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Δβα\Delta _{\beta \alpha}
The Δelta βeta αrchivesThe \ \Delta elta \ \beta eta \ \alpha rchives
CHAPTER 01
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Linear functions

CHAPTER 02
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Quadratic functions

under construction
CHAPTER 03
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Functions

under construction
CHAPTER 04
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Trigonometry

under construction
CHAPTER 05
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Trigonometric functions

under construction
CHAPTER 06
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Linear algebra

under construction
CHAPTER 07
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Vectors

under construction
CHAPTER 08
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Logarithms

under construction
CHAPTER 09
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Complex numbers

under construction
CHAPTER 10
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Differentiation

under construction
CHAPTER 11
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Integration

under construction

What are linear functions?01_01A

We generally think of linear functions in the form
y=mx+cy=mx+c
, where
mm
is the gradient and
cc
is the
yy
-intercept. There are, however, other ways including:
ax+by+c=0ax+by+c=0
,
ax+by=cax+by=c
, and
y−y1=m(x−x1)y-y_1=m(x-x_1)
(which is line-slope form). All of these are functions for finding straight lines, and should be middle-school knowledge. In all examples, we input an
xx
value, and get a
yy
value as an output, and this correlation is linear (hence,
linear
equation or function).
When given cartesian coordinates
A(x1,y1)A(x_1,y_1)
and
B(x2,y2)B(x_2,y_2)
, we're able to get the gradient,
mm
, using the following formula:
m=y2−y1x2−x1m=\frac{y_2-y_1}{x_2-x_1}
, which takes the
Δy\Delta y
and divides it by
Δx\Delta x
. We colloquially say it as "change in
yy
by change in
xx
", which happens to be the definition of the gradient!

Solutions01_02A

We can get either:
∞\infty
, 1, or 0 solutions. The number of solutions when dealing with two linear functions abides by these rules:

Solution occurrences

∞\infty
solutions: when
m1=m2m_1=m_2
and when both functions share common solutions
1 solution: when
m1≠m2m_1\neq m_2
0 solutions: when
m1=m2m_1=m_2
and when there's no common solution(s)
In simultaneous equations, these solutions represent where two functions intersect, and in intercepts, where functions(s) intercept the
xx
or
yy
-axis.
When dealing with linear equations in general, you can either get solutions by: plotting on a graph, solving for a variable, or using simultaneous equations (explanation below).

Perpendicular lines01_03A

Now that we've got the basics of linear functions, we can apply them to find out things like perpendiculars, and perpendicular bisectors. When talking about perpendicular lines, we're referring to those who intersect each other at 90 degree angles, and in order to find said perpendicular line, all you have to do is find the negative reciprocal of the gradient! This is defined as
m′=−1mm'=-\frac{1}{m}
.

Solutions and perpendicular lines

1) Determine how many solutions
{y=3x−4y=−2x+11\left\{ {y=3x-4 \atop y=-2x+11} \right.
has, and any
xx
and
yy
intercepts for both lines.
2) Without a GDC, conclude why
{y=100x−64y=100x−63.6\left\{ {y=100x-64 \atop y=100x-63.6} \right.
produces 0 solutions.
3) Find solutions on the
xx
axis, of the perpendicular line produced from
y=23x−28y=\frac{2}{3}x-28
with the intersection at
(0,−28)(0,-28)
.
4) The equation of a line is given as
y−a4x=1a(x−5)+2y-\frac{a}{4}x=\frac{1}{a}(x-5)+2
. If
m=−1m=-1
, find
aa
.

Midpoints and perpendicular bisectors01_04A

When faced with a question like "Find the perpendicular bisector of
f(x)f(x)
with the endpoints
[AB][AB]
at
A(2,5)A(2,5)
and
B(−10,3)B(-10,3)
", we need to note that the perpendicular
bisector
is a line (perpendicular to
f(x)f(x)
) created on the midpoint between two points
A(x1,y1)A(x_1,y_1)
and
B(x2,y2)B(x_2,y_2)
.
To find the midpoint, we use
(x1+x22,y1+y22)\left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right)
, which is the average (mean, if you would like to be specific) of the distances of
AA
and
BB
. This logic can also be applied to finding the centroid of a triangle, which is found by
(x1+x2+x33,y1+y2+y33)\left( \frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3} \right)
, however we don't have to worry about that for now.
If we want to find the function of a perpendicular bisector line (line perpendicular to
f(x)f(x)
at a midpoint of a segment
[AB][AB]
), we can just insert the reciprocal of the gradient (
m′=−1mm'=-\frac{1}{m}
) and the coordinates of the midpoint into the following equation:
y−y1=m(x−x1)y-y_1=m(x-x_1)
, where
x1,y1x_1,y_1
are the midpoint coordinates,
mm
is the gradient, and
x,yx,y
form the function itself. Then, just do some re-arranging to get the form required by the question.

