Module 16 of 47Lesson 1 of 6 · AI 2.1 — Equations of straight lines

Course map · Topic 2: Functions

2.1 Equations of straight lines

SL & HL12 worked questions

A straight line is the simplest model there is: something that goes up (or down) by the same amount every step. You must be able to find its gradient, write its equation in three different forms, and spot parallel and perpendicular lines.

✏️ 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[4 marks]

The points

A(2, 3)

and

B(6, 11)

lie on the 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

Part (a)

Step 1🧠 ThinkGradient = rise ÷ run

Going from

A

B

, how far do we go up (rise) and how far do we go across (run)? Up: from 3 to 11. Across: from 2 to 6. Always take the same point first, top and bottom.

Step 2✍️ By handUse the gradient formula

m = \dfrac{11 - 3}{6 - 2}

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

Part (b)

Step 1🧠 ThinkFind c using one point

We know

y = 2x + c

. The point

A(2, 3)

is on the line, so putting

x = 2

must give

y = 3

. That lets us find

c

Step 2✍️ By handSubstitute A(2, 3)

3 = 2\times2 + c

3 = 4 + c \Rightarrow c = -1

y = 2x - 1

Step 3🖩 On the GDCCheck: graph it and see it goes through both points

tab💻 Tab›2x-1›enter💻 Enter

On a Graphs page, type the right-hand side on the entry line. The line should pass through

(2, 3)

and

(6, 11)

✅ Check it Point

B

2\times6 - 1 = 11

. ✔

Final answer (a)

m = 2

(b)

y = 2x - 1

Question 2●○○ Warm-up✍️ By hand · Paper 1[3 marks]

The line

L

has equation

3x + 2y - 12 = 0

. Write down the gradient of

L

and the coordinates of the point where

L

crosses the

y

-axis.

Show the full step-by-step solution

Step 1🧠 ThinkGet y on its own

In the form

y = mx + c

we can simply read the gradient

m

and the

y

-intercept

c

. So first move everything except

2y

to the other side, then divide by 2.

Step 2✍️ By handRearrange

2y = -3x + 12

y = -\dfrac{3}{2}x + 6

gradient

m = -\dfrac{3}{2}

y

-intercept:

(0, 6)

Step 3🖩 On the GDCSee it on the graph

The line falls 3 units for every 2 units to the right, and crosses the

y

-axis at 6.

Final answer Gradient

-\dfrac{3}{2}

(

= -1.5

); crosses the

y

-axis at

(0, 6)

⚠️ Common mistake Saying the gradient is 3 (or

\dfrac{3}{2}

). The sign changes when

3x

moves across the equals sign.

Question 3●○○ Warm-up✍️ By hand · Paper 1[3 marks]

A line has gradient

-4

and passes through the point

(1, 5)

. Find its equation, 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🧠 ThinkOne point + gradient → point–gradient form

We are given exactly what

y - y_1 = m(x - x_1)

needs:

m = -4

x_1 = 1

y_1 = 5

. “

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

” just means a, b, d must be whole numbers (integers).

Step 2✍️ By handSubstitute, expand, tidy up

y - 5 = -4(x - 1)

y - 5 = -4x + 4

y = -4x + 9

4x + y - 9 = 0

A1A1

✅ Check it Put in

(1, 5)

4 + 5 - 9 = 0

. ✔ Gradient from

4x + y - 9 = 0

y = -4x + 9

, so

m = -4

. ✔

Final answer

4x + y - 9 = 0

(any multiple, e.g.

-4x - y + 9 = 0

, is also correct)

Question 4●●○ Standard✍️ By hand · Paper 1[3 marks]

Find the equation of the line that is parallel to

y = 3x - 2

and passes through the point

(-1, 4)

. Give your answer in the form

y = mx + c

Show the full step-by-step solution

Step 1🧠 ThinkParallel means same steepness

Parallel lines never meet, like railway tracks, so they climb at the same rate. The new line therefore also has gradient

m = 3

. Only the intercept is different.

