Module 16 of 46Lesson 1 of 12 · AA 2.1 — Equations of straight lines

Course map · Topic 2: Functions

2.1 Equations of straight lines

SL & HL11 worked questions

Straight lines are the simplest graphs, and they appear everywhere in AA: tangents and normals in calculus, perpendicular bisectors in geometry, linear models in context. You must be able to find a line's equation quickly and exactly, by hand.

✏️ Start with the questions. Try each one on paper, then open the solution and compare.

Stuck? Open the notes ↓

✏️ Practice questions

Try each question on paper first. Then open the solution and compare every step.

Question 1●○○ Warm-up✍️ By hand · Paper 1 · no calculator[4 marks]

The points

A(2,-1)

and

B(6,7)

lie on a straight line

L

(a) Find the gradient of $L$ .[2]
(b) Find the equation of $L$ in the form $y=mx+c$ .[2]

Show the full step-by-step solution

Step 1🧠 ThinkWhat is a gradient?

The gradient tells us how much

y

goes up when

x

goes up by 1. We find it from two points: “change in

y

” divided by “change in

x

”.

Part (a)

Step 1✍️ By handRise over run

m=\dfrac{7-(-1)}{6-2}

m=\dfrac{8}{4}=2

Part (b)

Step 1🧠 ThinkUse one point to find c

We know

y=2x+c

. Every point on the line fits this equation, so put in

A(2,-1)

x=2

y=-1

Step 2✍️ By handSubstitute A

-1=2(2)+c

c=-1-4=-5

y=2x-5

✅ Check it Check with

B

2(6)-5=7

. ✔ Both points fit.

Final answer (a)

m=2

(b)

y=2x-5

Question 2●○○ Warm-up✍️ By hand · Paper 1 · no calculator[4 marks]

The line

L

has equation

3x-4y+12=0

(a) Find the gradient of $L$ .[2]
(b) Find the coordinates of the points where $L$ meets the $x$ -axis and the $y$ -axis.[2]

Show the full step-by-step solution

Part (a)

Step 1🧠 ThinkMake y the subject

The gradient is the number in front of

x

only when the equation looks like

y=mx+c

. So first we get

y

on its own.

Step 2✍️ By handRearrange

4y=3x+12

y=\dfrac{3}{4}x+3

gradient

=\dfrac{3}{4}

Part (b)

Step 1🧠 ThinkAxis crossings

On the

y

-axis,

x=0

. On the

x

-axis,

y=0

. Put each zero into the original equation.

Step 2✍️ By handPut x = 0, then y = 0

x=0:\ -4y+12=0\Rightarrow y=3

, so

(0,3)

y=0:\ 3x+12=0\Rightarrow x=-4

, so

(-4,0)

Final answer (a)

\dfrac{3}{4}

(b)

(-4,0)

and

(0,3)

⚠️ Common mistake Saying the gradient is 3 (the coefficient of

x

) or

-\dfrac{3}{4}

. Rearrange carefully.

Question 3●○○ Warm-up✍️ By hand · Paper 1 · no calculator[4 marks]

The line

L_1

has equation

y=-\dfrac{2}{3}x+5

. The line

L_2

is perpendicular to

L_1

and passes through

(4,1)

. Find the equation of

L_2

, giving your answer in the form

ax+by+d=0

where

a,b,d\in\mathbb{Z}

Show the full step-by-step solution

Step 1🧠 ThinkPerpendicular means “flip and change the sign”

Perpendicular lines meet at a right angle. Their gradients multiply to

-1

. The gradient of

L_1

-\dfrac{2}{3}

; flip it to

\dfrac{3}{2}

and change the sign to

+\dfrac{3}{2}

Step 2✍️ By handGradient of L₂

m_2=-\dfrac{1}{-\frac{2}{3}}=\dfrac{3}{2}

Step 3✍️ By handPoint–gradient form, then clear the fraction

y-1=\dfrac{3}{2}(x-4)

2y-2=3x-12

(multiply both sides by 2)

3x-2y-10=0

A1A1

✅ Check it Put in

(4,1)

12-2-10=0

. ✔ And

\dfrac{3}{2}\times\left(-\dfrac{2}{3}\right)=-1

. ✔

Final answer

3x-2y-10=0

(or any integer multiple)

Question 4●●○ Standard✍️ By hand · Paper 1 · no calculator[3 marks]

Show that the points

A(-1,4)

B(2,-2)

and

C(5,-8)

lie on the same straight line.

