Straight-line graphs

1. Overview

Straight-line graphs are a fundamental topic in Additional Mathematics, crucial for success in both Paper 1 (without a calculator) and Paper 2. This topic covers essential coordinate geometry skills: finding equations of lines, understanding parallel and perpendicular relationships, and calculating midpoints and lengths. A key application is "linearization," where non-linear relationships are transformed into straight lines, allowing you to determine unknown constants from a graph's gradient and intercept. Mastery of straight-line graphs provides a foundation for more advanced topics like calculus and vectors. Expect to see these concepts tested in various problem-solving contexts.

Key Definitions

• Gradient ($m$): The measure of the steepness of a line, defined as the change in $y$ over the change in $x$.
• $y$-intercept ($c$): The point where the line crosses the $y$-axis (where $x = 0$).
• Parallel Lines: Lines with the same gradient that never intersect.
• Perpendicular Lines: Lines that meet at a $90^\circ$ angle; the product of their gradients is $-1$.
• Collinear Points: A set of points that all lie on the same single straight line.
• Perpendicular Bisector: A line that passes through the midpoint of a line segment at a right angle ($90^\circ$).

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Core Content

3.1 The Equation of a Straight Line

There are two primary forms used in Additional Mathematics:

1. Gradient-Intercept Form: $y = mx + c$
2. Point-Gradient Form: $y - y_1 = m(x - x_1)$ (Highly recommended for speed and accuracy in Paper 1).

Worked Example 1 — Finding the equation given two points

Find the equation of the line passing through the points $(1, -1)$ and $(3, 5)$, giving your answer in the form $ay + bx + c = 0$, where $a$, $b$, and $c$ are integers.

Step 1: Calculate the gradient, $m$. $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{5 - (-1)}{3 - 1} = \frac{6}{2} = 3$ Reason: Apply the gradient formula.

Step 2: Use the point-gradient form with the point $(1, -1)$. $y - (-1) = 3(x - 1)$ Reason: Substitute the gradient and coordinates into the point-gradient formula.

Step 3: Simplify and rearrange to the required form. $y + 1 = 3x - 3$ Reason: Expand the brackets. $y - 3x + 4 = 0$ Reason: Rearrange the equation. $\boxed{y - 3x + 4 = 0}$

3.2 Parallel and Perpendicular Lines

• Parallel: $m_1 = m_2$
• Perpendicular: $m_1 \times m_2 = -1$ or $m_2 = -\frac{1}{m_1}$ (the negative reciprocal).

Worked Example 2 — Finding the equation of a parallel line

A line $L_1$ has the equation $2y + 4x = 7$. Find the equation of the line $L_2$ that is parallel to $L_1$ and passes through the point $(2, 3)$. Give your answer in the form $y = mx + c$.

Step 1: Rearrange the equation of $L_1$ to find its gradient. $2y = -4x + 7$ Reason: Isolate the $y$ term. $y = -2x + \frac{7}{2}$ Reason: Divide by 2 to get the gradient-intercept form. Therefore, the gradient of $L_1$ is $m_1 = -2$.

Step 2: Since $L_2$ is parallel to $L_1$, it has the same gradient. $m_2 = m_1 = -2$ Reason: Parallel lines have equal gradients.

Step 3: Use the point-gradient form with the point $(2, 3)$ and the gradient $m_2 = -2$. $y - 3 = -2(x - 2)$ Reason: Substitute the gradient and coordinates into the point-gradient formula.

Step 4: Simplify and rearrange to the required form. $y - 3 = -2x + 4$ Reason: Expand the brackets. $y = -2x + 7$ Reason: Rearrange the equation. $\boxed{y = -2x + 7}$

3.3 Midpoint and Length of a Line

• Midpoint $M$: $\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$
• Length $d$: $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
• Note: In Paper 1, always leave lengths in exact surd form (e.g., $\sqrt{13}$) unless otherwise stated.

Worked Example 3 — Finding the length of a line segment

Points $A$ and $B$ have coordinates $(-2, 3)$ and $(4, -1)$ respectively. Calculate the length of the line segment $AB$, giving your answer in exact form.

Step 1: Apply the distance formula. $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(4 - (-2))^2 + (-1 - 3)^2}$ Reason: Substitute the coordinates into the distance formula.

Step 2: Simplify the expression. $d = \sqrt{(6)^2 + (-4)^2} = \sqrt{36 + 16} = \sqrt{52}$ Reason: Evaluate the squares.