Step 2✍️ By handUse the point (−1, 4)

m = 3

y - 4 = 3(x - (-1)) = 3(x + 1)

y - 4 = 3x + 3

y = 3x + 7

Step 3🖩 On the GDCBoth lines on one screen

💡 Remember Careful with the double minus:

x - (-1) = x + 1

Final answer

y = 3x + 7

Question 5●●○ Standard✍️ By hand · Paper 1[5 marks]

The line

L_1

has equation

2x - 5y + 10 = 0

. The line

L_2

is perpendicular to

L_1

and passes through the point

P(4, 1)

. Find the equation of

L_2

in the form

ax + by + d = 0

, where

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

Show the full step-by-step solution

Step 1🧠 ThinkPlan

Step 1: find the gradient of

L_1

(make

y

the subject). Step 2: the perpendicular gradient is “flip the fraction and change the sign”. Step 3: use the point

P

Step 2✍️ By handGradient of L₁

5y = 2x + 10

y = \dfrac{2}{5}x + 2 \Rightarrow m_1 = \dfrac{2}{5}

Step 3✍️ By handPerpendicular gradient

Flip

\dfrac{2}{5}

to get

\dfrac{5}{2}

, then change the sign.

m_2 = -\dfrac{5}{2}

Check:

\dfrac{2}{5}\times\left(-\dfrac{5}{2}\right) = -1

✔

Step 4✍️ By handEquation through P(4, 1)

y - 1 = -\dfrac{5}{2}(x - 4)

Multiply by 2:

2y - 2 = -5(x - 4) = -5x + 20

5x + 2y - 22 = 0

A1A1

Step 5🖩 On the GDCCheck the right angle on the GDC

menu💻 click menu on keypad›4: Window / Zoom›B: Zoom – Square

Graph

f1(x) = 0.4x + 2

and

f2(x) = -2.5x + 11

. On a normal window the angle can look wrong because the axes have different scales. Zoom – Square makes 1 unit across equal 1 unit up, and the 90° angle appears.

Final answer

5x + 2y - 22 = 0

⚠️ Common mistake Using

m_2 = \dfrac{5}{2}

(forgot the sign) or

m_2 = -\dfrac{2}{5}

(forgot to flip).

Question 6●●○ Standard✍️ + 🖩 · Paper 1[4 marks]

Find the coordinates of the point where the lines

y = 2x - 3

and

x + 3y = 12

intersect.

Show the full step-by-step solution

Step 1🧠 ThinkIntersect = the point that is on both lines

At the crossing point, the

x

and

y

values satisfy both equations at once. We can solve by hand (substitution) or let the GDC find the crossing.

Step 2✍️ By handBy hand: substitute y = 2x − 3 into the second equation

x + 3(2x - 3) = 12

x + 6x - 9 = 12

7x = 21 \Rightarrow x = 3

y = 2\times3 - 3 = 3

Step 3🖩 On the GDCWith the GDC: rearrange, graph, then Intersection

tab💻 Tab›2x-3›enter💻 Enter›tab💻 Tab›(12-x)/3›enter💻 Enter›menu💻 click menu on keypad›6: Analyze Graph›4: Intersection

The GDC needs

y = \ldots

, so rewrite

x + 3y = 12

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

. Then choose Intersection, click to the left of the crossing (lower bound) and to the right (upper bound).

menu → 6: Analyze Graph → 4: Intersection

Final answer

(3, 3)

Question 7●●○ Standard✍️ + 🖩 · Paper 1[6 marks]

The cost of a taxi ride in Gaborone is a linear function of the distance travelled. A ride of 8 km costs P 52 and a ride of 15 km costs P 83.50. Let

C

be the cost in pula and

d

the distance in km.

(a) Find an equation for $C$ in terms of $d$ .[3]
(b) Interpret, in context, the gradient and the $C$ -intercept.[2]
(c) Lesedi has P 70. Find the greatest distance she can travel.[1]

Show the full step-by-step solution

Part (a)

Step 1🧠 ThinkTwo facts = two points

“8 km costs P 52” is the point

(8, 52)

. “15 km costs P 83.50” is the point

(15, 83.5)

. A line through two points: gradient first, then the intercept.

