In the plane,
after designating an origin point \(O\)
and two perpendicular cardinal axes,
conventionally an \(x\)-axis whose positive direction points eastward
and a \(y\)-axis whose positive direction points northward,
we can describe the location of a point
with its rectangular coordinates,
an ordered pair \((a,b)\)
where \(a\) is its position in the \(x\) direction
and \(b\) is its position in the \(y\) direction.
Altogether this constitutes a rectangular coordinate system,
also called a Cartesian coordinate system.
The \(x\)- and \(y\)-axis partition the plane
into four quadrants conventionally labelled I, II, III, and IV,
starting in the upper-right and enumerated counterclockwise.
Once equipped with this coordinate system
the space is now referred to as the \(xy\)-plane.
For two points \(A\) and \(B\) in the plane let \(\operatorname{d}(A,B)\)
denote the distance between them.
In terms of their coordinates,
for the two points \(A(x_1,y_1)\) and \(B(x_2,y_2)\),
the distance between them is calculated by the formula
\( \operatorname{d}(A,B) = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\,. \)
For the same two points \(A(x_1,y_1)\) and \(B(x_2,y_2)\) in the plane,
their midpoint, the point half-way between them,
has coordinates \((\overline{x}, \overline{y})\)
given by the formulas
\[ \displaystyle \overline{x} = \frac{x_1+x_2}{2}
\qquad \displaystyle \overline{y} = \frac{y_1+y_2}{2}\,. \]
Given an equation involving two variables \(x\) and \(y\)
its graph curve is the set of all points
\((x,y)\) in the plane that satisfy that equation.
The best example to focus on now is a circle.
A circle is the curve consisting of all points
that are a fixed distance, its radius,
from a given point, its center.
An equation corresponding to a circle
with radius \(r\) and center \((h,k)\) is
\( (x-h)^2 +(y-k)^2 = r^2\,. \)
A circle of radius \(r\) has area \(\pi r^2\) and circumference \(2\pi r,\) where \(\pi \approx 3.1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679821480865132823066470938446095505822317253594081284811174502841027019385211055596446229489549303819644288109756659334461284756482337867831652712019091456485669234603486104543266482133936072602491412737245870066063155881748815209209628292540917153643678925903600113305305488204665213841469519415116094330572703657595919530921861173819326117931051185480744623799627495673518857527248912279381830119491298336733624406566430860213949463952247371907021798609437027705392171762931767523846748184676694051320005681271452635608277857713427577896091736371787214684409012249534301465495853710507922796892589235420199561121290219608640344181598136297747713099605187072113499999983729780499510597317328160963185950244594553469083026425223082533446850352619311881710100031378387528865875332083814206171776691473035982534904287554687311595628638823537875937519577818577805321712268066130019278766111959092164201988.\)
There is a unique line that passes through
any two points \(A\bigl(x_1, y_2\bigr)\) and \(B\bigl(x_2,y_2\bigr).\)
The slope of that line is
\[ m = \frac{\text{rise}}{\text{run}} = \frac{y_2-y_1}{x_2-x_1}; \]
this is the “rate” at which the line is increasing/decreasing.
It’s also referred to as the pitch or grade.
The “point-slope” form, “slope-intercept” form,
and “general” form of a line are respectively are
\[ (y-y_1) = m(x-x_1) \qquad\qquad y = mx+b \qquad\qquad Ax + By = C \,. \]
Horizontal and vertical lines have special forms, \(y=b\) and \(x=a\) respectively.
Two lines are parallel if they never intersect, in which case they must have the same slope. Two lines are perpendicular (or normal, or orthogonal) if they intersect at a right angle, so one’s slope will be the negative reciprocal of the other. E.g. if a line has slope \(\frac{p}{q}\) another line perpendicular to that one will have slope \(-\frac{q}{p}.\)
The unit circle is the circle of radius 1 centered at the origin,
which corresponds with the equation \(x^2+y^2 = 1.\)
The line \(y = mx\) will intersect the top-half of the unit circle at the point
\[ (x,y) = \biggl(\frac{1}{\sqrt{1+m^2}}, \frac{m}{\sqrt{1+m^2}}\biggr)\,. \]