nyuuzyou/wikis huggingface dataset:
Permutation Test/Examples/Arbitrary Example 1

== Examples of Permutation Tests ==
Let the null hypothesis be that two independent samples of $5$ and $7$ observations are from identical populations.
Let the alternative hypothesis be that the probability distributions of these populations differ only by their arithmetic means.
Let the statistic $d$ be the absolute difference between the arithmetic means of the samples.
Then a test may be constructed based on $d$.
A large $d$ supports rejection of the null hypothesis.
Suppose $d = 3.5$.
If the null hypothesis holds, the $5 + 7 = 12$ observations can be treated as a pooled sample from one population.
That value of $d$ is calcuated for all $\dbinom {12} 5 = 792$ equally likely sample pairs of $5$ and $7$ obtained by random sampling without replacement from that pooled sample.
If $d \ge 3.5$ for $30$ of these $7792$ pairs, for example, then $p = \dfrac {30} {792} = 0.0379$ is the $p$-value that gives the exact probability of making an error of the first kind if the null hypothesis is rejected when the observed $d = 3.5$.
A $p$-value may be calculated for any observed $d$, or any sample sizes $m$ and $n$.
Exact $p$-values can be evaluated using appropriate computer software.
== Sources ==
* 
* 
Category:Examples of Permutation Tests
Permutation is Cyclic iff At Most One Non-Trivial Orbit

== Theorem ==
Let $S$ be a set.
Let $\rho: S \to S$ be a permutation on $S$.
Then:
:$\rho$ is a cyclic permutation
:
:$S$ has no more than one orbit under $\rho$ with more than one element.
== Proof ==
=== Necessary Condition ===
=== Sufficient Condition ===
Recall the definition of cyclic permutation.
Let:
:$\rho = \begin {bmatrix} i & \map \rho i & \ldots & \map {\rho^{k - 1} } i \end {bmatrix}$
Note that the orbit $\set {i, \map \rho i, \ldots, \map {\rho^{k - 1} } i}$ is the only non-trivial orbit.
(If $k = 1$, then the above-mentioned orbit is also trivial.)
== Also see ==
Some sources use this result as a definition for a cyclic permutation.
== Sources ==
* : $\S 1.6$
Category:Cyclic Permutations‎
Category:Permutations
Permutation is Product of Transpositions

== Theorem ==
Let $S_n$ denote the symmetric group on $n$ letters.
Every element of $S_n$ can be expressed as a product of transpositions.
== Proof ==
Let $\pi \in S_n$.
From Existence and Uniqueness of Cycle Decomposition, $\pi$ can be uniquely expressed as a cycle decomposition, up to the order of factors.
From K-Cycle can be Factored into Transpositions, each one of the cyclic permutations that compose this cycle decomposition can be expressed as a product of transpositions.
The result follows.
== Sources ==
* : Chapter $2$: The Symmetric Groups: $\S 80$: Corollary
* 
* 
Category:Permutations
Category:Transpositions
Permutation of Cosets

== Theorem ==
Let $G$ be a group and let $H \le G$.
Let $\mathbb S$ be the set of all distinct left cosets of $H$ in $G$.
Then:
:$(1): \quad$ For any $g \in G$, the mapping $\theta_g: \mathbb S \to \mathbb S$ defined by $\map {\theta_g} {x H} = g x H$ is a permutation of $\mathbb S$.
:$(2): \quad$ The mapping $\theta$ defined by $\map \theta g = \theta_g$ is a homomorphism from $G$ into the symmetric group on $\mathbb S$.
:$(3): \quad$ The kernel of $\theta$ is the subgroup $\ds \bigcap_{x \mathop \in G} x H x^{-1}$.
=== Corollary 1 ===
=== Corollary 2 ===
== Proof ==
First we need to show that $\theta_g$ is well-defined and injective.
Thus $\theta_g$ is well-defined and injective.
Then we see that $\forall x H \in \mathbb S: \map {\theta_g} {g^{-1} x H} = x H$, so $\theta_g$ is surjective.
Thus $\theta_g$ is a well-defined bijection on $\mathbb S$, and therefore a permutation on $\mathbb S$.
Next we see:
This shows that $\theta_{u v} = \theta_u \theta_v$, and thus:
:$\map \theta {u v} = \map \theta u \, \map \theta v$
Thus $\theta$ is a homomorphism.
Now to calculate $\map \ker \theta$:
as required.
== Sources ==
* : Chapter $9$: Permutations: Proposition $9.22$
Category:Symmetric Groups
Category:Cosets
Category:Permutations
Permutation of Cosets/Corollary 1

