nyuuzyou/wikis huggingface dataset:
Equation of Tractrix/X-Axis Asymptote/Cartesian Form

== Definition ==
Let $S$ be a cord of length $a$ situated as a (straight) line segment whose endpoints are $P$ and $T$.
Let $S$ be aligned along the $y$-axis of a cartesian plane with $T$ at the origin and $P$ therefore at the point $\tuple {0, a}$.
Let $T$ be dragged along the $x$-axis.
The equation of the tractrix along which $P$ travels is:
:$x = a \map \ln {\dfrac {a \pm \sqrt {a^2 - y^2} } y} \mp \sqrt {a^2 - y^2}$
== Proof ==
Consider $P$ when it is at the point $\tuple {x, y}$.
:
The cord $S$ is tangent to the locus of $P$.
Thus from Pythagoras's Theorem:
:$\dfrac {\d y} {\d x} = -\dfrac y {\sqrt {a^2 - y^2} }$
Hence:
When $x = 0$ we have $y = a$.
Thus:
Hence the result.
== Linguistic Note ==
Category:Equations of Tractrices
Equation of Tractrix/X-Axis Asymptote/Parametric Form

== Definition ==
Let $S$ be a cord of length $a$ situated as a (straight) line segment whose endpoints are $P$ and $T$.
Let $S$ be aligned along the $y$-axis of a cartesian plane with $T$ at the origin and $P$ therefore at the point $\tuple {0, a}$.
Let $T$ be dragged along the $x$-axis.
=== Formulation 1 ===
=== Formulation 2 ===
Category:Equations of Tractrices
Equation of Tractrix/X-Axis Asymptote/Parametric Form/Formulation 1

== Definition ==
Let $S$ be a cord of length $a$ situated as a (straight) line segment whose endpoints are $P$ and $T$.
Let $S$ be aligned along the $y$-axis of a cartesian plane with $T$ at the origin and $P$ therefore at the point $\tuple {0, a}$.
Let $T$ be dragged along the $x$-axis.
The equation of the tractrix along which $P$ travels can be expressed in parametric form as:
== Proof ==
Consider $P$ when it is at the point $\tuple {x, y}$.
:
Consider the upper part of the tractrix.
The cord $S$ is tangent to the locus of $P$.
Let the angle formed by cord $S$ and the $x$-axis be $\theta$.
Then $y = a \sin \theta$.
Substituting this into the Cartesian Form of the equation of the tractrix:
== Linguistic Note ==
== Sources ==
* : $\S 11$: Special Plane Curves: Tractrix: $11.21$
* : $\S 9$: Special Plane Curves: Tractrix: $9.21.$
Category:Equations of Tractrices
Equation of Tractrix/X-Axis Asymptote/Parametric Form/Formulation 2

== Definition ==
Let $S$ be a cord of length $a$ situated as a (straight) line segment whose endpoints are $P$ and $T$.
Let $S$ be aligned along the $y$-axis of a cartesian plane with $T$ at the origin and $P$ therefore at the point $\tuple {0, a}$.
Let $T$ be dragged along the $x$-axis.
The equation of the tractrix along which $P$ travels can be expressed in parametric form as:
== Proof ==
Consider $P$ when it is at the point $\tuple {x, y}$.
:
Let $y = a \sech u$.
Substituting this into the Cartesian Form of the equation of the tractrix:
== Linguistic Note ==
== Sources ==
* 
* 
* 
Category:Equations of Tractrices
Equation of Tractrix/Y-Axis Asymptote

== Theorem ==
Let $S$ be a cord of length $a$ situated as a (straight) line segment whose endpoints are $P$ and $T$.
Let $S$ be aligned along the $x$-axis of a cartesian plane with $T$ at the origin and $P$ therefore at the point $\tuple {a, 0}$.
Let $T$ be dragged along the $y$-axis.
:
=== Cartesian Form ===
=== Parametric Form ===
Category:Equations of Tractrices
Equation of Tractrix/Y-Axis Asymptote/Cartesian Form

