Straight-Line Graphs

• Two numbers describe a line: gradient and intercept. Geometry asks you to build a line from conditions; linearisation asks you to read constants out of one that has already been drawn. Same two numbers, opposite direction of travel.
• Same order top and bottom · a vertical line has no gradient · negative and reciprocal · a bisector needs both conditions · exponentiate a logarithmic intercept.
• Label the points first, subtract in one consistent order, and check the sign of your gradient against the picture before you use it. A wrong gradient is the most expensive error in this topic because everything downstream inherits it.
• Point–gradient form is the workhorse: the instant you have a gradient and a point, you can write the line. Convert to \(y=mx+c\) to read off information, and to \(ax+by=c\) to present the answer.
• Parallel: copy the gradient. Perpendicular: invert it and negate it, then multiply the two together and confirm you get \(-1\). If either line is vertical, put the formula down and look at the picture instead.
• Midpoint adds and halves. Length subtracts, squares, adds and roots. Leave the root exact and simplified — \(2\sqrt{61}\), not \(15.6\).
• Midpoint, gradient, negative reciprocal, point–gradient form — then substitute the midpoint back into your answer to prove you met both conditions. “Equidistant from two points” is always a perpendicular-bisector question.
• Write the straight-line form first, then name \(X\) and \(Y\). Power model: \(\ln y\) against \(\ln x\), gradient \(n\). Exponential model: \(\ln y\) against \(x\), gradient \(\ln b\). In both, the intercept is \(\ln A\) — exponentiate it.

• Calculate a gradient from two points, keeping the subtraction order the same in numerator and denominator.
• Write the equation of a line from a gradient and one point using \(y-y_1=m(x-x_1)\).
• Write the equation of a line through two given points.
• Convert freely between \(y=mx+c\), \(y-y_1=m(x-x_1)\) and \(ax+by=c\).
• State the equation of a horizontal line and of a vertical line, and give the gradient of each — including saying that a vertical gradient is undefined rather than infinite or zero.
• Read a gradient and a vertical intercept off a general linear equation by rearranging it.
• Decide whether two given lines are parallel, perpendicular or neither.
• Find the line through a given point parallel to a given line.
• Find the line through a given point perpendicular to a given line, using the negative reciprocal.
• Explain why \(m_1m_2=-1\) applies only when both lines are non-vertical, and treat a horizontal–vertical pair as a separate case.
• Find the midpoint of a segment by averaging the coordinates.
• Find the length of a segment, and leave it as an exact surd in simplest form.
• Find the equation of a perpendicular bisector and verify that it satisfies both defining conditions.
• Use a perpendicular bisector inside a larger coordinate problem, such as locating a point equidistant from two others.
• Convert \(y=Ax^{n}\) into \(\ln y=\ln A+n\ln x\) and name the variables actually plotted.
• Convert \(y=Ab^{x}\) into \(\ln y=\ln A+x\ln b\) and name the variables actually plotted.
• Find \(A\), \(n\) and \(b\) from the gradient and intercept of a transformed graph, exponentiating where the intercept is \(\ln A\).
• Work backwards from a straight-line graph of constructed variables, such as \(y^{2}\) against \(x^{3}\), to the original relationship.
• Handle the required forms \(y^{2}=Ax^{3}+B\), \(\mathrm{e}^{2y}=Ax^{2}+B\) and \(y^{3}=A\ln x+B\).
• State the positivity conditions that a logarithmic transformation requires.

Where this topic goes next. Chapter 8 puts a circle on the same grid, and the perpendicular bisector you meet here is exactly how the centre of a circle through three points is found. The gradient idea returns in Chapter 14 as the derivative — the gradient of a curve at a point is the gradient of the straight line that just touches it. Straight-line graphs are not a side topic; they are the coordinate language the rest of the course is written in.

• 1. Subtracting the \(x\)-coordinates and the \(y\)-coordinates in different orders The error For \((-2,5)\) and \((4,-1)\), writing \(m=\dfrac{-1-5}{-2-4}=\dfrac{-6}{-6}=+1\). Why it fails The numerator was taken second-point-minus-first, but the denominator first-point-minus-second. Reversing one difference multiplies it by \(-1\), so the quotient comes out with the wrong sign. The correct value is \(-1\). Fix Label \((x_1,y_1)\) and \((x_2,y_2)\) explicitly before substituting, and read the formula left to right in one direction. Then sanity-check the sign against the points: here \(y\) falls as \(x\) rises, so the gradient must be negative.
