Cartesian coordinates

Basics

Cartesian plane denoted $\R^{2}$

Axes and Quadrants

\begin{aligned} x - a x i s & = {(x, y) \in \R^{2} : y = 0} \\ y - a x i s & = {(x, y) \in \R^{2} : x = 0} \\ f i r s t q u a d r a n t & = {(x, y) \in \R^{2} : x > 0, y > 0} \\ s e c o n d q u a d r a n t & = {(x, y) \in \R^{2} : x < 0, y > 0} \\ t h i r d q u a d r a n t & = {(x, y) \in \R^{2} : x < 0, y < 0} \\ f o u r t h q u a d r a n t & = {(x, y) \in \R^{2} : x > 0, y < 0} \end{aligned}

Calculating distance

Pythagoras

z = \sqrt{x^{2} + y^{2}}

Distance formula

Given three points:

\begin{aligned} A & = (x_{a}, y_{a}) \\ B & = (x_{b}, y_{b}) \\ C & = (x_{c}, y_{c}) \\ d i s t (A, B) & = \sqrt{(x_{a} - x_{b})^{2} + (y_{a} - y_{b})^{2}} \\ d i s t (A, C) & = \sqrt{(x_{a} - x_{c})^{2} + (y_{a} - y_{c})^{2}} \end{aligned}

Clustering

If $A$ and $B$ are in Cluster $I$
and $C$ is in Cluster $I I$

Then

$d i s t (A, B) << d i s t (A, C)$

Lines

Slope of line segment

Given
$A = (a, b)$
$B = (c, d)$

\begin{aligned} S l o p e o f \vec{A B} & = \\ m & = \frac{d - b}{c - a} \end{aligned}

Basic line formula

$y - y_{0} = m (x - x_{0})$ "Point-slope" form
$y = m x + b$ "slope-intercept" form

Given a line that passes through $(2, 1)$ and $(3, 2)$ , i.e. a slope of 1,
for any point $(x, y)$ to be on the line, the slope from $(2, 1)$ to $(x, y)$ must be $1$

\begin{aligned} 1 & = \frac{y - 1}{x - 2} \end{aligned}

rearranging...

\begin{aligned} y - 1 & = 1 (x - 2) \end{aligned}

so a definition of the line is:

\begin{aligned} ℓ = {(x, y) \in \R^{2} : y - 1 & = 1 (x - 2)} \end{aligned}

Point-slope formula for line

Above generalises to

If a line $ℓ$ has slope $m$ , and if $(x_{0}, y_{0})$ is any point on $ℓ$ , then $ℓ$ has the equation:

\begin{array}{r} y - y_{0} = m (x - x_{0}) \end{array}

Slope-intercept formula for line

Considering same line defined as

\begin{aligned} ℓ = {(x, y) \in \R^{2} : y - 1 & = 1 (x - 2)} \end{aligned}

This line intercepts the $y$ -axis at a point $(0, b)$ i.e.

\begin{aligned} b - 1 & = 1 (0 - 2) \\ b & = - 1 \end{aligned}

Substituting this intercept point back into the Point-Intercept formula:

\begin{aligned} y - (- 1) & = 1 (x - 0) \\ y + 1 & = x \\ y & = 1 x - 1 \end{aligned}

This generalises into the Slope-Intercept formula for the line:

If $ℓ$ has slope $m$ , and $ℓ$ hits the y-axis at $(0, b)$ , then

\begin{aligned} y & = m x + b \end{aligned}

is an equation for $ℓ$ , where $m$ is the slope and $b$ is the y-intercept.

Based on course material by Paul Bendich and Daniel Egger from Data Science Maths Skills