== Definition ==
Let $S$ be a cord of length $a$ situated as a (straight) line segment whose endpoints are $P$ and $T$.
Let $S$ be aligned along the $x$-axis of a cartesian plane with $T$ at the origin and $P$ therefore at the point $\tuple {a, 0}$.
Let $T$ be dragged along the $y$-axis.
The equation of the tractrix along which $P$ travels is:
:$y = a \map \ln {\dfrac {a \pm \sqrt {a^2 - x^2} } x} \mp \sqrt {a^2 - x^2}$
== Proof ==
Consider $P$ when it is at the point $\tuple {x, y}$.
:
The cord $S$ is tangent to the locus of $P$.
Thus from Pythagoras's Theorem:
:$\dfrac {\d y} {\d x} = -\dfrac {\sqrt {a^2 - x^2} } x$
Hence:
When $y = 0$ we have $x = a$.
Thus:
Hence the result.
== Linguistic Note ==
== Sources ==
* : $\S 2.12$: The Hanging Chain. Pursuit Curves: Example $(2)$
Category:Equations of Tractrices
Equation of Tractrix/Y-Axis Asymptote/Parametric Form

== Definition ==
Let $S$ be a cord of length $a$ situated as a (straight) line segment whose endpoints are $P$ and $T$.
Let $S$ be aligned along the $x$-axis of a cartesian plane with $T$ at the origin and $P$ therefore at the point $\tuple {a, 0}$.
Let $T$ be dragged along the $y$-axis.
The equation of the tractrix along which $P$ travels can be expressed in parametric form as:
== Proof ==
Consider $P$ when it is at the point $\tuple {x, y}$.
:
Consider the upper part of the tractrix.
The cord $S$ is tangent to the locus of $P$.
Let the angle formed by cord $S$ and the $y$-axis be $\theta$.
Then $x = a \sin \theta$.
Substituting this into the Cartesian Form of the equation of the tractix:
== Linguistic Note ==
== Sources ==
* : $\S 2.12$: The Hanging Chain. Pursuit Curves: Example $(2)$
Category:Equations of Tractrices
Equation of Trochoid

== Theorem ==
Consider a circle $C$ of radius $a$ rolling without slipping along the x-axis of a cartesian plane.
Consider the point $P$ on on the line of a radius of $C$ at a distance $b$ from the center of $C$.
Let $P$ be on the y-axis when the center of $C$ is also on the y-axis.
Consider the trochoid traced out by the point $P$.
Let $\tuple {x, y}$ be the coordinates of $P$ as it travels over the plane.
The point $P = \tuple {x, y}$ is described by the equations:
:$x = a \theta - b \sin \theta$
:$y = a - b \cos \theta$
== Proof ==
Let $C$ have rolled so that the radius to the point $P = \tuple {x, y}$ is at angle $\theta$ to the vertical.
The center of $C$ is at $\tuple {a \theta, a}$.
Then it follows from the definition of sine and cosine that:
:$x = a \theta - b \sin \theta$
:$y = a - b \cos \theta$
whence the result.
== Sources ==
* : $\S 11$: Special Plane Curves: Trochoid: $11.20$
* 
* 
* : $\S 9$: Special Plane Curves: Trochoid: $9.20.$
Category:Trochoids
Equation of Unit Circle

== Theorem ==
Let the unit circle have its center at the origin of the Cartesian plane.
Its equation is given by:
:$x^2 + y^2 = 1$
== Proof ==
From Equation of Circle, the equation of a circle with radius $R$ and center $\tuple {a, b}$ is:
:$\paren {x - a}^2 + \paren {y - b}^2 = R^2$
Substituting $\tuple {0, 0}$ for $\tuple {a, b}$ and $1$ for $R$ gives the result.
== Sources ==
* : $\S 3$. Roots of Unity
* 
Category:Unit Circles
Category:Equation of Circle
Equation of Unit Circle in Complex Plane