== Theorem ==
Let $G$ be a group.
Let $H \le G$ such that $\index G H = n$ where $n \in \Z$.
Then:
:$\exists N \lhd G: N \lhd H: n \divides \index G H \divides n!$
== Proof ==
Apply Permutation of Cosets to $H$ and let $N = \map \ker \theta$.
Then:
:$N \lhd G$ and $N \lhd H$
so from the Correspondence Theorem:
:$H / N \le G / N$
such that:
:$\index {G / N} {H / N} = n$
Thus:
:$n \divides \index G N$
Also by Permutation of Cosets:
:$\exists K \in S_n: G / N \cong K$
Thus:
:$\index G N \divides n!$
as required.
== Sources ==
* : Chapter $9$: Permutations: Corollary $9.23$
Category:Symmetric Groups
Category:Cosets
Permutation of Cosets/Corollary 2

== Corollary to Permutation of Cosets ==
Let $G$ be a group.
Let $p$ be the smallest prime such that:
: $p \divides \order G$
where $\divides$ denotes divisibility.
Let $\exists H: H \le G$ such that $\order H = p$.
Then $H$ is a normal subgroup of $G$.
== Proof ==
Apply Permutation of Cosets: Corollary 1 to $H$ to find some $N \lhd G$ such that: 
:$\index G N \divides p!$
Because $\index G N \divides \order G$, it divides $\gcd \set {\order G, p!}$.
Because $p$ is the smallest prime dividing $\order G$, it follows that:
:$\gcd \set {\order G, p!} = p$
Thus:
:$\index G N = p = \index G H$
Because $N \subseteq H$, it must follow that $N = H$.
== Sources ==
* : Chapter $9$: Permutations: Corollary $9.25$
Category:Symmetric Groups
Category:Normal Subgroups
Permutation of Determinant Indices

== Theorem ==
Let $\mathbf A = \sqbrk a_n$ be a square matrix of order $n$ over a field.
Let $\lambda: \N_{> 0} \to \N_{> 0}$ be any fixed permutation on $\N_{> 0}$.
Let $\map \det {\mathbf A}$ be the determinant of $\mathbf A$.
Let $\struct {S_n, \circ}$ be the symmetric group of $n$ letters. 
Then:
:$\ds \map \det {\mathbf A} = \sum_{\mu \mathop \in S_n} \paren {\map \sgn \mu \map \sgn \lambda \prod_{k \mathop = 1}^n a_{\map \lambda k, \map \mu k} }$
:$\ds \map \det {\mathbf A} = \sum_{\mu \mathop \in S_n} \paren {\map \sgn \mu \map \sgn \lambda \prod_{k \mathop = 1}^n a_{\map \mu k, \map \lambda k} }$
where:
:the summation $\ds \sum_{\mu \mathop \in S_n}$ ranges over all the $n!$ permutations of $\set {1, 2, \ldots, n}$
:$\map \sgn \mu$ is the sign of the permutation $\mu$.
== Proof ==
First it is shown that:
:$\ds \map \det {\mathbf A} = \sum_{\mu \mathop \in S_n} \paren {\map \sgn \mu \map \sgn \lambda \prod_{k \mathop = 1}^n a_{\map \lambda k, \map \mu k} }$
Let $\nu: \N_{> 0} \to \N_{> 0}$ be a permutation on $\N_{> 0}$ such that $\nu \circ \lambda = \mu$.
The product can be rearranged as:
:$\ds \prod_{k \mathop = 1}^n a_{\map \lambda k, \map \mu k} = a_{\map \lambda 1, \map \mu 1} a_{\map \lambda 2, \map \mu 2} \cdots a_{\map \lambda n, \map \mu n} = a_{1, \map \nu 1} a_{2, \map \nu 2} \cdots a_{n, \map \nu n} = \prod_{k \mathop = 1}^n a_{k, \map \nu k}$
from .
By Parity Function is Homomorphism:
:$\map \sgn \mu \map \sgn \lambda = \map \sgn \lambda \map \sgn \nu \map \sgn \lambda = \map {\sgn^2} \lambda \map \sgn \nu = \map \sgn \nu$
and so:
:$\ds \map \det {\mathbf A} = \sum_{\nu \mathop \in S_n} \paren {\map \sgn \nu \prod_{k \mathop = 1}^n a_{k, \map \nu k} }$
which is the usual definition for the determinant.
Next it is to be shown:
:$\ds \map \det {\mathbf A} = \sum_{\mu \mathop \in S_n} \paren {\map \sgn \mu \map \sgn \lambda \prod_{k \mathop = 1}^n a_{\map \mu k, \map \lambda k} }$
Let $\nu: \N_{> 0} \to \N_{> 0}$ be a permutation on $\N_{> 0}$ such that $\nu \circ \mu = \lambda$.
The result follows via a similar argument.
Category:Determinants
Permutation of Indices

