight line, equal to the change in y divided by the corresponding change in x between any two points on the line. On the Core route it is read from a grid: draw a gradient triangle between two points where the line crosses grid intersections, count the change in y and the change in x, and divide. On the Extended route it is also calculated from the coordinates of two points as (y2 minus y1) divided by (x2 minus x1), with the same subtraction order used in numerator and denominator. A positive gradient rises from left to right, a negative gradient falls, a horizontal line has gradient zero, and a vertical line has an undefined gradient because its change in x is zero.

Drawing Linear Graphs

The process of converting a linear equation into an accurately ruled straight line on a coordinate grid, by tabulating at least two correct points and plotting them to within half of the smallest square, or by finding the two axis intercepts. A linear equation is one in which x and y appear only to the first power, so its graph is always a straight line, and two correct points are enough to determine it. On the Core route the equation is given in the form y = mx + c, or a table of values is supplied; the Extended route may be given any linear form, such as 3x + 2y = 5, and must rearrange first. The vertical line x = k is the one straight line that cannot be written in the form y = mx + c.

Equations of Linear Graphs

An algebraic relationship satisfied by the coordinates of every point on a straight line and by no other point. On the Core route two forms are needed: y = mx + c, which shows the gradient m and the y-intercept c directly, and x = k, which describes a vertical line and is the only straight line that cannot be written as y = mx + c. The Extended route adds other forms of the same line, in particular the integer form ax + by = c, and the point-gradient form y - y1 = m(x - x1) built from one known point and the gradient.

• “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.