== Theorem ==
Consider the unit circle $C$ whose center is at $\tuple {0, 0}$ on the complex plane.
Its equation is given by:
:$\cmod z = 1$
where $\cmod z$ denotes the complex modulus of $z$.
=== Corollary 1 ===
=== Corollary 2 ===
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $1$: Complex Numbers: The $n$th Roots of Unity
Category:Geometry of Complex Plane
Category:Equation of Unit Circle in Complex Plane
Equation of Unit Circle in Complex Plane/Corollary 1

== Corollary to Equation of Unit Circle in Complex Plane ==
Consider the unit circle $C$ whose center is at $\left({0, 0}\right)$ on the complex plane.
The equation of $C$ can be given by:
:$z \overline z = 1$
where $\overline z$ denotes the complex conjugate of $z$.
== Proof ==
From Equation of Unit Circle in Complex Plane, the equation of $C$ can also be given by:
:$\cmod z = 1$
where $\cmod z$ denotes the complex modulus of $z$.
Thus:
== Sources ==
* : $\S 3$. Roots of Unity
Category:Equation of Unit Circle in Complex Plane
Equation of Unit Circle in Complex Plane/Corollary 2

== Corollary to Equation of Unit Circle in Complex Plane ==
Consider the unit circle $C$ whose center is at $\left({0, 0}\right)$ on the complex plane.
The equation of $C$ can be given by:
:$\overline z = \dfrac 1 z$
where $\overline z$ denotes the complex conjugate of $z$.
== Proof ==
From Equation of Unit Circle in Complex Plane: Corollary 1, the equation of $C$ can also be given by:
:$z \overline z = 1$
The result follows by multiplying by $\dfrac 1 z$.
== Sources ==
* : $\S 3$. Roots of Unity
Category:Equation of Unit Circle in Complex Plane
Equation of Unit Circle in Complex Plane/Proof 2

== Theorem ==
== Proof ==
Let $C$ be the set of points on the unit circle defined above.
Let $P$ be the set:
:$\set {z \in \C: \cmod z = 1}$
Let $z = x + i y \in C$.
$z$ can be expressed in polar form as:
:$z = \polar {r, \theta}$
where:
:$x = r \cos \theta$
:$y = r \sin \theta$
By definition of unit circle, $z$ is $1$ unit away from $\tuple {0, 0}$.
By Pythagoras's Theorem:
:$\sqrt {\paren {r \cos \theta}^2 + \paren {r \sin \theta}^2} = 1$
from which:
:$\sqrt {x^2 + y^2} = 1$
By definition of complex modulus:
:$\cmod z = 1$
and so $z \in P$.
Thus $C \subseteq P$.
Let $z \in P$.
Then by definition $\cmod z = 1$.
By Modulus of Complex Number equals its Distance from Origin, $z$ is $1$ unit away from $\tuple {0, 0}$.
By definition of $C$, it follows that $z \in C$.
Thus $P \subseteq C$.
The result follows by definition of set equality.
Category:Equation of Unit Circle in Complex Plane
Equation of Vertical Line

== Theorem ==
Let $\LL$ be a vertical line embedded in the Cartesian plane $\CC$.
Then the equation of $\LL$ can be given by:
:$x = a$
where $\tuple {a, 0}$ is the point at which $\LL$ intersects the $x$-axis.
:
== Proof ==
From the Normal Form of Equation of Straight Line in Plane, a general straight line can be expressed in the form:
:$x \cos \alpha + y \sin \alpha = p$
where:
:$p$ is the length of a perpendicular $\PP$ from $\LL$ to the origin.
:$\alpha$ is the angle made between $\PP$ and the $x$-axis.
As $\LL$ is vertical, then by definition $\PP$ is horizontal.
By definition, the horizontal line through the origin is the $x$-axis itself.
Thus $\alpha = 0$ and $p = a$
Hence the equation of $\LL$ becomes:
Hence the result.
== Also see ==
* Equation of Horizontal Line
Category:Equations of Straight Lines in Plane
Equation of Wave with Constant Velocity