Step 3: Simplify the surd. $d = \sqrt{4 \times 13} = 2\sqrt{13}$ Reason: Simplify the surd to its simplest form. $\boxed{d = 2\sqrt{13}}$

3.4 The Perpendicular Bisector

To find the equation of a perpendicular bisector of segment $AB$:

1. Find the midpoint of $AB$.
2. Find the gradient of $AB$ ($m_{AB}$).
3. Calculate the perpendicular gradient ($m_\perp = -1/m_{AB}$).
4. Use the midpoint and $m_\perp$ to find the equation.

Worked Example 4: Perpendicular Bisector Find the equation of the perpendicular bisector of the line joining $P(-3, 8)$ and $Q(9, 4)$. Give your answer in the form $ax + by + c = 0$, where $a$, $b$, and $c$ are integers.

Step 1: Find the midpoint of $PQ$. $M = \left( \frac{-3 + 9}{2}, \frac{8 + 4}{2} \right) = (3, 6)$ Reason: Apply the midpoint formula.

Step 2: Find the gradient of $PQ$. $m_{PQ} = \frac{4 - 8}{9 - (-3)} = \frac{-4}{12} = -\frac{1}{3}$ Reason: Apply the gradient formula.

Step 3: Find the perpendicular gradient. $m_\perp = -\frac{1}{m_{PQ}} = -\frac{1}{-\frac{1}{3}} = 3$ Reason: The gradient of a perpendicular line is the negative reciprocal.

Step 4: Find the equation using $M(3, 6)$ and $m = 3$. $y - 6 = 3(x - 3)$ Reason: Use the point-gradient form. $y - 6 = 3x - 9$ Reason: Expand the brackets. $y - 3x + 3 = 0$ Reason: Rearrange to the required form. $\boxed{y - 3x + 3 = 0}$

3.5 Linear Law (Transformations)

Non-linear equations can be transformed into the form $Y = mX + c$, where $Y$ and $X$ are functions of $x$ and $y$.

Original Equation	$Y$-axis (vertical)	$X$-axis (horizontal)	Gradient ($m$)	Intercept ($c$)
$y = Ax^n$	$\ln y$	$\ln x$	$n$	$\ln A$
$y = Ab^x$	$\ln y$	$x$	$\ln b$	$\ln A$
$y^2 = Ax^3 + B$	$y^2$	$x^3$	$A$	$B$
$y = \frac{A}{x} + B$	$y$	$1/x$	$A$	$B$

Worked Example 5: Transforming $y = Ab^x$ Variables $x$ and $y$ are related such that when $\ln y$ is plotted against $x$, a straight line passing through $(0, \ln 3)$ and $(2, \ln 12)$ is obtained. Find the exact values of $A$ and $b$.

Step 1: Linearize the equation. $\ln y = \ln(Ab^x) = \ln A + x \ln b$ Reason: Apply the logarithm product and power rules. This matches $Y = mX + c$ where $Y = \ln y$, $X = x$, $m = \ln b$, and $c = \ln A$.

Step 2: Find the intercept ($c$). The line passes through $(0, \ln 3)$, so $c = \ln 3$. Reason: The intercept is the value of $Y$ when $X = 0$. $\ln A = \ln 3 \Rightarrow A = 3$. Reason: If $\ln a = \ln b$, then $a = b$.

Step 3: Find the gradient ($m$). $m = \frac{\ln 12 - \ln 3}{2 - 0} = \frac{\ln(12/3)}{2} = \frac{\ln 4}{2}$ Reason: Apply the gradient formula and the logarithm quotient rule. Using log laws: $m = \frac{\ln(2^2)}{2} = \frac{2 \ln 2}{2} = \ln 2$. Reason: Apply the logarithm power rule.

Step 4: Solve for $b$. $\ln b = \ln 2 \Rightarrow b = 2$. Reason: If $\ln a = \ln b$, then $a = b$. $\boxed{A = 3, b = 2}$

Worked Example 6: Transforming to Straight Line Form

The variables $x$ and $y$ are related by the equation $y = px^q$, where $p$ and $q$ are constants. When $\ln y$ is plotted against $\ln x$, a straight line is obtained with gradient 0.5 and intercept 1.6. Find the values of $p$ and $q$.

Step 1: Linearize the equation. $\ln y = \ln(px^q) = \ln p + q \ln x$ Reason: Apply the logarithm product and power rules. This matches $Y = mX + c$ where $Y = \ln y$, $X = \ln x$, $m = q$, and $c = \ln p$.