Show the full step-by-step solution

Step 1🧠 ThinkWhat does “collinear” need?

Three points lie on one line if the slope from

A

B

is the same as the slope from

B

C

. Because both pieces share the point

B

, equal gradients means they are the same line, not two parallel lines.

Step 2✍️ By handTwo gradients

m_{AB}=\dfrac{-2-4}{2-(-1)}=\dfrac{-6}{3}=-2

M1A1

m_{BC}=\dfrac{-8-(-2)}{5-2}=\dfrac{-6}{3}=-2

m_{AB}=m_{BC}

and

B

is a common point, so

A

B

C

are collinearR1 AG

💡 Remember In a “show that” question the last line must state the conclusion in words. Mentioning the common point

B

is what earns the reasoning mark.

Final answer

m_{AB}=m_{BC}=-2

and the lines share

B

, so the points are collinear.

Question 5●●○ Standard✍️ By hand · Paper 1 · no calculator[5 marks]

Consider the lines

L_1:\ kx+2y=7

and

L_2:\ 3x-6y=1

, where

k\in\mathbb{R}

(a) Find the value of $k$ for which $L_1$ and $L_2$ are parallel.[2]
(b) Find the value of $k$ for which $L_1$ and $L_2$ are perpendicular.[3]

Show the full step-by-step solution

Step 1🧠 ThinkGet both gradients first

Write each line as

y=\ldots

. The gradient of

L_1

will contain

k

; then we set up an equation for

k

Step 2✍️ By handGradients

L_1:\ 2y=-kx+7\Rightarrow m_1=-\dfrac{k}{2}

L_2:\ 6y=3x-1\Rightarrow m_2=\dfrac{3}{6}=\dfrac{1}{2}

Part (a)

Step 1✍️ By handParallel: equal gradients

-\dfrac{k}{2}=\dfrac{1}{2}

k=-1

Part (b)

Step 1✍️ By handPerpendicular: product = −1

-\dfrac{k}{2}\times\dfrac{1}{2}=-1

-\dfrac{k}{4}=-1

k=4

✅ Check it With

k=4

m_1=-2

m_2=\dfrac{1}{2}

, product

-1

. ✔

Final answer (a)

k=-1

(b)

k=4

Question 6●●○ Standard✍️ By hand · Paper 1 · no calculator[5 marks]

The points

P(-2,3)

and

Q(4,-1)

are the ends of a line segment. Find the equation of the perpendicular bisector of

PQ

, giving your answer in the form

ax+by+d=0

where

a,b,d\in\mathbb{Z}

Show the full step-by-step solution

Step 1🧠 ThinkTwo words, two facts

“Bisector” = it cuts

PQ

exactly in half, so it passes through the midpoint. “Perpendicular” = it meets

PQ

at a right angle, so its gradient is the negative reciprocal of

m_{PQ}

Step 2✍️ By handMidpoint and gradient of PQ

M=\left(\dfrac{-2+4}{2},\dfrac{3+(-1)}{2}\right)=(1,1)

m_{PQ}=\dfrac{-1-3}{4-(-2)}=\dfrac{-4}{6}=-\dfrac{2}{3}

perpendicular gradient

=\dfrac{3}{2}

Step 3✍️ By handEquation through M(1, 1)

y-1=\dfrac{3}{2}(x-1)

2y-2=3x-3

3x-2y-1=0

A1A1

✅ Check it Every point on a perpendicular bisector is the same distance from

P

and

Q

. Try

(1,1)

: distance

^2

P

9+4=13

, to

Q

9+4=13

. ✔

Final answer

3x-2y-1=0

Question 7●●○ Standard✍️ + 🖩 · Paper 2[4 marks]

Find the coordinates of the point of intersection of the lines

2x+3y=12

and

y=0.8x-1.5

Show the full step-by-step solution

Step 1🧠 ThinkWhere two lines cross

The crossing point is on both lines, so its

x

and

y

fit both equations at the same time. On Paper 2 the quickest safe route is the GDC: graph both lines and use Intersection.