== Theorem ==
Let $\phi$ be a wave which is propagated along the $x$-axis in the positive direction with constant velocity $c$ and without change of shape.
Let $\paren {\map \phi x}_{t \mathop = 0} = \map f x$ be the wave profile of $\phi$.
Then the disturbance of $\phi$ at point $x$ and time $t$ can be expressed using the equation:
:$\map \phi {x, t} = \map f {x - c t}$
where:
:$x$ denotes the displacement from the origin along the $x$-axis
:$t$ denotes the time.
=== Corollary ===
== Proof ==
Let us imagine a snapshot of $\phi$ at the time $t = 0$.
Then, , the wave $\phi$ is described by the equation:
:$\phi = \map f x$
Also , $\phi$ is propagated with no change of shape.
Hence, an imagined snapshot of $\phi$ at the general time $t$ will be:
:identical with that at $t = 0$
:moved along the $x$-axis a distance $s$.
By SUVAT:
:$s = c t$
in the positive direction along the $x$-axis.
Let us set the origin at the point $x = c t$.
Let the displacement measured from this new origin be $X$.
Then we have:
:$X = x + c t$
Hence the new equation for the wave profile of $\phi$ is:
:$\phi = \map f X$
But referred to that original fixed origin, this means:
:$\phi = \map f {x - c t}$
== Sources ==
* : Chapter $\text {I}$: The Equation of Wave Motion: $\S 2$: $(1)$
Category:Equation of Wave with Constant Velocity
Category:Examples of Use of Wave Equation
Equation of Wave with Constant Velocity/Corollary

== Theorem ==
Let $\phi$ be a wave which is propagated along the $x$-axis in the negative direction with constant velocity $c$ and without change of shape.
Let $\paren {\map \phi x}_{t \mathop = 0} = \map f x$ be the wave profile of $\phi$.
Then the disturbance of $\phi$ at point $x$ and time $t$ can be expressed using the equation:
:$\map \phi {x, t} = \map f {x + c t}$
where:
:$x$ denotes the displacement from the origin along the $x$-axis
:$t$ denotes the time.
== Proof ==
We have that the velocity of $\phi$ in the negative direction is $c$.
Hence the velocity of $\phi$ in the positive direction is $-c$.
By Equation of Wave with Constant Velocity:
:$\phi = \map f {x - \paren {-c} t}$
Hence the result.
== Sources ==
* : Chapter $\text {I}$: The Equation of Wave Motion: $\S 2$: $(2)$
Category:Equation of Wave with Constant Velocity
Equation of Wavefront of Plane Wave/Direction Cosine Form

== Theorem ==
Let $\phi$ be a plane wave.
Let an arbitrary wavefront of $\phi$ be denoted $P$.
Let the direction of propagation of $\phi$ be expressed as:
:$x : y : z = l : m : n$
where $l$, $m$ and $n$ are the direction cosines of the normal to $P$.
Then:
:$l x + m y + n z = K$
where $K$ is constant for a given plane wave $\phi$.
== Proof ==
== Sources ==
* : Chapter $\text {I}$: The Equation of Wave Motion: $\S 4$: $(12)$
Category:Equation of Wavefront of Plane Wave
Equation of Witch of Agnesi/Cartesian