Midpoints and perpendicular bisectors

1) The midpoint of a line segment
[AB][AB]
is
(2,3)(2,3)
. Point
AA
has the coordinate
(4,4)(4,4)
. Find the coordinate of point
BB
, and hence find the equation of the line.
2) Consider the function
f(x)=x2f(x)=x^2
. Find the midpoint of the line segment connecting the points
(1,1)(1,1)
and
(3,9)(3,9)
on the graph of
f(x)f(x)
.
3) A quadrilateral
ABCDABCD
has vertices
A(0,0)A(0,0)
,
B(4,0)B(4,0)
,
C(6,2)C(6,2)
, and
D(2,2)D(2,2)
. Consider a point
MM
, lying in the center of
ABAB
and
BCBC
; find the midpoint in the line segment
[MC][MC]
.
4) Find the perpendicular bisector of the line
f(x)=−7x+5f(x)=-7x+5
with the points
A(1,−2)A(1,-2)
and
A(2,−9)A(2,-9)
.
5) Consider the midpoint generated from question 3. Form an function for that line, and then find the function of the perpendicular bisector.

Simultaneous equations01_05A

When solving simultaneous equations, especially in IB, the general elimination method isn"t enough. Gaussian elimination using augmented matrices provides a more efficient approach. This involves representing equations as matrices and applying row operations to solve. Don't worry about understanding matrices fully now; they'll be introduced in a later chapter.
In order to solve them using matrices, we need to turn the formed matrix into what's called "reduced row form" (don't worry, you don't have to memorize that word). This involves us getting the following forms of a matrix:

Row forms

1 solution:
(1□□01□)\begin{pmatrix} 1 & \square & \square \\ 0 & 1 & \square \\ \end{pmatrix}
0 solutions:
(1□□000)\begin{pmatrix} 1 & \square & \square \\ 0 & 0 & 0 \\ \end{pmatrix}
Infinite solutions:
(1□□00□)\begin{pmatrix} 1 & \square & \square \\ 0 & 0 & \square \\ \end{pmatrix}
Note that in exams, you might not explicitly be asked 'give the value of
aa
that produces a single solution'. Instead, exams might say: "consistent systems" (systems have at least one solution), "inconsistent systems" (systems have no solutions), and "unique solution" (systems have exactly one solution), and then the obvious "infinite solutions" (equations are linearly dependent of each other [hint hint], meaning there's an infinite number of solutions).
We can input equations into a matrix, and manipulate the contents of the matrix to get our required solutions. An example of said manipulation is below:

Solving systems using Gaussian elimination

Consider the system
{5x+y=210x−2y=1\left\{ {5x+y=2 \atop 10x-2y=1} \right.
.
Transition it to an augmented matrix (by taking the coefficients/constants, and placing them in a matrix):
(51210−21)\begin{pmatrix} 5 & 1 & 2 \\ 10 & -2 & 1 \\ \end{pmatrix}
Multiply everything in row 1 by 2 and take row 2 away from that result (also noted as
R2=2R1−R2R_2=2R_1-R_2
):
(512043)\begin{pmatrix} 5 & 1 & 2 \\ 0 & 4 & 3 \\ \end{pmatrix}
Then divide row 1 by 5 and row 2 by 4 (noted as
R1=R15R_1=\frac{R_1}{5}
and
R2=R24R_2=\frac{R_2}{4}
):
(115250134)\begin{pmatrix} 1 & \frac{1}{5} & \frac{2}{5} \\ 0 & 1 & \frac{3}{4} \\ \end{pmatrix}
(see how this looks like the form of a singular-solution RRM?)
The convert the matrix back into equation form:
x+15y=25x+\frac{1}{5}y=\frac{2}{5}
for the top row and
y=34y=\frac{3}{4}
for the bottom row.
Simply do some substitution to arrive to the solution
(14,34)(\frac{1}{4},\frac{3}{4})
, which you can check on Desmos, is in fact the answer!
Having a solution to two sets of equations is nice, however we can get infinite solutions if they never intersect (remember Note N01.0a from the top). With this, you have to "introduce" a variable yourself, and express the answer as that variable. For now, you won't have to worry about this. I found it quite hard to find questions that worked to produce infinite solutions, and always found ones that produce
00
. Think about why that might be!
Try some yourself. Remember, there are different reduced-row-matrix forms that could be achieved!

Solving systems using Gaussian elimination

1) Consider the system
{5x+2y=55x+2y=6\left\{ {5x+2y=5 \atop 5x+2y=6} \right.
. Show that there's no solutions to this system.
2) Consider the system
{x+3y=42x+6y=8\left\{ {x+3y=4 \atop 2x+6y=8} \right.
. Explain why there are infinitely many solutions.
3) Discuss the solutions to
{x−5y=82x−10y=a\left\{ {x-5y=8 \atop 2x-10y=a} \right.
for all
a∈Ra \in \mathbb{R}
, including values that provide: infinite, one, or no solutions.

Practical applications01_06A

Common examples of applications involves establishing relationships (in physics or chemistry) and finding break-even points (in finance). Generally because they're linear equations, the applications are quite easy to solve, so if you just want to glance over this, you don't have to solve them if you don't want to.

Practical applications

1) A repair specialist charges an initial inspection fee of
200200
dollars. Then when he starts fixing things, he charges
5050
dollars an hour. Figure out how many hours he has worked, to earn
1,0001,000
dollars.
2) The freezing temperature of water is
0C0C
and
32F32F
. The boiling point is
100C100C
and
212F212F
. Deduce a relationship, and hence convert
30C30C
to
FF
.