9662e.dev

\Delta _{\beta \alpha}

\Delta _{\beta \alpha}

Linear equations

\Delta _{\beta \alpha}

Quadratic equations

\Delta _{\beta \alpha}

Functions

\Delta _{\beta \alpha}

Linear algebra

\Delta _{\beta \alpha}

Vectors

\Delta _{\beta \alpha}

Logarithms

\Delta _{\beta \alpha}

Complex numbers

Useful resources

Download formula booklet

Download subject guide

\Delta _{\beta \alpha}

Linear equations

Starter

List out everything you know about linear equations. This can be stuff ranging from equations, to proofs. This will get you thinking about the basics, which we'll either develop, or build upon, i.e. with perpendicular bisectors.

What are linear equations?

We generally think of linear equations in the form

y=mx+c

, where

m

is the gradient and

c

is the

y

-intercept. There are, however, other ways including:

ax+by+c=0

ax+by=c

, and

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

x

value, and get a

y

value as an output, and this correlation is linear (hence,

linear

equation).

When given cartesian coordinates

A(x_1,y_1)

and

B(x_2,y_2)

, we're able to get the gradient,

m

, using the following formula:

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

, which takes the

\Delta y

and divides it by

\Delta x

. We colloquially say it as "change in

y

by change in

x

", which happens to be the definition of the gradient!

Solutions

We can get either:

\infty

, 1, or 0 solutions. The number of solutions when dealing with two linear equations abides by these rules:

Note N01.0a - Solution occurrences

\infty

solutions: when

m_1=m_2

and when both equations share common solutions

1 solution: when

m_1\neq m_2

0 solutions: when

m_1=m_2

and when there's no common solution(s)

In simultaneous equations, these solutions represent where two equations intersect, and in intercepts, where equation(s) intercept the

x

y

-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 lines

Now that we've got the basics of line equations, 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'=-\frac{1}{m}

Coursework C01.0a - Simple solutions and perpendicular lines

1) Determine how many solutions

\left\{ {y=3x-4 \atop y=-2x+11} \right.

has, and any

x

and

y

intercepts for both lines.

2) Without a GDC, conclude why

\left\{ {y=100x-64 \atop y=100x-63.6} \right.

produces 0 solutions.

3) Find solutions on the

x

axis, of the perpendicular line produced from

y=\frac{2}{3}x-28

with the intersection at

(0,-28)

4) The equation of a line is given as

y-\frac{a}{4}x=\frac{1}{a}(x-5)+2

. If

m=-1

, find

a

Midpoints and perpendicular bisectors

When faced with a question like "Find the perpendicular bisector of

f(x)

with the endpoints

[AB]

A(2,5)

and

B(-10,3)

", we need to note that the perpendicular

bisector

is a line (perpendicular to

f(x)

) created on the midpoint between two points

A(x_1,y_1)

and

B(x_2,y_2)

To find the midpoint, we use

\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

A

and

B

. This logic can also be applied to finding the centroid of a triangle, which is found by

\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 equation of a perpendicular bisector line (line perpendicular to

f(x)

at a midpoint of a segment

[AB]

), we can just insert the reciprocal of the gradient (

m'=-\frac{1}{m}

) and the coordinates of the midpoint into the following equation:

y-y_1=m(x-x_1)

, where

x_1,y_1

are the midpoint coordinates,

m

is the gradient, and

x,y

form the function itself. Then, just do some re-arranging to get the form required by the question.

Coursework C01.1a - Midpoints/perpendicular bisectors

1) The midpoint of a line segment

[AB]

(2,3)

. Point

A

has the coordinate

(4,4)

. Find the coordinate of point

B

, and hence find the equation of the line.

2) Consider the function

f(x)=x^2

. Find the midpoint of the line segment connecting the points

(1,1)

and

(3,9)

on the graph of

f(x)

3) A quadrilateral

ABCD

has vertices

A(0,0)

B(4,0)

C(6,2)

, and

D(2,2)

. Consider a point

M

, lying in the center of

AB

and

BC

; find the midpoint in the line segment

[MC]

4) Find the perpendicular bisector of the line

f(x)=-7x+5

with the points

A(1,-2)

and

A(2,-9)

5) Consider the midpoint generated from question 3. Form an equation for that line, and then find the equation of the perpendicular bisector.

Simultaneous equations

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:

Note N01.1a - Reduced row form "forms"

1 solution:

\begin{pmatrix} 1 & \square & \square \\ 0 & 1 & \square \\ \end{pmatrix}

0 solutions:

\begin{pmatrix} 1 & \square & \square \\ 0 & 0 & 0 \\ \end{pmatrix}

Infinite solutions:

\begin{pmatrix} 1 & \square & \square \\ 0 & 0 & \square \\ \end{pmatrix}

Note that in exams, you might not explicitly be asked 'give the value of

a

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:

Example E01.0a - Solving systems with Gaussian elimination

Consider the system

\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):

\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

R_2=2R_1-R_2

\begin{pmatrix} 5 & 1 & 2 \\ 0 & 4 & 3 \\ \end{pmatrix}

Then divide row 1 by 5 and row 2 by 4 (noted as

R_1=\frac{R_1}{5}

and

R_2=\frac{R_2}{4}

\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+\frac{1}{5}y=\frac{2}{5}

for the top row and

y=\frac{3}{4}

for the bottom row.

Simply do some substitution to arrive to the solution

(\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

0

. Think about why that might be!

Try some yourself. Remember, there are different reduced-row-matrix forms that could be achieved!

Coursework C01.2a - Solving systems using Gaussian elimination

1) Consider the system

\left\{ {5x+2y=5 \atop 5x+2y=6} \right.

. Show that there's no solutions to this system.

2) Consider the system

\left\{ {x+3y=4 \atop 2x+6y=8} \right.

. Explain why there are infinitely many solutions.

3) Discuss the solutions to

\left\{ {x-5y=8 \atop 2x-10y=a} \right.

for all

a \in \mathbb{R}

, including values that provide: infinite, one, or no solutions.

Practical applications

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.

Coursework C01.3a - Practical applications

1) A repair specialist charges an initial inspection fee of

200

dollars. Then when he starts fixing things, he charges

50

dollars an hour. Figure out how many hours he has worked, to earn

1,000

dollars.

2) The freezing temperature of water is

0C

and

32F

. The boiling point is

100C

and

212F

. Deduce a relationship, and hence convert

30C

F

ECQ01 direct link