IB Maths AA SL · Unit 2: Functions
IB Maths AA SL Equations of Straight Lines Questions
Exam-style IB Maths AA SL equations of straight lines questions with worked solutions. Start with the 3 fully worked examples below — each is solved step by step the way an IB examiner expects — then try the practice questions and check your working against the mark scheme.
- 19 questions
- Paper 1: 13
- Paper 2: 6
- 5 easy
- 6 medium
- 4 hard
- 3 very hard
- 1 starter
- 3 worked examples
Practise Equations of Straight Lines questions →
AA SL formula booklet
What's examined in AA SL equations of straight lines
The question bank covers these equations of straight lines question types (number of questions in brackets):
- Parallel and perpendicular lines (7)
- Line equations and properties (6)
- Midpoints and intersections (6)
Key formulas
- Equation of a straight line
- \(y - y_1 = m(x - x_1)\)
- Gradient
- \(m = \dfrac{y_2 - y_1}{x_2 - x_1}\)
In the same notation as the IB formula booklet. All AA SL formulas →
Equations of Straight Lines worked examples
Worked example 1: Finding gradients and intercepts · easy
The equation of a line $L_1$ is given by $3x + 4y - 12 = 0$. Find the exact gradient of $L_1$ and the coordinates of its $x$ and $y$-intercepts.
1. Rearrange the equation into the standard $y = mx + c$ form: $4y = -3x + 12$.
2. Divide every term by $4$ to isolate $y$: $y = -\frac{3}{4}x + 3$.
3. State the exact gradient, which is the coefficient of $x$: $\mathbf{m = -\frac{3}{4}}$.
4. Identify the $y$-intercept from the constant $c = 3$, giving coordinates $\mathbf{(0, 3)}$.
5. Calculate the $x$-intercept by setting $y = 0$: $3x + 4(0) - 12 = 0 \implies x = 4$, giving $\mathbf{(4, 0)}$.
Examiner tip: Always remember to divide every single term by the coefficient of $y$ when rearranging into $y = mx + c$. Forgetting to divide the constant term is a frequent careless error that ruins the $y$-intercept.
Worked example 2: Perpendicular bisectors · medium
Find the equation of the perpendicular bisector of the line segment joining the points $A(-2, 5)$ and $B(4, -7)$. Give your answer in the form $ax + by + d = 0$, where $a, b, d \in \mathbb{Z}$.
1. Calculate the midpoint $M$: $M = \left(\frac{-2 + 4}{2}, \frac{5 - 7}{2}\right) = (1, -1)$.
2. Calculate the gradient of $AB$: $m_{AB} = \frac{-7 - 5}{4 - (-2)} = \frac{-12}{6} = -2$.
3. Determine the perpendicular gradient: $m_{\perp} = \frac{1}{2}$.
4. Substitute $M$ and $m_\perp$ into point-slope form: $y + 1 = \frac{1}{2}(x - 1)$.
5. Rearrange into integer form: $2y + 2 = x - 1 \implies \mathbf{x - 2y - 3 = 0}$.
Examiner tip: Students frequently forget to use the midpoint for the perpendicular bisector, mistakenly substituting one of the original endpoints ($A$ or $B$) into the line equation instead.
Worked example 3: Intersection and perpendicular lines · hard
Line $L_1$ has the equation $5y - 2x + 1 = 0$. Point $A$ lies on $L_1$ and has a $y$-coordinate of $3$. A second line $L_2$ is perpendicular to $L_1$ and passes through $A$. Find the equation of $L_2$ in the form $y = mx + c$.
1. Substitute $y = 3$ into $L_1$: $5(3) - 2x + 1 = 0 \implies 2x = 16 \implies x = 8$. So $A = (8, 3)$.
2. Rearrange $L_1$: $y = \frac{2}{5}x - \frac{1}{5}$, so $m_1 = \frac{2}{5}$.
3. Perpendicular gradient: $m_2 = -\frac{5}{2} = -2.5$.
4. Substitute $A$ and $m_2$ into point-slope: $y - 3 = -2.5(x - 8)$.
5. Simplify: $\mathbf{y = -2.5x + 23}$.
Examiner tip: When finding the perpendicular gradient, remember to both flip the fraction and change the sign (the negative reciprocal). Doing only one of these steps is a classic Paper 1 trap.
Try these IB Maths AA SL equations of straight lines questions
Three questions from the bank, easiest first. Mark schemes and AI marking of your written working are in the practice area.
Question 1 · easy · 3 marks · Paper 1
The coordinates of point \(A\) are \((1, 3)\) and the coordinates of point \(B\) are \((4, 9)\).
Find the equation of the line passing through \(A\) and \(B\). Give your answer in the form \(y = mx + c\).
Question 2 · medium · 5 marks · Paper 2
Taxi Company A charges a fixed fee of \(\text{€}4.00\) plus \(\text{€}1.50\) per kilometre travelled. Taxi Company B charges a fixed fee of \(\text{€}2.50\) plus \(\text{€}2.00\) per kilometre travelled.
Defining suitable variables, write down the linear equations for the cost of travelling with Company A and Company B.
Use your graphic display calculator to determine the distance for which both taxi companies cost exactly the same amount.
Question 3 · hard · 4 marks · Paper 1
The line passing through the points \(A(k, 4)\) and \(B(6, k)\) has a gradient of \(3\). Find the exact value of \(k\).
All 19 equations of straight lines questions with mark schemes →
FAQ
How many IB Maths AA SL equations of straight lines questions are there?
There are 19 exam-style equations of straight lines questions in the AA SL question bank (Paper 1: 13 · Paper 2: 6), graded 5 easy, 6 medium, 4 hard, 3 very hard, 1 starter. Every question has a full IB-style mark scheme (M, A and R marks).
Is equations of straight lines on Paper 1 or Paper 2?
Both. In the bank, Paper 1: 13 · Paper 2: 6. Paper 1 is non-calculator, so practise the exact-value algebra as well as the GDC methods.
Where can I get the mark schemes?
Open the AA SL Unit 2 practice page: every question has a step-by-step IB-style mark scheme, and you can photograph your working for instant AI marking. The worked examples on this page are free.
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