• 2. Giving a vertical line a finite gradient The error For \((5,1)\) and \((5,9)\), writing \(m=\dfrac{8}{0}=0\), or \(m=\infty\), or simply picking a large number. Why it fails Division by zero is not a large quantity; it is an operation with no result. The gradient of \(x=5\) does not exist, so no numerical answer of any size is correct. “Infinity” does not rescue the situation either, because it is not a number: the honest and complete statement is that the gradient is undefined. Fix Before dividing, check whether \(x_2=x_1\). If it is, stop: write “the gradient is undefined” and give the line as \(x=5\).
• 3. Believing that “perpendicular” simply means “negative” The error Claiming the perpendicular to \(m=\tfrac34\) has gradient \(-\tfrac34\). Why it fails Test it against the condition itself: \(\tfrac34\times\left(-\tfrac34\right)=-\tfrac{9}{16}\), not \(-1\). Negating alone reflects the line's direction in the horizontal, producing a mirror image rather than a right angle. Fix Learn the rule as two operations in order: invert, then negate. Verify by multiplying.
• 4. Changing the sign but not turning the fraction over The error From \(m_1=-\tfrac{2}{5}\), giving \(m_2=\tfrac{2}{5}\) — the sign has changed, but the fraction has not been inverted. Why it fails \(-\tfrac25\times\tfrac25=-\tfrac{4}{25}\ne -1\). This is the procedural half of mistake 3: the candidate knows two things must happen and performs only one of them. Fix The check distinguishes the two failure modes precisely. A product of \(+1\) means you inverted but forgot the sign; a product that is neither \(\pm1\) means you did not invert. The correct answer here is \(\tfrac52\).
• 5. Applying \(m_1m_2=-1\) to a vertical line The error Asked for the line perpendicular to \(x=4\) through \((4,7)\), substituting an invented gradient for \(x=4\) and grinding out an answer. Why it fails The condition is a statement about two numbers. If one of them does not exist, the equation cannot be written down, let alone solved. Any answer obtained this way rests on a value that was never valid. Fix Handle the case geometrically: the perpendicular to a vertical line is horizontal, so the answer is \(y=7\). Similarly the perpendicular to \(y=k\) is vertical.
• 6. Averaging the coordinate differences instead of the coordinate values for a midpoint The error For \(A(-3,8)\) and \(B(9,-2)\), computing \(\left(\dfrac{12}{2},\dfrac{-10}{2}\right)=(6,-5)\). Why it fails The gradient formula has leaked into the midpoint formula. The result is not even on the segment: at \(x=6\) the segment \(AB\) is at \(y=0.5\), nowhere near \(-5\). The midpoint is \(\left(\dfrac{-3+9}{2},\dfrac{8+(-2)}{2}\right)=(3,3)\). Fix Say it while writing: “gradient subtracts, midpoint adds”. Then check that both coordinates of your midpoint lie strictly between those of \(A\) and \(B\).
• 7. Forgetting the square root in the distance formula The error Reporting \(AB=244\) instead of \(AB=\sqrt{244}\). Why it fails Pythagoras gives \(AB^{2}\), not \(AB\). The final step of taking the root is part of the method, not a formality. Fix Estimate before you finish. The longer leg is \(12\), so the hypotenuse must be a little more than \(12\) — a value of \(244\) is impossible on inspection. Write the working as \(AB^{2}=\ldots\) on one line and \(AB=\ldots\) on the next, so the missing step becomes visible.
• 8. Rounding an exact surd unnecessarily The error Writing \(AB=15.6\) when \(AB=2\sqrt{61}\) was available and no accuracy was specified. Why it fails \(2\sqrt{61}\) is the length; \(15.6\) is an approximation to it. Where a question asks for an exact value, a rounded decimal does not answer it; and even where it does not, replacing an exact value with a rounded one discards information. It also propagates: an area computed from \(15.6\) carries that error forward. Fix Leave surds exact and simplified. Give a decimal only when the question asks for one, or as a supplementary comment after the exact answer.