== Theorem ==
Let $R: \Z \to \set {\T, \F}$ be a propositional function on the set of integers.
Let the fiber of truth of $R$ be finite.
Let $\pi$ is a permutation on the fiber of truth of $R$.
Then:
=== Permutation of Indices of Summation ===
=== Permutation of Indices of Product ===
Category:Summations
Category:Continued Products
Permutation of Indices of Product

== Theorem ==
Let $R: \Z \to \set {\T, \F}$ be a propositional function on the set of integers.
Let the fiber of truth of $R$ be finite.
Then:
:$\ds \prod_{\map R j} a_j = \prod_{\map R {\map \pi j} } a_{\map \pi j}$
where:
:$\ds \prod_{\map R j} a_j$ denotes the continued product over $a_j$ for all $j$ that satisfy the propositional function $\map R j$
:$\pi$ is a permutation on the fiber of truth of $R$.
== Proof ==
== Sources ==
* : $\S 1.2.3$: Sums and Products
Category:Continued Products
Permutation of Indices of Summation

== Theorem ==
Let $R: \Z \to \set {\T, \F}$ be a propositional function on the set of integers.
Let the fiber of truth of $R$ be finite.
Then:
:$\ds \sum_{\map R j} a_j = \sum_{\map R {\map \pi j} } a_{\map \pi j}$
where:
:$\ds \sum_{\map R j} a_j$ denotes the summation over $a_j$ for all $j$ that satisfy the propositional function $\map R j$
:$\pi$ is a permutation on the fiber of truth of $R$.
=== Infinite Series ===
== Proof ==
== Also known as ==
The operation of permutation of indices of a summation can be seen referred to as a permutation of the range.
However, as the term range is ambiguous in the literature, and as its use here is not strictly accurate (it is the fiber of truth of $R$, not its range, which is being permuted, its use on is discouraged.
== Sources ==
* : $\S 1.2.3$: Sums and Products: $(5)$
Category:Summations
Permutation of Indices of Summation/Infinite Series

== Theorem ==
Let $R: \Z \to \set {\T, \F}$ be a propositional function on the set of integers.
Let the fiber of truth of $R$ be infinite.
Let $\ds \sum_{\map R i} a_i$ be absolutely convergent.
Then:
:$\ds \sum_{\map R j} a_j = \sum_{\map R {\map \pi j} } a_{\map \pi j}$
where:
:$\ds \sum_{\map R j} a_j$ denotes the summation over $a_j$ for all $j$ that satisfy the propositional function $\map R j$
:$\pi$ is a permutation on the fiber of truth of $R$.
== Proof ==
This is a restatemtent of Manipulation of Absolutely Convergent Series: Permutation in the context of summations.
== Sources ==
* : $\S 1.2.3$: Sums and Products
Category:Summations
Permutation of Indices of Supremum

== Theorem ==
Let $\family {a_i}_{i \mathop \in I}$ be a family of elements of the non-negative real numbers $\R_{\ge 0}$ indexed by $I$.
Let $\map R i$ be a propositional functions of $i \in I$.
Let $\ds \sup_{\map R i} a_i$ be the indexed supremum on $\family {a_i}$.
Then:
:$\ds \sum_{\map R i} a_i = \sum_{\map R {\map \pi i} } a_{\map \pi i}$
where $\pi$ is a permutation on the fiber of truth of $R$.
== Proof ==
== Sources ==
* : $\S 1.2.3$: Sums and Products: Exercise $35$
Category:Suprema
Permutation of Set is Automorphism of Set under Left Operation