== Theorem ==
:
The equation of the Witch of Agnesi is given in cartesian coordinates as:
:$y = \dfrac {8 a^3} {x^2 + 4 a^2}$
== Proof ==
Let $P = \tuple {x, y}$ and $A = \tuple {d, y}$.
We have that:
:$\dfrac {OM} {MN} = \dfrac {2 a} x = \dfrac y d$.
Also, by Pythagoras's Theorem:
:$\paren {a - y}^2 + d^2 = a^2 \implies y \paren {2 a - y} = d^2$
Eliminating $d$ gives us:
:$\dfrac {y^2} {y \paren {2 a - y} } = \dfrac {\paren {2 a}^2} {x^2}$
Hence:
:$\dfrac y {2 a - y} = \dfrac {4 a^2} {x^2}$
from which:
:$y = \dfrac {8 a^3} {x^2 + 4 a^2}$
== Also presented as ==
== Sources ==
* : $\S 11$: Special Plane Curves: Witch of Agnesi: $11.22$
* 
* : $\S 9$: Special Plane Curves: Witch of Agnesi: $9.22.$
Category:Witch of Agnesi
Equation of Witch of Agnesi/Parametric Form

== Theorem ==
:
The equation of the Witch of Agnesi can be presented in parametric form as:
:$\begin {cases} x = 2 a \cot \theta \\ y = a \paren {1 - \cos 2 \theta} \end {cases}$
== Proof ==
Let $P = \tuple {x, y}$ and $A = \tuple {d, y}$.
Let $\theta$ be the angle that $ON$ makes with the horizontal.
We have by definition of cotangent:
:$\dfrac {OM} {MN} = \dfrac {2 a} x = \cot \theta$
By Thales' Theorem $\angle OAM$ is a right angle.
Hence $\angle OMA = \theta$ and so:
:$OA = 2 a \cos \theta$
Thus:
:$2 a - y = 2 a \cos^2 \theta$
== Sources ==
* : $\S 11$: Special Plane Curves: Witch of Agnesi: $11.23$
* : $\S 9$: Special Plane Curves: Witch of Agnesi: $9.23.$
Category:Witch of Agnesi
Equation relating Points of Parallelogram in Complex Plane

== Theorem ==
Let $ABVU$ be a parallelogram in the complex plane whose vertices correspond to the complex numbers $a, b, v, u$ respectively.
Let $\angle BAU = \alpha$.
Let $\cmod {UA} = \lambda \cmod {AB}$.
:
Then:
:$u = \paren {1 - q} a + q b$
:$v = -q a + \paren {1 + q} b$
where:
:$q = \lambda e^{i \alpha}$
== Proof ==
From Geometrical Interpretation of Complex Subtraction, the four sides of $UABC$ can be defined as:
Thus: 
Then we have:
== Sources ==
* : $\S 2$. Geometrical Representations: Exercise $7$.
Category:Parallelograms
Category:Geometry of Complex Plane
Equation relating Points of Parallelogram in Complex Plane/Mistake

== Source Work ==
:
: Chapter $2$: Geometrical Representations
:: Exercise $7$
This mistake can be seen in the $1960$ edition as published by Routledge & Kegan Paul.
== Mistake ==
:The vertices parallelogram $ABVU$ are represented by the complex numbers $a, b, v, u$ respectively. The angle $UAB$ is equal to $\alpha$ and $\cmod {UA} = \lambda \cmod {AB}$. Prove that $u = \paren {1 - q} a + q b$ and $v = -q a + \paren {1 + q} b$, where $q = \lambda e^{i \alpha}$.
The configuration as described is:
:
However, because of the sense of $UAB$, this leads to:
:$u = \paren {1 - q} a + q b$ and $v = -q a + \paren {1 + q} b$, where $q = \lambda e^{-i \alpha}$.
The question as it was intended to be stated should contain:
:The angle $BAU$ is equal to $\alpha$ ...
See Equation relating Points of Parallelogram in Complex Plane for a solution to the correctly-stated exercise.
== Sources ==
* : $\S 2$. Geometrical Representations: Exercise $7$
Category:Mistakes/Complex Numbers/Walter Ledermann
Equations defining Plane Reflection/Cartesian