Step 2: Identify the gradient and intercept. The gradient is given as 0.5, so $q = 0.5$. Reason: The gradient of the line is equal to $q$. The intercept is given as 1.6, so $\ln p = 1.6$. Reason: The intercept of the line is equal to $\ln p$.

Step 3: Solve for $p$. $p = e^{1.6}$ Reason: Take the exponential of both sides to solve for $p$. $\boxed{p = e^{1.6}, q = 0.5}$

Extended Content (Extended Only)

Additional Mathematics is a single-tier syllabus — all content above applies to all students.

Key Equations

Gradient: $m = \frac{y_2 - y_1}{x_2 - x_1}$ (Not on formula sheet)

Distance: $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ (Not on formula sheet)

Midpoint: $M = \left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right)$ (Not on formula sheet)

Perpendicularity: $m_1 m_2 = -1$ (Vital for bisector questions, not on formula sheet)

Log Transformation (for $y=Ax^n$): $\ln y = n \ln x + \ln A$ (Not on formula sheet - derived from log laws)

Common Mistakes to Avoid

• ❌ Wrong: Prematurely rounding $\sqrt{20}$ to $4.47$ in Paper 1.
• ✅ Right: Keep answers as exact surds ($2\sqrt{5}$) or fractions unless the question specifies "3 significant figures."
• ❌ Wrong: Cubing terms individually in Linear Law (e.g., transforming $y^2 = Ax^3 + B$ into $y = A^{1/2}x^{3/2} + B^{1/2}$).
• ✅ Right: Identify the variables for the axes first (e.g., $Y = y^2$ and $X = x^3$), then rearrange correctly.
• ❌ Wrong: Forgetting the negative sign when finding the perpendicular gradient, calculating $m_\perp = 1/m$ instead of $m_\perp = -1/m$.
• ✅ Right: Always take the negative reciprocal to find the perpendicular gradient.
• ❌ Wrong: Mixing up the Midpoint formula (addition) with the Gradient formula (subtraction).
• ✅ Right: Midpoint is an average (add and divide by 2), gradient is a rate of change (difference in y over difference in x).
• ❌ Wrong: Failing to check for domain restrictions when using logarithms. For example, assuming $\ln(x)$ is defined for all $x$ values.
• ✅ Right: Remember that $\ln(x)$ is only defined for $x > 0$.

Exam Tips

• State Formulas First: Even if you make a calculation error, stating $m = \frac{y_2-y_1}{x_2-x_1}$ can earn you a Method mark.
• "Show That" Questions: In coordinate geometry, if you are asked to "show that" a line is a perpendicular bisector, you must calculate the midpoint, the gradient, and show the $m_1 m_2 = -1$ calculation explicitly.
• Paper 2 Shortcuts: Use your calculator’s "Table" mode or "Linear Regression" mode to check gradients and intercepts if the question allows, but always show the algebraic working.
• Command Words: If a question says "Hence," you must use your previous answer (e.g., using a midpoint you just calculated to find a bisector).
• Check Domain: In Linear Law problems involving logarithms, remember that $\ln(y)$ is only defined for $y > 0$. Check if your constants $A$ or $b$ must be positive.

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Exam-Style Questions

Practice these original exam-style questions to test your understanding. Each question mirrors the style, structure, and mark allocation of real Cambridge 0606 papers.

Exam-Style Question 1 — Paper 1 (No Calculator Allowed) [7 marks]

Question:

The line $L_1$ has equation $3x + 4y = 11$. The line $L_2$ is perpendicular to $L_1$ and passes through the point $A(2, 3)$.

(a) Find the gradient of $L_1$. [2]

(b) Find the equation of $L_2$ in the form $ax + by = c$, where $a$, $b$, and $c$ are integers. [3]

(c) Find the coordinates of the point of intersection of $L_1$ and $L_2$. [2]

Worked Solution:

(a)

1. Rearrange the equation of $L_1$ into the form $y = mx + c$ to identify the gradient. $4y = -3x + 11$ $y = -\frac{3}{4}x + \frac{11}{4}$ [Isolate y to find gradient]
2. State the gradient of $L_1$. $m_1 = -\frac{3}{4}$ [Identify the coefficient of x]

$\boxed{-\frac{3}{4}}$

How to earn full marks: Rearrange the equation to the form y = mx + c and clearly state the gradient, m.