== Theorem ==
Let $S$ be a set.
Let $\struct {S, \gets}$ be the algebraic structure formed from $S$ under the left operation.
Let $f$ be a permutation on $S$.
Then $f$ is an automorphism of $f$.
== Proof ==
We have that $f$ is a permutation and so a fortiori a bijection.
It remains to show that $f$ is a homomorphism.
So, let $a, b \in S$ be arbitrary.
We have:
The result follows.
== Also see ==
* Permutation of Set is Automorphism of Set under Right Operation
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 6$: Isomorphisms of Algebraic Structures: Exercise $6.11 \ \text {(a)}$
Category:Examples of Automorphisms
Category:Left Operation
Permutation of Set is Automorphism of Set under Right Operation

== Theorem ==
Let $S$ be a set.
Let $\struct {S, \to}$ be the algebraic structure formed from $S$ under the right operation.
Let $f$ be a permutation on $S$.
Then $f$ is an automorphism of $f$.
== Proof ==
We have that $f$ is a permutation and so a fortiori a bijection.
It remains to show that $f$ is a homomorphism.
So, let $a, b \in S$ be arbitrary.
We have:
The result follows.
== Also see ==
* Permutation of Set is Automorphism of Set under Left Operation
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 6$: Isomorphisms of Algebraic Structures: Exercise $6.11 \ \text {(a)}$
Category:Examples of Automorphisms
Category:Right Operation
Permutation of Variables of Primitive Recursive Function

== Theorem ==
Let $f: \N^k \to \N$ be a primitive recursive function.
Let $\sigma$ be a permutation of $\set {1, 2, \ldots, k}$.
Then the function $h: \N^k \to \N$ defined as:
:$\map h {n_1, n_2, \ldots, n_k} = \map f {n_{\map \sigma 1}, n_{\map \sigma 2}, \ldots, n_{\map \sigma k} }$
is also primitive recursive.
== Proof ==
We have that:
:$\forall j \in \set {1, 2, \ldots, k}: n_{\map \sigma j} = \pr^k_{\map \sigma j}$.
Thus $h$ is obtained by substitution from $f$ and the projection functions $\pr^k_{\map \sigma j}$.
The result follows.
It follows that if a function $h$ can be obtained from known primitive recursive functions by primitive recursion where a variable other than the last one is taken as the recursion variable, then $h$ is primitive recursive.
Category:Primitive Recursive Functions
Permutation on Polynomial/Examples/Polynomial on 3 Variables

== Examples of Permutations on Polynomials ==
Consider the polynomial on $3$ variables:
:$\map f {x_1, x_2, x_3} = {x_1}^2 + 2 x_1 x_2 = 4 x_1 x_2 {x_3}^2$
Let $\rho := \begin{pmatrix} 1 & 2 & 3 \end{pmatrix}$ be a permutation on the Symmetric Group on 3 Letters $S_3$.
Then:
:$\rho \circ f = {x_2}^2 + 2 x_2 x_3 = 4 x_2 x_3 {x_1}^2$
== Sources ==
* : Chapter $9$: Permutations: Example $9.12$
Category:Examples of Permutations on Polynomials
Permutation on Polynomial is Group Action

== Theorem ==
Let $n \in \Z: n > 0$.
Let $F_n$ be the set of all polynomials in $n$ variables $x_1, x_2, \ldots, x_n$:
:$F = \set {\map f {x_1, x_2, \ldots, x_n}: f \text{ is a polynomial in $n$ variables} }$
Let $S_n$ denote the symmetric group on $n$ letters.
Let $*: S_n \times F \to F$ be the mapping defined as:
:$\forall \pi \in S_n, f \in F: \pi * \map f {x_1, x_2, \ldots, x_n} = \map f {x_{\map \pi 1}, x_{\map \pi 2}, \ldots, x_{\map \pi n} }$
Then $*$ is a group action.
== Proof ==
Let $\pi, \rho \in S_n$.
Let $\pi * f$ be the permutation on the polynomial $f$ by $\pi$.
Let $e \in S_n$ be the identity of $S_n$.
From Symmetric Group is Group:
:$e * f = f$
thus fulfilling .
Then we have that:
thus fulfilling 
== Also see ==
* Stabilizer of Polynomial
== Sources ==
* : Chapter $6$: Cosets: Exercise $10$
* : Chapter $9$: Permutations: Proposition $9.13$
* : Chapter $10$: The Orbit-Stabiliser Theorem: Example $10.3$
Category:Group Actions
Category:Permutations
Perpendicular Axes Theorem