Step 2✍️ By handGradient, then intercept

m = \dfrac{83.5 - 52}{15 - 8} = \dfrac{31.5}{7} = 4.5

M1A1

52 = 4.5\times8 + c \Rightarrow c = 52 - 36 = 16

C = 4.5d + 16

Part (b)

Step 1✍️ By handSay what the numbers mean, with units

Gradient 4.5: each extra km costs P 4.50.A1

Intercept 16: a fixed charge of P 16 before any distance (the starting fee).A1

Part (c)

Step 1✍️ By handSet C = 70 and solve

4.5d + 16 = 70

4.5d = 54

d = 12

kmA1

Step 2🖩 On the GDCOr: intersect with the line y = 70

Graph

f1(x) = 4.5x + 16

and

f2(x) = 70

, then Intersection. Use a window that fits the context:

x

from 0 to 20,

y

from 0 to 110.

Final answer (a)

C = 4.5d + 16

(b) P 4.50 per km; P 16 fixed starting charge (c) 12 km

⚠️ Common mistake In (b), writing “the gradient is 4.5” without saying what it means. Interpret = give the meaning with units.

Question 8●●○ Standard✍️ By hand · Paper 1[3 marks]

The line through the points

(k, 7)

and

(3, -1)

has gradient

4

. Find the value of

k

Show the full step-by-step solution

Step 1🧠 ThinkWrite the gradient formula with the unknown in it

We don't know one

x

-coordinate, but we know the answer to the gradient calculation. So write the formula, set it equal to 4, and solve for

k

Step 2✍️ By handForm an equation and solve

\dfrac{7 - (-1)}{k - 3} = 4

\dfrac{8}{k - 3} = 4

8 = 4(k - 3) = 4k - 12

4k = 20 \Rightarrow k = 5

✅ Check it

\dfrac{7 - (-1)}{5 - 3} = \dfrac{8}{2} = 4

. ✔

Final answer

k = 5

Question 9●●○ Standard✍️ By hand · Paper 1[3 marks]

The line

kx + 3y = 6

is perpendicular to the line

y = 1.5x + 2

. Find the value of

k

Show the full step-by-step solution

Step 1🧠 ThinkFind both gradients, then multiply to −1

The second line already shows its gradient:

1.5 = \dfrac{3}{2}

. For the first line we must make

y

the subject — its gradient will contain

k

Step 2✍️ By handGradient of kx + 3y = 6

3y = -kx + 6

y = -\dfrac{k}{3}x + 2 \Rightarrow m_1 = -\dfrac{k}{3}

Step 3✍️ By handPerpendicular condition

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

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

k = 2

✅ Check it With

k = 2

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

, and

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

. ✔

Final answer

k = 2

Question 10●●● Exam-level✍️ By hand · Paper 1[7 marks]

The points

A(-2, 1)

and

B(4, 5)

are shown on a map grid.

(a) Find the gradient of $AB$ .[2]
(b) Find the coordinates of $M$ , the midpoint of $AB$ .[1]
(c) Find the equation of the line through $M$ that is perpendicular to $AB$ . Give your answer in the form $ax + by + d = 0$ .[3]
(d) This line crosses the $x$ -axis at $Q$ . Find the coordinates of $Q$ .[1]

Show the full step-by-step solution

Part (a)

Step 1✍️ By handGradient formula

m_{AB} = \dfrac{5 - 1}{4 - (-2)} = \dfrac{4}{6}

m_{AB} = \dfrac{2}{3}

Part (b)

Step 1🧠 ThinkMidpoint = the average of the coordinates

Halfway between the

x

-values, and halfway between the

y

-values.

Step 2✍️ By handAverage the coordinates

M = \left(\dfrac{-2 + 4}{2}, \dfrac{1 + 5}{2}\right) = (1, 3)

Part (c)

Step 1✍️ By handPerpendicular gradient, then point M

m = -\dfrac{3}{2}

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

2y - 6 = -3x + 3

3x + 2y - 9 = 0

Part (d)

Step 1🧠 ThinkOn the x-axis, y = 0

Every point on the

x

-axis has

y = 0

. Put

y = 0

into the equation.

