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Cartesian coordinates

Basics

Cartesian plane denoted R2\R^2

Axes and Quadrants

xaxis={(x,y)R2:y=0}yaxis={(x,y)R2:x=0}first quadrant={(x,y)R2:x>0,y>0}second quadrant={(x,y)R2:x<0,y>0}third quadrant={(x,y)R2:x<0,y<0}fourth quadrant={(x,y)R2:x>0,y<0}\begin{alignedat}{1} x-axis &= \lbrace (x,y) \in \R^2 : y=0 \rbrace \\ y-axis &= \lbrace (x,y) \in \R^2 : x=0 \rbrace \\ first \nobreakspace quadrant &= \lbrace (x,y) \in \R^2 : x>0, y>0 \rbrace \\ second \nobreakspace quadrant &= \lbrace (x,y) \in \R^2 : x<0, y>0 \rbrace \\ third \nobreakspace quadrant &= \lbrace (x,y) \in \R^2 : x<0, y<0 \rbrace \\ fourth \nobreakspace quadrant &= \lbrace (x,y) \in \R^2 : x>0, y<0 \rbrace \\ \end{alignedat}

Calculating distance

Pythagoras
pythagoras drawio

z=x2+y2z = \sqrt{x^2 + y^2}

Distance formula

Given three points:

A=(xa,ya)B=(xb,yb)C=(xc,yc)dist(A,B)=(xaxb)2+(yayb)2dist(A,C)=(xaxc)2+(yayc)2\begin{alignedat}{1} A &= (x_a, y_a) \\ B &= (x_b, y_b) \\ C &= (x_c, y_c) \\ \\ dist(A,B) &= \sqrt{(x_a-x_b)^2+(y_a-y_b)^2} \\ \\ dist(A,C) &= \sqrt{(x_a-x_c)^2+(y_a-y_c)^2} \end{alignedat}

Clustering

If AA and BB are in Cluster II
and CC is in Cluster IIII

Then

dist(A,B)<<dist(A,C)dist(A,B)<< dist(A,C)

Lines

Slope of line segment

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

SlopeofAB=m=dbca\begin{alignedat}{1} Slope\enspace of \enspace \overrightarrow{AB} &= \\ m &= \frac{d-b}{c-a} \end{alignedat}

Basic line formula

yy0=m(xx0)y-y_0 = m(x-x_0) "Point-slope" form
y=mx+by = mx + b "slope-intercept" form

Given a line that passes through (2,1)(2,1) and (3,2)(3,2), i.e. a slope of 1,
for any point (x,y)(x,y) to be on the line, the slope from (2,1)(2,1) to (x,y)(x,y) must be 11

1=y1x2\begin{alignedat}{1} 1 &= \frac {y-1}{x-2} \\ \end{alignedat}

rearranging...

y1=1(x2)\begin{alignedat}{1} y − 1 &= 1(x − 2) \end{alignedat}

so a definition of the line is:

={(x,y)R2:y1=1(x2)}\begin{alignedat}{1} \ell = \big\lbrace (x,y) \in \R^2 : y − 1 &= 1(x − 2) \big\rbrace \\ \end{alignedat}

Point-slope formula for line

Above generalises to

If a line \ell has slope mm, and if (x0,y0)(x_0, y_0) is any point on \ell, then \ell has the equation:

yy0=m(xx0)\begin{alignedat}{1} y-y_0 = m(x-x_0) \end{alignedat}

Slope-intercept formula for line

Considering same line defined as

={(x,y)R2:y1=1(x2)}\begin{alignedat}{1} \ell = \big\lbrace (x,y) \in \R^2 : y − 1 &= 1(x − 2) \big\rbrace \\ \end{alignedat}

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

b1=1(02)b=1\begin{alignedat}{1} b-1 &= 1(0-2) \\ b &= -1 \end{alignedat}

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

y(1)=1(x0)y+1=xy=1x1\begin{alignedat}{1} y-(-1) &= 1(x-0) \\ y+1 &= x \\ y &= 1x-1 \end{alignedat}

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

If \ell has slope mm, and \ell hits the y-axis at (0,b)(0, b), then

y=mx+b\begin{alignedat}{1} y &= mx + b \\ \end{alignedat}

is an equation for \ell, where mm is the slope and bb is the y-intercept.

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

Referred in


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Cartesian coordinates