== Theorem ==
Let $\LL$ be a lamina.
Let $O_x$ and $O_y$ be axes of rotation in the plane of $\LL$ perpendicular to each other and intersecting at $O$.
Let $O_z$ be an axis of rotation of $\LL$ through $O$ perpendicular to both $O_x$ and $O_y$.
Let $I_x$, $I_y$ and $I_z$ be the moments of inertia of $\LL$ about $O_x$, $O_y$ and $O_z$ respectively.
Then:
:$I_z = I_x + I_y$
== Proof ==
== Also see ==
* Parallel Axes Theorem, also known as the Huygens-Steiner Theorem
== Sources ==
* 
* 
Category:Moments of Inertia
Category:Named Theorems
Perpendicular Bisector is Locus of Points Equidistant from Endpoints

== Theorem ==
Let $AB$ be a straight line segment.
The locus of points which are equidistant from $A$ and $B$ is the perpendicular bisector of $AB$.
== Proof ==
Let $E$ denote the set of points equidistant from $A$ and $B$.
Let $F$ denote the set of points on the perpendicular bisector of $AB$.
We are to show that $E = F$.
:
First we show that $F \subseteq E$.
Let $P \in F$.
Hence by definition $P$ is on the perpendicular bisector of $AB$.
Let $C$ be the intersection of $AB$ and the perpendicular bisector.
We have:
:$\angle PCA = \angle PCB = 90 \degrees$
:$AC = CB$
:$PC$ is common
By Pythagoras's Theorem:
:$AP^2 = AC^2 + PC^2 = BC^2 + PC^2 = BP^2$
and so:
:$AP = BP$
That is, $P$ is equidistant from $A$ and $B$, and so:
:$P \in E$
Hence:
:$F \subseteq E$
Let $P \in E$ be arbitrary.
That is, let $P$ be equidistant from $A$ and $B$.
Drop a perpendicular $PC$ to $AB$.
We inspect the triangles $\triangle PCA$ and $\triangle PCA$
We have:
:$PA = PB$
:$\angle PCA = \angle PCB = 90 \degrees$
:$PC$ is common
By Pythagoras's Theorem:
:$AC^2 = PA^2 - PC^2$
:$BC^2 = PB^2 - PC^2$
But as $PA = PB$ it follows that:
:$AC = BC$
Thus $PC$ is the perpendicular bisector of $AB$.
That is:
:$P \in F$
Hence by definition of subset:
:$E \subseteq F$
We have:
:$E \subseteq F$
and:
:$F \subseteq E$
and so by definition of set equality:
:$E = F$
Hence the result.
== Sources ==
* : $\text {IV}$. Pure Geometry: Plane Geometry: The circumcentre
* 
Category:Perpendicular Bisectors
Perpendicular Bisector of Chord Passes Through Center

== Theorem ==
The perpendicular bisector of any chord of any given circle must pass through the center of that circle.
== Proof ==
:
Let $F$ be the center of the circle in question.
Draw any chord $AB$ on the circle.
Bisect $AB$ at $D$.
Construct $CE$ perpendicular to $AB$ at $D$, where $D$ and $E$ are where this perpendicular meets the circle.
Then the center $F$ lies on $CE$.
The proof is as follows.
Join $FA, FD, FB$.
As $F$ is the center, $FA = FB$.
Also, as $D$ bisects $AB$, we have $DA = DB$.
As $FD$ is common, then from Triangle Side-Side-Side Congruence, $\triangle ADF = \triangle BDF$.
In particular, $\angle ADF = \angle BDF$; both are right angles.
From :
:
So $\angle ADF$ and $\angle BDF$ are both right angles.
Thus, by definition, $F$ lies on the perpendicular bisector of $AB$.
Hence the result.
== Historical Note ==
== Sources ==
* : Book $\text{III}$. Propositions
Category:Perpendicular Bisectors
Category:Circles
Perpendicular Bisector of Chord Passes Through Center/Historical Note

== Historical Note on Perpendicular Bisector of Chord Passes Through Center ==
The argument for this particular result originates from Proposition $1$ of Book $\text{III}$ of .
However, the result itself is due to , who reasoned that this result was more fundamental.
This theorem is the converse of .
== Sources ==
* : Book $\text{III}$. Propositions
Category:Historical Notes
Perpendicular Bisector of Triangle is Altitude of Medial Triangle