Step 2✍️ By handSet y = 0

3x + 0 - 9 = 0

x = 3 \Rightarrow Q(3, 0)

Step 3🖩 On the GDCPicture of the whole question

The red line (f2) is the perpendicular bisector: it cuts

AB

in half at a right angle (Zoom – Square used).

Final answer (a)

\dfrac{2}{3}

(b)

M(1, 3)

(c)

3x + 2y - 9 = 0

(d)

Q(3, 0)

Question 11●●● Exam-level✍️ + 🖩 · Paper 2[7 marks]

Two water tanks are filled and emptied at a farm near Maun. Tank A starts with 1200 litres and loses water at a constant 75 litres per hour. Tank B starts with 300 litres and gains water at a constant 45 litres per hour. Let

V

be the volume in litres after

t

hours.

(a) Write down an equation for the volume in each tank.[2]
(b) Find the time when the two tanks contain the same volume, and find this volume.[3]
(c) Find the time it takes for tank A to become empty.[2]

Show the full step-by-step solution

Part (a)

Step 1🧠 ThinkStarting value + rate × time

The starting amount is the intercept

c

. The rate is the gradient

m

: negative when the water goes down, positive when it goes up.

Step 2✍️ By handWrite the two models

V_A = 1200 - 75t

V_B = 300 + 45t

Part (b)

Step 1✍️ By handSame volume: set them equal

1200 - 75t = 300 + 45t

900 = 120t

t = 7.5

hoursA1

V = 300 + 45\times7.5 = 637.5

litresA1

Step 2🖩 On the GDCSame answer from the graphs

menu💻 click menu on keypad›4: Window / Zoom›1: Window Settings…›menu💻 click menu on keypad›6: Analyze Graph›4: Intersection

Set the window to fit the story:

x

(time) 0 to 18,

y

(volume) 0 to 1300. Then use Intersection.

Part (c)

Step 1✍️ By handEmpty means V = 0

1200 - 75t = 0

t = \dfrac{1200}{75} = 16

hoursA1

Final answer (a)

V_A = 1200 - 75t

V_B = 300 + 45t

(b) after 7.5 hours, 637.5 litres (c) 16 hours

Question 12●●● Exam-level✍️ + 🖩 · Paper 2[9 marks]

The line

L_1

has equation

3x - 4y + 12 = 0

. It meets the

x

-axis at

P

and the

y

-axis at

Q

O

is the origin.

(a) Find the coordinates of $P$ and of $Q$ .[2]
(b) Find the area of triangle $OPQ$ .[1]
(c) The line $L_2$ passes through $O$ and is perpendicular to $L_1$ . Find the equation of $L_2$ .[2]
(d) Find the coordinates of the point $R$ where $L_1$ and $L_2$ meet.[2]
(e) Hence find the shortest distance from $O$ to $L_1$ .[2]

Show the full step-by-step solution

Part (a)

Step 1🧠 ThinkAxis crossings

On the

x

-axis

y = 0

; on the

y

-axis

x = 0

. Put each one into the equation.

Step 2✍️ By handTwo substitutions

y = 0: 3x + 12 = 0 \Rightarrow x = -4,\ P(-4, 0)

x = 0: -4y + 12 = 0 \Rightarrow y = 3,\ Q(0, 3)

Part (b)

Step 1✍️ By handRight-angled triangle: half × base × height

The triangle has its right angle at

O

. Base

OP = 4

, height

OQ = 3

(lengths are positive).

Area

= \dfrac{1}{2}\times4\times3 = 6

Part (c)

Step 1✍️ By handGradient of L₁, then the perpendicular

4y = 3x + 12 \Rightarrow y = \dfrac{3}{4}x + 3

m_2 = -\dfrac{4}{3}

Through the origin, so

c = 0

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

Part (d)

Step 1🖩 On the GDCIntersection on the GDC

tab💻 Tab›0.75x+3›enter💻 Enter›tab💻 Tab›-4x/3›enter💻 Enter›menu💻 click menu on keypad›6: Analyze Graph›4: Intersection

Use Zoom – Square so the right angle at

R

is visible.