• 9. A “perpendicular bisector” that does not pass through the midpoint The error Finding the perpendicular gradient correctly, then using endpoint \(A\) rather than the midpoint \(M\). For \(A(-3,8)\), \(B(9,-2)\) this produces \(6x-5y=-58\). Why it fails That line genuinely is perpendicular to \(AB\), but it crosses \(AB\) at \(A\), so it bisects nothing. Half of the definition has been satisfied and half ignored. Fix Substitute the midpoint into your final answer. Here \(6(3)-5(3)=3\ne-58\), which exposes the error in one line. The correct bisector is \(6x-5y=3\).
• 10. A line through the midpoint that is not perpendicular The error Using the midpoint but keeping the gradient of \(AB\) itself, giving \(y-3=-\tfrac56(x-3)\). Why it fails This is the line \(AB\) all over again — it passes through \(A\), \(B\) and \(M\). A bisector that lies along the segment it is supposed to bisect is a contradiction, and the perpendicular step was simply skipped. Fix Check that your bisector's gradient multiplied by \(m_{AB}\) gives \(-1\). If it gives \(m_{AB}^{2}\), you reused the original gradient.
• 11. Plotting \(x\) and \(y\) when the axes should carry transformed quantities The error Given \(y=Ax^{n}\) and a table of values, plotting \(y\) against \(x\) and trying to read a gradient off the resulting curve. Why it fails A curve has no single gradient and no meaningful intercept, so there is nothing to read. The whole purpose of the transformation is to produce a graph that has two measurable features. Fix Before plotting anything, write the model in the form \(Y=mX+c\) and state what \(X\) and \(Y\) are. Add the transformed values as extra rows of the table.
• 12. Reversing which transformed variable is horizontal and which is vertical The error Reading “plot \(y^{2}\) against \(x^{3}\)” as \(x^{3}\) vertically and \(y^{2}\) horizontally. Why it fails Both numbers you are about to read change. The line \(Y=4X+9\) becomes \(X=\tfrac14Y-\tfrac94\): the gradient becomes the reciprocal \(\tfrac14\) and the intercept becomes \(-\tfrac94\), so the recovered relationship is wrong in both constants. Fix “\(P\) against \(Q\)” always means \(P\) vertical, \(Q\) horizontal. Write \(Y=\ldots\) and \(X=\ldots\) as your first line and label any sketch with those quantities.
• 13. Reading \(A\) directly from an intercept that is \(\ln A\) The error A plot of \(\ln y\) against \(\ln x\) has intercept \(1.7\); concluding \(A=1.7\). Why it fails The straight-line form is \(\ln y=n\ln x+\ln A\), so the constant term is \(\ln A\). An intercept of \(1.7\) therefore says \(\ln A=1.7\), giving \(A=\mathrm{e}^{1.7}\approx5.47\) — more than three times the value reported. Fix Write “intercept \(=\ln A\)” before substituting any number, so the exponentiation is unavoidable.
• 14. Reading \(b\) directly from a gradient that is \(\ln b\) The error A plot of \(\ln y\) against \(x\) has gradient \(1.099\); concluding \(b=1.099\). Why it fails For \(y=Ab^{x}\), the straight-line form is \(\ln y=x\ln b+\ln A\), so the gradient is \(\ln b\). The actual base is \(b=\mathrm{e}^{1.099}=3\). Reporting \(1.099\) describes a relationship that barely grows, where the true one triples at every step. Fix Sanity-check against the data. If \(y\) roughly triples as \(x\) increases by \(1\), then \(b\) must be near \(3\), so a \(b\) close to \(1\) is immediately suspect.
• 15. Taking logarithms of quantities that are not positive The error Linearising a data set that contains \(x=0\) or a negative \(y\), and plotting those points anyway. Why it fails \(\ln x\) is defined only for \(x>0\). At \(x=0\) it is undefined, and for \(x<0\) there is no real value at all. Such points cannot appear on a logarithmic axis, so any line drawn through them is meaningless. Fix State the conditions when you state the transformation: for \(y=Ax^{n}\), “valid for \(x>0\) and \(y>0\)”. If the data violates them, the model or the transformation is the wrong one.
• 16. Stopping before transforming back to the original variables The error Asked to express \(y\) in terms of \(x\), finishing at \(\ln y=2.5\ln x+1.7\), or at \(\mathrm{e}^{2y}=5x^{2}-3\). Why it fails The question asked for \(y\), and \(\ln y\) is not \(y\). The transformation was a tool for finding the constants; leaving the answer in transformed form leaves the job half done. Fix Read the command word again at the end. “Express \(y\) in terms of \(x\)” means the last line begins “\(y=\)”. From \(\mathrm{e}^{2y}=5x^{2}-3\) that means \(y=\tfrac12\ln\!\left(5x^{2}-3\right)\).