== Theorem ==
Let $\LL$ be a straight line through the origin $O$ of a cartesian plane.
Let the angle between $\LL$ and the $x$-axis be $\alpha$.
Let $\phi_\alpha$ denote the reflection in the plane whose axis is $\LL$.
Let $P = \tuple {x, y}$ be an arbitrary point in the plane.
Then:
:$\map {\phi_\alpha} P = \tuple {x \cos 2 \alpha + y \sin 2 \alpha, x \sin 2 \alpha - y \cos 2 \alpha}$
== Proof ==
:
Let $\LL$ reflect $P = \tuple {x, y}$ onto $P' = \tuple {x', y'}$.
Let $OP$ form an angle $\theta$ with the $x$-axis.
We have:
:$OP = OP'$
Thus:
Then:
and:
The result follows.
== Examples ==
== Sources ==
* 
* 
* 
Category:Geometric Reflections
Category:Equations defining Plane Reflection
Equations defining Plane Reflection/Examples/X-Axis

== Theorem ==
Let $\phi_x$ denote the reflection in the plane whose axis is the $x$-axis.
Let $P = \tuple {x, y}$ be an arbitrary point in the plane
Then:
:$\map {\phi_x} P = \tuple {x, -y}$
== Proof ==
From Equations defining Plane Reflection:
:$\map {\phi_\alpha} P = \tuple {x \cos 2 \alpha + y \sin 2 \alpha, x \sin 2 \alpha - y \cos 2 \alpha}$
where $\alpha$ denotes the angle between the axis and the $x$-axis.
By definition, the $x$-axis, being coincident with itself, is at a zero angle with itself.
Hence $\phi_x$ can be expressed as $\phi_\alpha$ in the above equations such that $\alpha = 0$.
Hence we have:
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations: Example $28.4$
Category:Equations defining Plane Reflection
Equations defining Plane Reflection/Examples/Y-Axis

== Theorem ==
Let $\phi_y$ denote the reflection in the plane whose axis is the $y$-axis.
Let $P = \tuple {x, y}$ be an arbitrary point in the plane
Then:
:$\map {\phi_y} P = \tuple {-x, y}$
== Proof ==
From Equations defining Plane Reflection:
:$\map {\phi_\alpha} P = \tuple {x \cos 2 \alpha + y \sin 2 \alpha, x \sin 2 \alpha - y \cos 2 \alpha}$
where $\alpha$ denotes the angle between the axis and the $x$-axis.
By definition, the $y$-axis, is perpendicular to the $x$-axis
Hence $\phi_y$ can be expressed as $\phi_\alpha$ in the above equations such that $\alpha = \dfrac \pi 2$ radians.
Hence we have:
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations: Example $28.4$
* 
* 
Category:Equations defining Plane Reflection
Equations defining Plane Rotation/Cartesian

== Theorem ==
Let $r_\alpha$ be the rotation of the plane about the origin through an angle of $\alpha$.
Let $P = \tuple {x, y}$ be an arbitrary point in the plane.
Then:
:$\map {r_\alpha} P = \tuple {x \cos \alpha - y \sin \alpha, x \sin \alpha + y \cos \alpha}$
== Proof ==
:
Let $r_\alpha$ rotate $P = \tuple {x, y}$ onto $P' = \tuple {x', y'}$.
Let $OP$ form an angle $\theta$ with the $x$-axis.
We have:
:$OP = OP'$
Thus:
Then:
and:
The result follows.
== Examples ==
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations: Example $28.2$
* : $2$: Continuity generalized: metric spaces: Exercise $2.6: 21$
Category:Geometric Rotations
Category:Equations defining Plane Rotation
Equations defining Plane Rotation/Examples/Right Angle

== Theorem ==
Let $r_\Box$ be the rotation of the plane about the origin through a right angle.
Let $P = \tuple {x, y}$ be an arbitrary point in the plane
Then:
:$\map {r_\Box} P = \tuple {y, -x}$
== Proof ==
From Equations defining Plane Rotation:
:$\map {r_\alpha} P = \tuple {x \cos \alpha - y \sin \alpha, x \sin \alpha + y \cos \alpha}$
where $\alpha$ denotes the angle of rotation.
Hence $r_\Box$ can be expressed as $r_\alpha$ in the above equations such that $\alpha = \dfrac \pi 2$.
Hence we have:
Category:Equations defining Plane Rotation
Equations defining Projection in Plane/Cartesian