Step 2✍️ By handOr by hand

3x - 4\left(-\dfrac{4}{3}x\right) + 12 = 0 \Rightarrow \dfrac{25}{3}x = -12

x = -1.44,\ y = -\dfrac{4}{3}\times(-1.44) = 1.92

Part (e)

Step 1🧠 ThinkWhy “hence”?

The shortest way from a point to a line is the path that meets the line at 90°. That path is exactly

OR

. So the distance we want is the length

OR

Step 2✍️ By handDistance from O(0, 0) to R(−1.44, 1.92)

OR = \sqrt{(-1.44)^2 + 1.92^2}

OR = \sqrt{5.76} = 2.4

✅ Check it Area of

OPQ

another way:

\dfrac{1}{2}\times PQ\times OR = \dfrac{1}{2}\times5\times2.4 = 6

, the same as (b). ✔

Final answer (a)

P(-4, 0)

Q(0, 3)

(b) 6 (c)

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

(d)

R(-1.44, 1.92)

(e) 2.4

🆘 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 $m$ = how steep the line is = $\dfrac{\text{rise}}{\text{run}} = \dfrac{y_2-y_1}{x_2-x_1}$ . Positive: goes up to the right. Negative: goes down. Zero: flat (horizontal).
Gradient–intercept form $y = mx + c$ : $m$ is the gradient and $c$ is the $y$ -intercept (where the line crosses the $y$ -axis).
General form $ax + by + d = 0$ : everything on one side, usually with whole numbers. To read the gradient, rearrange to $y = \ldots$ first.
Point–gradient form $y - y_1 = m(x - x_1)$ : use it when you know the gradient and one point $(x_1, y_1)$ .
Parallel lines have the same gradient: $m_1 = m_2$ .
Perpendicular lines (meet at 90°) have gradients that multiply to $-1$ : $m_1 \times m_2 = -1$ . So flip the fraction and change the sign: $\dfrac{2}{3} \to -\dfrac{3}{2}$ .
In a context, the gradient is a rate (e.g. pula per km) and the $y$ -intercept is the starting value (when $x = 0$ ).

Gradient

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

📘 In the formula booklet

Gradient–intercept form

y = mx + c

📘 In the formula booklet

General form

ax + by + d = 0

📘 In the formula booklet

Point–gradient form

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.

Draw a line on a Graphs page: show the entry line, type the right-hand side, press enter

tab💻 Tab›2x-3›enter💻 Enter

Press tab to move to the next box.
Type 2x-3.
Press enter.

Worked example: Question 1 — Check: graph it and see it goes through both points

On a Graphs page, type the right-hand side on the entry line. The line should pass through

(2, 3)

and

(6, 11)

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

Worked example: Question 6 — With the GDC: rearrange, graph, then Intersection

The GDC needs

y = \ldots

, so rewrite

x + 3y = 12

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

. Then choose Intersection, click to the left of the crossing (lower bound) and to the right (upper bound).

Make right angles look like right angles (equal scales)

menu💻 click menu on keypad›4: Window / Zoom›B: Zoom – Square

Start on a Graphs page (ctrl doc adds one).
Press menu.
Choose 4: Window / Zoom.
Choose B: Zoom – Square.

The menu path

Worked example: Question 5 — Check the right angle on the GDC

Graph

f1(x) = 0.4x + 2

and

f2(x) = -2.5x + 11

. On a normal window the angle can look wrong because the axes have different scales. Zoom – Square makes 1 unit across equal 1 unit up, and the 90° angle appears.

⚠️ Where students lose marks

Subtracting the coordinates in a different order on top and bottom: $\dfrac{11-3}{2-6}$ is wrong — keep the same point first in both.
Reading the gradient straight from $ax + by + d = 0$ as $a$ . You must make $y$ the subject first: $3x + 2y - 12 = 0$ has gradient $-\dfrac{3}{2}$ , not 3.
For a perpendicular gradient, only flipping the fraction or only changing the sign. You must do both.
When asked for the form $ax + by + d = 0$ , leaving fractions in. Multiply through so $a$ , $b$ , $d$ are integers.

IB DP Mathematics: Applications and Interpretation (AI) — SL and HL AI 2.1 — Equations of straight lines

IB DP Mathematics: Applications and Interpretation (AI) — 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