Step 2🖩 On the GDCRewrite as y = …, graph, then Intersection

ctrl+doc💻 Ctrl+I›tab💻 Tab›(12-2x)/3›enter💻 Enter›tab💻 Tab›0.8x-1.5›enter💻 Enter›menu💻 click menu on keypad›6: Analyze Graph›4: Intersection

The Graphs page only accepts

y=\ldots

, so

2x+3y=12

becomes

y=\dfrac{12-2x}{3}

. In Intersection click to the left of the crossing (lower bound), then to the right (upper bound).

menu → 6: Analyze Graph → 4: Intersection

Intersection: (3.75, 1.5)

Step 3🖩 On the GDCOr: simult( on a Calculator page

simult([[2,3][0.8,-1]],[[12][1.5]])›enter💻 Enter

📍 Where to find it on the TI-Nspire

simult(is inmenu💻 click menu on keypad›7: Matrix & Vector›6: Simultaneous

Write both as

ax+by=c

2x+3y=12

and

0.8x-y=1.5

. The first matrix holds the coefficients, the second the right-hand sides.

x = 3.75, y = 1.5

Step 4✍️ By handBy hand (to check)

2x+3(0.8x-1.5)=12

4.4x=16.5\Rightarrow x=3.75

y=0.8(3.75)-1.5=1.5

Final answer

(3.75,\ 1.5)

Question 8●●○ Standard🖩 GDC · Paper 2[5 marks]

Water drains from a tank at a constant rate. After 4 minutes the tank holds 1350 litres; after 10 minutes it holds 870 litres. Let

V

be the volume in litres after

t

minutes.

(a) Find an equation for $V$ in terms of $t$ .[3]
(b) Write down the volume at $t=0$ .[1]
(c) Find the time at which the tank is empty.[1]

Show the full step-by-step solution

Step 1🧠 ThinkConstant rate = straight line

A constant rate means a straight-line graph of

V

against

t

. The two readings are two points:

(4,1350)

and

(10,870)

Part (a)

Step 1✍️ By handGradient, then the equation

m=\dfrac{870-1350}{10-4}=\dfrac{-480}{6}=-80

M1A1

V-1350=-80(t-4)

V=-80t+1670

💡 Remember The gradient

-80

has a meaning: the tank loses 80 litres every minute.

Part (b)

Step 1✍️ By handThe intercept

V=1670

litresA1

Part (c)

Step 1🖩 On the GDCGraph V and use Zero

tab💻 Tab›-80x+1670›enter💻 Enter›menu💻 click menu on keypad›4: Window / Zoom›1: Window Settings…›menu💻 click menu on keypad›6: Analyze Graph›1: Zero

Set the window to fit the story:

x

(time) from 0 to 25,

y

(volume) from 0 to 1800. Then Zero: click left, then right of the crossing.

Step 2✍️ By handAnswer

t=20.875\approx20.9

minutesA1

Final answer (a)

V=-80t+1670

(b) 1670 litres (c)

t=20.875\approx20.9

minutes

Question 9●●● Exam-level✍️ By hand · Paper 1 · no calculator[9 marks]

The triangle

ABC

has vertices

A(1,5)

B(-3,-3)

and

C(5,3)

(a) Show that angle $BAC$ is a right angle.[3]
(b) Find the equation of the line through $C$ that is parallel to $AB$ .[2]
(c) Find the $x$ -coordinate where this line meets the $x$ -axis.[1]
(d) Find the exact area of triangle $ABC$ .[3]

Show the full step-by-step solution

Part (a)

Step 1🧠 ThinkRight angle at A

The angle at

A

is between the sides

AB

and

AC

. It is

90^\circ

exactly when these two sides are perpendicular: their gradients multiply to

-1

Step 2✍️ By handTwo gradients and their product

m_{AB}=\dfrac{-3-5}{-3-1}=\dfrac{-8}{-4}=2

m_{AC}=\dfrac{3-5}{5-1}=\dfrac{-2}{4}=-\dfrac{1}{2}

2\times\left(-\dfrac{1}{2}\right)=-1

, so

AB\perp AC

and

\angle BAC=90^\circ

R1 AG

Part (b)

Step 1✍️ By handSame gradient as AB, through C

y-3=2(x-5)

y=2x-7

Part (c)

Step 1✍️ By handPut y = 0

0=2x-7\Rightarrow x=\dfrac{7}{2}

Part (d)

Step 1🧠 ThinkRight-angled triangle: half base × height

Because the right angle is at

A

, the two sides

AB

and

AC

are the base and the height. Find their lengths with Pythagoras and keep surds exact.

