SL 2.1 Use equations and features of straight lines | International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

Functions

IB Mathematics: Analysis and Approaches SL — Study topic SL 2.1

Straight lines model relationships with a constant rate of change. For example, if a taxi fare has a fixed starting charge and then increases by the same amount for each kilometre, the relationship between distance and cost can be represented by a straight line. This lesson develops the algebra and graph features needed to describe and interpret such lines. It assumes familiarity with coordinates, substitution, and solving simple linear equations.

What you will learn

Identify a straight line’s gradient and intercepts from its equation or graph.
Use common equation forms to describe a straight line.
Find the equation of a line from suitable information, such as a point and gradient.
Determine where two lines meet and recognize parallel and perpendicular lines.
Use graphing technology to check, rather than replace, analytical work.

1. Prior knowledge: coordinates and gradient

A point on a graph is written as

(x,y)

:

x

is the horizontal coordinate and

y

is the vertical coordinate. The gradient, often called the slope, measures the change in

y

for each unit change in

x

. For two points

(x_1,y_1)

and

(x_2,y_2)

, calculate change in

y

divided by change in

x

. Keep the order consistent in both differences.

A positive gradient means the line rises as you move from left to right; a negative gradient means it falls. A horizontal line has gradient

0

. A vertical line has no defined gradient because its

x

-coordinate does not change, so the gradient formula would require division by zero.

An intercept is where a graph crosses an axis. At the vertical axis,

x=0

; at the horizontal axis,

y=0

. These substitutions are useful when finding intercepts from an equation.

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

Gradient compares vertical change with horizontal change.
A vertical line has equation $x=a$ for a constant $a$ ; it cannot be written as $y=mx+c$ .

2. Equations and graphical features

The form

y=mx+c

is useful when a line is not vertical. Here

m

is the gradient and

c

is the

y

-intercept. Setting

x=0

gives the intercept point

(0,c)

. To find the

x

-intercept, set

y=0

and solve for

x

, provided the line is not horizontal.

A second useful form is

y-y_1=m(x-x_1)

. It describes the line with gradient

m

through the known point

(x_1,y_1)

. Expanding the brackets can convert it to

y=mx+c

. Both equations describe the same set of points when they are equivalent.

A line may also be given in the form

Ax+By=C

, where

A

,

B

, and

C

are constants. If

B\ne0

, rearrange to make

y

the subject and read the gradient and vertical intercept. If

B=0

, the equation gives a vertical line. A graph shows the same information visually: its steepness and direction indicate gradient, and its crossings of the axes show intercepts.

For two non-vertical lines, equal gradients mean they are parallel or are the same line. If their gradients differ, they cross once. Two non-vertical lines are perpendicular when the product of their gradients is

-1

. Vertical and horizontal lines are also perpendicular. These relationships can help check a proposed equation.

y-y_1=m(x-x_1)

In $y=mx+c$ , $m$ is the gradient and $c$ is the vertical intercept.
Use a point-gradient form when a point and gradient are known.
Parallel distinct lines have equal gradients; perpendicular non-vertical lines have gradients whose product is $-1$ .

3. Intersections, context, and technology

The intersection of two lines is a point that satisfies both equations. Find it by solving the equations simultaneously: make the expressions for

y

equal, solve for

x

, then substitute to find

y

. On a graph, it is the crossing point. The algebra gives a more precise result than reading coordinates from a plotted image.

In a contextual model, identify what each variable represents and include units. For instance, if

x

is time in minutes and

y

is distance in kilometres, the gradient has units of kilometres per minute. A vertical intercept may represent a starting amount, but its meaning depends on the situation. The model is only appropriate over the range where its assumptions make sense.

A graphing calculator can check a line equation or intersection. Enter both equations, choose a window that includes the relevant values, and use the intersection feature if available. Check the result by substitution in both equations. A poorly chosen viewing window can make a correct line look flat or hide an intersection, so the display is not a substitute for reasoning.

y_1=y_2

An intersection must satisfy both line equations.
In context, interpret the gradient and intercept using the variables and units given.
Technology is a useful visual and numerical check; verify its result algebraically.

Worked example

Find a line from two points

Find the equation of the line through

A(2,5)

and

B(6,13)

, and state its intercepts.

Calculate the gradient
Use the change in the $y$ -coordinates divided by the change in the $x$ -coordinates. The order of subtraction is the same in numerator and denominator.
$m=\frac{13-5}{6-2}=2$
Use a point and the gradient
Substitute the gradient and point $A$ into the point-gradient form. Expand and rearrange to obtain the slope-intercept form.
$y-5=2(x-2)\quad\Rightarrow\quad y=2x+1$
Find the intercepts
The vertical intercept occurs when $x=0$ . The horizontal intercept occurs when $y=0$ ; solve the resulting equation.
$(0,1)\qquad\left(-\frac{1}{2},0\right)$

Answer: The equation is

y=2x+1

. Its vertical intercept is

(0,1)

and its horizontal intercept is

\left(-\frac{1}{2},0\right)

.