(b)

1. Find the gradient of $L_2$. Since $L_2$ is perpendicular to $L_1$, $m_1m_2 = -1$. $m_2 = \frac{-1}{m_1} = \frac{-1}{-\frac{3}{4}} = \frac{4}{3}$ [Use perpendicular gradient property]
2. Use the point-gradient form of a line to find the equation of $L_2$. $y - y_1 = m(x - x_1)$ $y - 3 = \frac{4}{3}(x - 2)$ [Apply formula $y - y_1 = m(x - x_1)$]
3. Rearrange into the required form. $3(y - 3) = 4(x - 2)$ $3y - 9 = 4x - 8$ $4x - 3y = -1$

$\boxed{4x - 3y = -1}$

How to earn full marks: Remember to use the negative reciprocal of the gradient for perpendicular lines and rearrange to the required integer form.

(c)

1. Solve the simultaneous equations: $3x + 4y = 11$ $4x - 3y = -1$ [Set up simultaneous equations]
2. Multiply the first equation by 3 and the second by 4 to eliminate $y$. $9x + 12y = 33$ $16x - 12y = -4$ [Prepare to eliminate variable]
3. Add the two equations. $25x = 29$ $x = \frac{29}{25}$ [Isolate x]
4. Substitute $x = \frac{29}{25}$ into $3x + 4y = 11$ $3(\frac{29}{25}) + 4y = 11$ $4y = 11 - \frac{87}{25} = \frac{275 - 87}{25} = \frac{188}{25}$ $y = \frac{47}{25}$ [Substitute to solve for y]

$\boxed{(\frac{29}{25}, \frac{47}{25})}$

How to earn full marks: Show your working clearly when solving simultaneous equations, especially when dealing with fractions.

Common Pitfall: When finding the equation of a perpendicular line, remember to take the negative reciprocal of the original gradient. Also, double-check your arithmetic when solving simultaneous equations to avoid errors with fractions.

Exam-Style Question 2 — Paper 1 (No Calculator Allowed) [8 marks]

Question:

The variables $x$ and $y$ are related by the equation $y = a x^n$, where $a$ and $n$ are constants. When $\ln y$ is plotted against $\ln x$, a straight line is obtained which passes through the points $(2, 5)$ and $(5, -1)$.

(a) Find the gradient of the straight line. [2]

(b) Find the equation of the straight line, giving your answer in the form $\ln y = p \ln x + q$. [2]

(c) Find the values of $a$ and $n$. [4]

Worked Solution:

(a)

1. Use the gradient formula to find the gradient. $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-1 - 5}{5 - 2} = \frac{-6}{3}$ [Apply gradient formula]
2. State the gradient. $m = -2$

$\boxed{-2}$

How to earn full marks: Apply the gradient formula correctly, ensuring you subtract the y and x coordinates in the same order.

(b)

1. Use the point-gradient form of a line, using the point (2, 5). $\ln y - 5 = -2(\ln x - 2)$ [Apply formula $y - y_1 = m(x - x_1)$]
2. Rearrange the equation into the required form. $\ln y = -2\ln x + 4 + 5$ $\ln y = -2\ln x + 9$

$\boxed{\ln y = -2\ln x + 9}$

How to earn full marks: Substitute one of the given points and the gradient into the point-gradient form, then rearrange to the required form.

(c)

1. Compare $\ln y = -2\ln x + 9$ with $\ln y = \ln(ax^n) = \ln a + n \ln x$. [Recognise how to relate back to original equation]
2. Equate coefficients. $n = -2$ $\ln a = 9$ [Match coefficients]
3. Solve for $a$. $a = e^9$

$\boxed{a = e^9, n = -2}$

How to earn full marks: Remember to equate the coefficients of the linear equation with the logarithmic form and use the exponential function to find a.

Common Pitfall: Remember that $\ln(ax^n)$ can be split into $\ln a + n \ln x$. Don't forget the properties of logarithms when transforming equations into straight-line form. Also, make sure you use the exponential function to find 'a' after finding ln a.

Exam-Style Question 3 — Paper 2 (Calculator Allowed) [7 marks]

Question:

The line $y = mx + c$ passes through the points $(1, 3.2)$ and $(5, 8.8)$.

(a) Find the values of $m$ and $c$. [4]

(b) The point $(p, 15)$ lies on the line. Find the value of $p$. [2]

(c) Find the equation of the line that is parallel to $y = mx + c$ and passes through the midpoint of the line segment joining $(1, 3.2)$ and $(5, 8.8)$. [1]

Worked Solution:

(a)

1. Use the gradient formula to find $m$. $m = \frac{8.8 - 3.2}{5 - 1} = \frac{5.6}{4} = 1.4$ [Apply gradient formula]
2. Substitute $m=1.4$ and the point $(1, 3.2)$ into $y = mx + c$. $3.2 = 1.4(1) + c$ $3.2 = 1.4 + c$ [Substitute values to find c]
3. Solve for $c$. $c = 3.2 - 1.4 = 1.8$

$\boxed{m = 1.4, c = 1.8}$

How to earn full marks: Show your working for both m and c, including the substitution step to find c.