Step 2✍️ By handLengths and area

AB=\sqrt{4^2+8^2}=\sqrt{80}=4\sqrt5

AC=\sqrt{4^2+2^2}=\sqrt{20}=2\sqrt5

Area

=\dfrac{1}{2}\times4\sqrt5\times2\sqrt5

=\dfrac{1}{2}\times8\times5=20

Final answer (a)

m_{AB}m_{AC}=2\times\left(-\dfrac12\right)=-1

(b)

y=2x-7

(c)

x=\dfrac{7}{2}

(d) 20 square units

Question 10●●● Exam-level✍️ + 🖩 · Paper 2[8 marks]

In Gaborone, Taxi A charges a fixed fee of P 12 plus P 6.50 per kilometre. Taxi B charges P 20 plus P 5.20 per kilometre. Let

C

be the cost in pula for a journey of

d

km.

(a) Write down an equation for $C$ for each taxi.[2]
(b) Find the distance for which both taxis cost the same, and that cost.[3]
(c) Thato travels 10 km. State which taxi is cheaper and by how much.[2]
(d) Interpret the gradient of the equation for Taxi B.[1]

Show the full step-by-step solution

Part (a)

Step 1🧠 ThinkFixed fee = intercept, per-km price = gradient

The fixed fee is paid even for 0 km, so it is the

C

-intercept. Each extra kilometre adds the same amount, so that amount is the gradient.

Step 2✍️ By handTwo models

C=6.5d+12

C=5.2d+20

Part (b)

Step 1🖩 On the GDCGraph both and find the Intersection

tab💻 Tab›6.5x+12›enter💻 Enter›tab💻 Tab›5.2x+20›enter💻 Enter›menu💻 click menu on keypad›6: Analyze Graph›4: Intersection

Window:

x

from 0 to 12,

y

from 0 to 100.

Step 2✍️ By handAnswer with units

d=6.15385\ldots\approx6.15

kmA1

C=\text{P }52

Check by hand:

6.5d+12=5.2d+20\Rightarrow1.3d=8\Rightarrow d=\dfrac{80}{13}

Part (c)

Step 1✍️ By handSubstitute d = 10

6.5(10)+12=77

5.2(10)+20=72

Taxi B is cheaper, by P 5A1A1

Part (d)

Step 1✍️ By handMeaning of 5.2

Taxi B charges P 5.20 for each extra kilometre.A1

Final answer (a) A:

C=6.5d+12

; B:

C=5.2d+20

(b)

d\approx6.15

km, cost P 52 (c) Taxi B, by P 5 (d) P 5.20 per km

Question 11●●● Exam-level✍️ By hand · Paper 1 · no calculator[10 marks]

The line

L_1

passes through the points

(-2,5)

and

(4,-4)

. The line

L_2

has equation

2x-3y+k=0

and passes through the point

(5,1)

(a) Find the equation of $L_1$ in the form $ax+by+d=0$ , where $a,b,d\in\mathbb{Z}$ .[3]
(b) Find the value of $k$ .[1]
(c) Show that $L_1$ and $L_2$ are perpendicular.[2]
(d) Find the point of intersection of $L_1$ and $L_2$ .[2]
(e) Find the area of the triangle formed by $L_1$ , $L_2$ and the $y$ -axis.[2]

Show the full step-by-step solution

Part (a)

Step 1✍️ By handGradient and equation

m_1=\dfrac{-4-5}{4-(-2)}=\dfrac{-9}{6}=-\dfrac{3}{2}

y-5=-\dfrac{3}{2}(x+2)

2y-10=-3x-6

3x+2y-4=0

Part (b)

Step 1✍️ By handThe point (5, 1) fits L₂

2(5)-3(1)+k=0

k=-7

Part (c)

Step 1✍️ By handGradient of L₂ and the product

3y=2x-7\Rightarrow m_2=\dfrac{2}{3}

m_1m_2=-\dfrac{3}{2}\times\dfrac{2}{3}=-1

, so

L_1\perp L_2

R1 AG

Part (d)

Step 1🧠 ThinkElimination

We have

3x+2y=4

and

2x-3y=7

. Make the

y

-terms cancel: multiply the first by 3 and the second by 2, then add.