== Theorem ==
Let $M$ and $N$ be distinct lines in the plane, both of which pass through the origin.
Let the angle between $M$ and the $x$-axis be $\theta$.
Let the angle between $N$ and the $x$-axis be $\phi$.
Let $\pr_{M, N}$ be the projection on $M$ along $N$:
:$\forall P \in \R^2: \map {\pr_{M, N} } P =$ the intersection of $M$ with the line through $P$ parallel to $N$.
Let $P = \tuple {x, y}$ be an arbitrary point in the plane.
Then:
:$\map {\pr_{M, N} } P = \begin {cases} 
\tuple {0, y - x \tan \phi} & : \theta = \dfrac \pi 2 \\
\tuple {x, x \tan \theta} & : \phi = \dfrac \pi 2 \\
\tuple {\dfrac {x \tan \phi} {\tan \phi - \tan \theta} - \dfrac y {\tan \phi - \tan \theta}, \dfrac {y \tan \theta} {\tan \theta - \tan \phi} - \dfrac {x \tan \theta \tan \phi} {\tan \theta - \tan \phi} } & : \text {otherwise} 
\end {cases}$
== Proof ==
Let $P = \tuple {x_1, y_1} \in \R^2$ be arbitrary.
:
From Equation of Straight Line in Plane: Slope-Intercept Form, we can express $M$ and $N$ as follows:
where:
Let $\pr_{M, N}$ map $P = \tuple {x_1, y_1}$ onto $P' = \tuple {x_2, y_2}$.
From Equation of Straight Line in Plane: Point-Slope Form, $PP'$ can be expressed as:
:$y - y_1 = n \paren {x - x_1}$
This holds also for $\tuple {x_2, y_2}$, and so:
Similarly:
The result follows by substituting for $m$ and $n$.
=== Special Cases ===
When either $M$ or $N$ are coincident with the $y$-axis, then either $\tan \theta$ or $\tan \phi$ is undefined.
Hence these cases are analysed as special cases.
Let $M$ be coincident with the $y$-axis.
:
By definition of the projection on $M$ along $N$:
:$x_2 = 0$
while from the definition of tangent:
:$\tan \phi = \dfrac {y_1 - y_2} {x_1}$
from which it follows that:
:$y_2 = y_1 - x_1 \tan \phi$
Hence if $P = \tuple {x, y}$ is an arbitrary point in the plane, then:
:$\map {\pr_{M, N} } P = \tuple {0, y - x \tan \phi}$
Let $N$ be coincident with the $y$-axis.
:
By definition of the projection on $M$ along $N$:
:$x_2 = x_1$
while from the definition of tangent:
:$\tan \theta = \dfrac {y_2} {x_2}$
from which it follows that:
:$y_2 = x_1 \tan \theta$
Hence if $P = \tuple {x, y}$ is an arbitrary point in the plane, then:
:$\map {\pr_{M, N} } P = \tuple {x, x \tan \theta}$
Category:Equations defining Projection in Plane
Equations defining Rotation of Axes of Cartesian Plane

== Theorem ==
Let $\CC$ be a Cartesian plane.
Let $P$ be the Cartesian coordinates of an arbitrary point in $\CC$.
Let $\tuple {x, y}$ be the Cartesian coordinates of $P$.
Let $\CC$ be subject to a rotation of axes about the origin through an angle $\theta$ clockwise.
Let $\tuple {x', y'}$ be the Cartesian coordinates of $P$ the coordinate axes after that rotation.
Then:
== Proof ==
== Sources ==
* 
* 
Category:Rotation of Axes
Category:Cartesian Plane