Check: Substitution confirms that both given points lie on the line: when

x=2

,

y=5

, and when

x=6

,

y=13

.

Worked example

Use a contextual rate and starting value

A container holds

18

litres of water and is filled at a constant rate of

2.5

litres per minute. Let

t

be time in minutes and

V

be volume in litres. Write a model and find when the volume reaches

43

litres.

Identify gradient and intercept
The constant filling rate is the gradient, measured in litres per minute. The starting volume is the vertical intercept at time zero.
$V=2.5t+18$
Solve for the required time
Set the volume equal to $43$ litres and solve the linear equation. The resulting time is non-negative, so it is consistent with the situation.
$43=2.5t+18\quad\Rightarrow\quad t=10$

Answer: The model is

V=2.5t+18

, and the volume reaches

43

litres after

10

minutes.

Check: Substituting

t=10

gives

V=2.5(10)+18=43

litres. The model assumes the filling rate remains constant.

Worked example

Find and interpret an intersection

Find the intersection of

y=3x-2

and

y=-x+10

. Then state whether the lines are perpendicular.

Set the expressions equal
At the intersection, both equations give the same $y$ -value for the same $x$ -value. Equating them gives one equation in $x$ .
$3x-2=-x+10\quad\Rightarrow\quad x=3$
Find the matching vertical coordinate
Substitute $x=3$ into either line equation. Both equations must give the same result.
$y=3(3)-2=7$
Check perpendicularity
The gradients are $3$ and $-1$ . Their product is not $-1$ , so the lines are not perpendicular.
$3(-1)=-3$

Answer: The lines intersect at

(3,7)

and are not perpendicular.

Check: At

x=3

, the second equation gives

y=-3+10=7

, confirming the intersection.

Common mistakes and how to avoid them

Reversing the subtraction order in only one part of the gradient calculation.

Correction: Use the same point order in both differences, such as

y_2-y_1

over

x_2-x_1

.

Treating the vertical intercept as the horizontal intercept.

Correction: For the vertical intercept set

x=0

; for the horizontal intercept set

y=0

.

Assuming every straight line can be written as

y=mx+c

.

Correction: A vertical line has equation

x=a

and has no defined gradient.

Reporting a calculator’s intersection without checking it.

Correction: Substitute the reported coordinates into both equations; both must be satisfied.

Lesson summary

Gradient is change in $y$ divided by change in $x$ .
Use $y=mx+c$ to read the gradient and vertical intercept directly.
Use point-gradient form when a gradient and a point are known.
Find intersections by solving the two line equations together.
Check parallelism or perpendicularity using gradients, and interpret contextual values with their units.

Check your understanding

Question 1

What is the gradient of the line through

(1,4)

and

(5,12)

?

$2$
$-2$
$\frac{1}{2}$
$4$

Show answer and explanation

2

The gradient is

(12-4)/(5-1)=8/4=2

.

Question 2

What is the vertical intercept of

y=-3x+7

?

$(0,-3)$
$(7,0)$
$(0,7)$
$(3,7)$

Show answer and explanation

(0,7)

At the vertical axis,

x=0

, so

y=7

and the intercept is

(0,7)

.

Question 3

Which line is perpendicular to

y=\frac{1}{2}x+4

?

$y=2x-1$
$y=-2x+1$
$y=-\frac{1}{2}x+1$
$y=\frac{1}{2}x-1$

Show answer and explanation

y=-2x+1

The first line has gradient

\frac{1}{2}

. A perpendicular non-vertical line has gradient

-2

, since their product is

-1

.

Key terms

Gradient: The ratio of the change in vertical coordinate to the change in horizontal coordinate along a line.
Intercept: A point where a graph crosses one of the coordinate axes.
Intersection: A point that lies on both of two graphs.
Point-gradient form: An equation form that specifies a line using its gradient and one point on the line.

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Published by DoAssignment. This AI-assisted lesson follows International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL, study topic SL 2.1. It is a study resource, not an official curriculum publication.

Before publication, the draft is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability, and then requires administrator approval. Errors can still occur, so corrections are welcomed.

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SL 2.1 · Use equations and features of straight lines

International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

IB Mathematics: Analysis and Approaches SL — Study topic SL 2.1

What you will learn

1. Prior knowledge: coordinates and gradient

2. Equations and graphical features

3. Intersections, context, and technology

Find a line from two points

Use a contextual rate and starting value

Find and interpret an intersection

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through IB AA SL

About this lesson and its review