Step 2✍️ By handSolve simultaneously

9x+6y=12

and

4x-6y=14

13x=26\Rightarrow x=2

3(2)+2y=4\Rightarrow y=-1

; point

(2,-1)

Part (e)

Step 1🧠 ThinkDraw it in your head

Two corners of the triangle are on the

y

-axis (the

y

-intercepts of the lines). The third corner is

(2,-1)

. The side on the

y

-axis is the base; the height is the horizontal distance from the

y

-axis to

(2,-1)

, which is 2.

Step 2✍️ By handIntercepts and area

L_1:\ x=0\Rightarrow y=2

;

L_2:\ x=0\Rightarrow y=-\dfrac{7}{3}

base

=2-\left(-\dfrac{7}{3}\right)=\dfrac{13}{3}

area

=\dfrac{1}{2}\times\dfrac{13}{3}\times2=\dfrac{13}{3}

Final answer (a)

3x+2y-4=0

(b)

k=-7

(c)

m_1m_2=-1

(d)

(2,-1)

(e)

\dfrac{13}{3}

🆘 Stuck on a question? Open the notes, GDC toolkit and common mistakes

💻 Using the TI-Nspire CX Student Software on a laptop

The screens in these notes are drawn exactly as the Student Software shows them. The keypad on the left of the software works like the handheld — click its keys with the mouse.
menu: click menu on the on-screen keypad (or Documents Toolbox ▸ 🔧). The menu numbers and letters are the same as in these notes.
enter = Enter · tab = Tab (next box) · esc = Esc · del = Backspace.
Powers: type ^ (Shift+6), the power, then press → to come back down before typing more. x² = ^2 then →.
Fraction: type / · √: type sqrt( · π: type pi · eˣ: type exp( · store: type :=. For EE (×10ⁿ) and i, click them on the keypad.
New page: Ctrl+I (or the Insert menu) · change page: Ctrl+← / Ctrl+→ · graph entry line: Ctrl+G · var: Ctrl+L.
Radian / Degree: click Settings in the status bar at the bottom of the window. The mode (RAD or DEG) shows in the top bar.
IB exam mode (Press-to-Test): everything in these notes works on the non-CAS TI-Nspire with the IB exam settings. In exam mode there are no dot-product, cross-product, unit-vector or eigenvalue keys and no Diff Eq graphs — the notes show the allowed way (type the components, use polyRoots on the characteristic equation, do Euler steps on a Calculator page).
If you type nDeriv(, the software shows it as centralDiff( — it is the same command.

🎯 What you must know

Gradient through $(x_1,y_1)$ and $(x_2,y_2)$ : $m=\dfrac{y_2-y_1}{x_2-x_1}$ = rise over run. Positive: up to the right; negative: down; zero: horizontal.
Three forms of a line: gradient–intercept $y=mx+c$ ; general $ax+by+d=0$ ; point–gradient $y-y_1=m(x-x_1)$ .
From $ax+by+d=0$ : rearrange to $y=-\dfrac{a}{b}x-\dfrac{d}{b}$ , so the gradient is $-\dfrac{a}{b}$ .
Parallel lines have equal gradients: $m_1=m_2$ . Perpendicular lines: $m_1\times m_2=-1$ , i.e. $m_2=-\dfrac{1}{m_1}$ (flip and change sign).
Vertical line: $x=k$ (gradient undefined). Horizontal line: $y=k$ (gradient 0).
“Give your answer in the form $ax+by+d=0$ where $a,b,d\in\mathbb{Z}$ ” means: clear all fractions, bring everything to one side.
Intersection of two lines: solve simultaneously (substitution or elimination) by hand; on Paper 2 use Graphs → Intersection or simult(.