• 17. Rounding transformed values too early The error Rounding \(\ln 5=1.609\) to \(1.6\) before exponentiating. Why it fails Exponentiating magnifies error. \(\mathrm{e}^{1.609}=4.998\), correctly rounding to \(5.00\), but \(\mathrm{e}^{1.6}=4.953\), which rounds to \(4.95\). A rounding that looked harmless in the logarithm has changed the answer in the second significant figure. Fix Carry at least three decimal places through the logarithmic working and round only the final constant. Where the intercept is given exactly, as \(\ln 2\), keep it exact and do not convert it to a decimal at all.
• 18. Trying to linearise a sum with logarithms The error Given \(y^{2}=Ax^{3}+B\), taking logarithms to get \(2\ln y=\ln A+3\ln x\). Why it fails That step silently discards \(B\) and treats the model as \(y^{2}=Ax^{3}\). There is no law for \(\ln(P+Q)\), so the logarithm of the right-hand side cannot be split at all. Logarithms linearise products and powers, never sums. Fix Look at the model first. If the unknowns already sit outside as a coefficient and a constant, plot the constructed variables directly — here \(y^{2}\) against \(x^{3}\) — and leave logarithms out of it.
• 19. Using an unequally scaled sketch as evidence of perpendicularity The error Drawing two lines on axes with different horizontal and vertical scales, observing that they “look” at right angles, and offering that as the justification. Why it fails Stretching one axis changes every angle in the picture. On axes where the vertical scale is twice the horizontal, the perpendicular pair \(y=2x\) and \(y=-\tfrac12x\) does not appear at right angles, while other, non-perpendicular pairs do. The appearance carries no information. Fix Perpendicularity is established by \(m_1m_2=-1\), or by the horizontal–vertical argument. Use a diagram to organise your thinking, never as the proof.

• Every candidate for Cambridge IGCSE Additional Mathematics 0606 takes two written papers of equal weight, each of two hours and 80 marks and each worth half the qualification, both compulsory and both requiring full method. They differ in one respect that matters a great deal for this topic: a calculator is not allowed in Paper 1, while a scientific calculator is required for Paper 2. Either paper may assess any part of the subject content, so Topic 7 can appear on either one. The examination is available in the June and November series, and in the March series in India, for this cycle, and the candidate grade range is A* to E. What follows describes the shapes of question this material takes; it is a guide to preparation, not a prediction of any particular paper.
• What the non-calculator paper means for outcome 7.4. Everything in outcomes 7.1 to 7.3 is exact arithmetic and is unaffected. Outcome 7.4 is not. Without a calculator a logarithm or an exponential cannot be turned into a decimal, so on Paper 1 the constants stay exact: an intercept handed over as \(\ln 2\) rather than as \(0.693\), and an answer left as \(A=\mathrm{e}^{1.7}\) rather than \(5.47\). That is a restriction on decimals, not on logarithms — the exact ones are ordinary non-calculator steps, and \(\ln 1=0\), \(\mathrm{e}^{0}=1\) and \(\ln\!\left(\mathrm{e}^{2}\right)=2\) all appear in this chapter's own working. With a calculator, a table of decimals and a decimal answer both become possible. Practise the exact route and the decimal route: the algebra is identical, and only the last line differs.
• Nothing in this topic is given to you. The List of formulas printed in the examination papers supplies the equation of a circle, four mensuration formulas, the quadratic formula, the binomial theorem, the arithmetic and geometric series formulas, the three trigonometric identities and the sine, cosine and triangle-area formulas. It contains no straight-line formula at all. The gradient formula, the midpoint formula, the distance formula, the point–gradient form and both the parallel and perpendicular conditions must be known and selected by you. Treat every formula card in this chapter as something to memorise, not something to look up.
• Points are given, then several parts follow: a gradient, a line, a perpendicular, a midpoint, an intersection, sometimes an area. Each part uses the previous answer.
• A perpendicular bisector is requested directly, or indirectly as “the point equidistant from \(A\) and \(B\) lying on…”.
• A non-linear law with unknown constants, plus either a transformed graph or a small table of values. You state what to plot, then extract the constants.