(b)

1. Substitute $y = 15$ and the values of $m$ and $c$ into $y = mx + c$. $15 = 1.4p + 1.8$ [Substitute known variables]
2. Solve for $p$. $1.4p = 13.2$ $p = \frac{13.2}{1.4} = 9.42857...$

$\boxed{p = 9.43}$ (to 3 s.f.)

How to earn full marks: Substitute the given y value and the calculated m and c values into the equation, and give your answer to the specified degree of accuracy.

(c)

1. Find the midpoint of the line segment joining $(1, 3.2)$ and $(5, 8.8)$. Midpoint $= (\frac{1+5}{2}, \frac{3.2+8.8}{2}) = (3, 6)$ [Apply midpoint formula]
2. The parallel line has the same gradient, $m = 1.4$. So its equation is $y = 1.4x + c'$.
3. Substitute the midpoint $(3, 6)$ into $y = 1.4x + c'$. $6 = 1.4(3) + c'$ [Substitute to find new value of c]
4. Solve for $c'$. $c' = 6 - 4.2 = 1.8$
5. The equation of the parallel line is $y = 1.4x + 1.8$.

$\boxed{y = 1.4x + 1.8}$

How to earn full marks: Remember that parallel lines have the same gradient, and clearly show the calculation of the midpoint.

Common Pitfall: Remember that parallel lines have the same gradient. When finding the equation of a parallel line, only the y-intercept will be different. Also, pay attention to the required degree of accuracy and round your final answer appropriately.

Exam-Style Question 4 — Paper 2 (Calculator Allowed) [7 marks]

Question:

Variables $x$ and $y$ are related by the equation $y = B e^{kx}$, where $B$ and $k$ are constants. When $\ln y$ is plotted against $x$, a straight line is obtained. The gradient of this line is $0.35$ and the intercept on the vertical axis is $2.01$.

(a) Find the values of $B$ and $k$. Give your answers to 3 significant figures. [4]

(b) Estimate the value of $y$ when $x = 4$. [2]

(c) Estimate the value of $x$ when $y = 5$. [1]

Worked Solution:

(a)

1. Take the natural logarithm of both sides of the equation $y = B e^{kx}$. $\ln y = \ln (B e^{kx}) = \ln B + \ln e^{kx} = \ln B + kx$ [Apply logarithms and simplify]
2. Compare $\ln y = kx + \ln B$ with the equation of a straight line $y = mx + c$. The gradient is $k$ and the y-intercept is $\ln B$. [Recognise connection to straight line graph]
3. We are given that the gradient is $0.35$, so $k = 0.35$. [State k]
4. We are given that the y-intercept is $2.01$, so $\ln B = 2.01$. $B = e^{2.01} = 7.4658...$ [Find B]
5. Round to 3 significant figures.

$\boxed{B = 7.47, k = 0.350}$

How to earn full marks: Show the logarithmic manipulation clearly, and remember to use the exponential function to find B from ln B.

(b)

1. Substitute $x = 4$, $B = 7.47$, and $k = 0.350$ into $y = B e^{kx}$. $y = 7.47 e^{0.350 \times 4} = 7.47 e^{1.4} = 29.987...$ [Substitute and evaluate]
2. State estimate.

$\boxed{y = 30.0}$ (to 3 s.f.)

How to earn full marks: Substitute the values correctly into the original equation and round your answer to the specified number of significant figures.

(c)

1. Substitute $y = 5$, $B = 7.47$, and $k = 0.350$ into $y = B e^{kx}$. $5 = 7.47 e^{0.350 x}$ [Substitute and rearrange]
2. Solve for $x$. $e^{0.350 x} = \frac{5}{7.47} = 0.66934...$ $0.350 x = \ln 0.66934...$ $x = \frac{\ln 0.66934...}{0.350} = \frac{-0.40118...}{0.350} = -1.1462...$

$\boxed{x=-1.15}$ (to 3 s.f.)

How to earn full marks: Show each step of your working when rearranging and solving for x, including taking the natural logarithm.

Common Pitfall: Make sure you take the natural logarithm correctly when linearizing the equation. Also, remember to use the exponential function to find B after finding ln B. Always pay attention to the required number of significant figures in the final answer.