Gradient

m=\dfrac{y_2-y_1}{x_2-x_1}

📘 In the formula booklet

Equations of a line

y=mx+c

;\ \ ax+by+d=0;\ \ y-y_1=m(x-x_1)\)

📘 In the formula booklet

Perpendicular gradients

m_1\times m_2=-1

🧠 Learn this one

🖩 Your GDC toolkit for this subtopic

Follow the steps on your own TI-Nspire CX II. The screens show what you should see.

Graph two lines (each written as y = …)

ctrl+doc💻 Ctrl+I›tab💻 Tab›(12-2x)/3›enter💻 Enter

Press ctrl then doc to add a new page.
Press tab to move to the next box.
Type (12-2x)/3.
Press enter.

Intersection: (3.75, 1.5)

Worked example: Question 7 — Rewrite as y = …, graph, then Intersection

The Graphs page only accepts

y=\ldots

, so

2x+3y=12

becomes

y=\dfrac{12-2x}{3}

. In Intersection click to the left of the crossing (lower bound), then to the right (upper bound).

Find where two lines cross

menu💻 click menu on keypad›6: Analyze Graph›4: Intersection

Start on a Graphs page (ctrl doc adds one).
Press menu.
Choose 6: Analyze Graph.
Choose 4: Intersection.
Move the pointer to the LEFT of the crossing point and press click, then to the RIGHT and press click. The intersection point and its coordinates appear.

menu → 6: Analyze Graph → 4: Intersection

Intersection: (3.75, 1.5)

Worked example: Question 7 — Rewrite as y = …, graph, then Intersection

The Graphs page only accepts

y=\ldots

, so

2x+3y=12

becomes

y=\dfrac{12-2x}{3}

. In Intersection click to the left of the crossing (lower bound), then to the right (upper bound).

Choose a window that fits a context

menu💻 click menu on keypad›4: Window / Zoom›1: Window Settings…

Start on a Graphs page (ctrl doc adds one).
Press menu.
Choose 4: Window / Zoom.
Choose 1: Window Settings….
Type the XMin, XMax, YMin and YMax values (press tab to move between boxes), then press enter. The graph redraws in the new window.

The menu path

Worked example: Question 8 — Graph V and use Zero

Set the window to fit the story:

x

(time) from 0 to 25,

y

(volume) from 0 to 1800. Then Zero: click left, then right of the crossing.

Solve two linear equations on a Calculator page

simult([[2,3][0.8,-1]],[[12][1.5]])›enter💻 Enter

📍 Where to find it on the TI-Nspire

simult(is inmenu💻 click menu on keypad›7: Matrix & Vector›6: Simultaneous

Type simult([[2,3][0.8,-1]],[[12][1.5]]).
Press enter.

x = 3.75, y = 1.5

Worked example: Question 7 — Or: simult( on a Calculator page

Write both as

ax+by=c

2x+3y=12

and

0.8x-y=1.5

. The first matrix holds the coefficients, the second the right-hand sides.

⚠️ Where students lose marks

Subtracting coordinates in a different order on top and bottom: $\dfrac{y_2-y_1}{x_1-x_2}$ gives the wrong sign.
Reading the gradient of $3x-4y+12=0$ as 3. Rearrange to $y=\ldots$ first: the gradient is $\dfrac{3}{4}$ .
Perpendicular gradient: only changing the sign ( $2\to-2$ ) or only flipping ( $2\to\dfrac12$ ). You must do both: $2\to-\dfrac{1}{2}$ .
Leaving fractions in the answer when the question asks for $ax+by+d=0$ with integer coefficients.
On the GDC: forgetting that a line like $2x+3y=12$ must be rewritten as $y=\dfrac{12-2x}{3}$ before it can be graphed.

IB DP Mathematics: Analysis and Approaches (AA) — SL and HL AA 2.1 — Equations of straight lines

IB DP Mathematics: Analysis and Approaches (AA) — SL and HL

2.1 Equations of straight lines

✏️ Practice questions

🎯 What you must know

🖩 Your GDC toolkit for this subtopic

⚠️ Where students lose marks