== Theorem ==
Let $\triangle ABC$ be a triangle.
Let $\triangle DEF$ be the medial triangle of $\triangle ABC$.
Let a perpendicular bisector be constructed on $AC$ at $F$ to intersect $DE$ at $P$.
Then $FP$ is an altitude of $\triangle DEF$.
== Proof ==
:
Consider the triangles $\triangle ABC$ and $\triangle DBE$.
By the Midline Theorem:
:$BA : BD = 2 : 1$
:$BC : BE = 2 : 1$
and:
:$AC \parallel DE$
From Parallelism implies Equal Alternate Angles:
:$\angle AFP = \angle FPE$
By construction, $\angle AFP$ is a right angle.
Thus $\angle FPE$ is also a right angle.
That is, $FP$ is perpendicular to $DE$.
By construction, $FP$ passes through the vertex $F$ of $\triangle DEF$.
The result follows by definition of altitude of triangle.
Category:Perpendicular Bisectors
Category:Medial Triangles
Perpendicular Bisectors of Triangle Meet at Point

== Theorem ==
Let $\triangle ABC$ be a triangle.
The perpendicular bisectors of $AB$, $BC$ and $AC$ all intersect at the same point.
== Proof ==
Let the perpendicular bisectors of $AC$ and $AB$ be constructed through $D$ and $E$ respectively to meet at $F$.
:
By definition of perpendicular bisector:
:$AE = EB$
and:
:$\angle AEF = \angle BEF$ are right angles.
From Triangle Side-Angle-Side Congruence:
:$\triangle AEF = \triangle BEF$
and so $AF = BF$.
Similarly, by definition of perpendicular bisector:
:$AD = DC$
and:
:$\angle ADF = \angle CDF$ are right angles.
From Triangle Side-Angle-Side Congruence:
:$\triangle ADF = \triangle CDF$
and so $AF = CF$.
Thus:
:$BF = CF$
Let $FG$ be the angle bisector of $\angle BFC$.
We have:
:$BF = CF$ from above
and:
:$\angle BFG = \angle CFG$ by construction
Thus by Triangle Side-Angle-Side Congruence:
:$\triangle BFG = \triangle CFG$
and so $BG = CG$.
Thus as $\angle BGF = \angle CGF$ it follows that they are both right angles.
Thus $FG$ is the perpendicular bisector of $BC$.
Thus we have all three perpendicular bisectors of the sides of $ABC$ meeting at the same point $G$.
== Also see ==
* Circumscribing Circle about Triangle, where it can be seen that $F$ is the circumcenter of $\triangle ABC$.
== Sources ==
* : $\text {IV}$. Pure Geometry: Plane Geometry: The circumcentre
* 
Category:Perpendicular Bisectors
Category:Triangles
Perpendicular Distance from Straight Line in Plane to Origin

== Theorem ==
Let $L$ be the straight line embedded in the cartesian plane whose equation is given as:
:$a x + b y = c$
Then the perpendicular distance $d$ between $L$ and $\tuple {0, 0}$ is given by:
:$d = \size {\dfrac c {\sqrt {a^2 + b^2} } }$
== Proof ==
From Perpendicular Distance from Straight Line in Plane to Point, the perpendicular distance $d$ between $L$ and the point $\tuple {x_0, y_0}$ is given by:
:$d = \dfrac {\size {a x_0 + b y_0 + c} } {\sqrt {a^2 + b^2} }$
The result follows by setting $x_0 = 0$ and $y_0 = 0$.
== Examples ==
Category:Perpendicular Distance from Straight Line in Plane to Point
Perpendicular Distance from Straight Line in Plane to Origin/Examples/Example 1

== Examples of Perpendicular Distance from Straight Line in Plane to Origin ==
Let $\LL$ be the straight line defined by the equation:
:$a x - b y = 1$
The perpendicular distance $d$ from $\LL$ to the origin $\tuple {0, 0}$ is given by:
:$d = \dfrac 1 {\sqrt {a^2 + b^2} }$
== Proof ==
From Perpendicular Distance from Straight Line in Plane to Origin, $d$ is given by:
:$d = \dfrac 1 {\sqrt {a^2 + \paren {-b}^2} }$
from which the result follows.
== Sources ==
* : $\text {2-3}$ The Linear Diophantine Equation: Exercise $7$
Category:Perpendicular Distance from Straight Line in Plane to Point
Perpendicular Distance from Straight Line in Plane to Point/General Form

== Theorem ==
Let $\LL$ be a straight line embedded in a cartesian plane, given by the equation:
:$a x + b y + c = 0$
Let $P$ be a point in the cartesian plane whose coordinates are given by:
:$P = \tuple {x_0, y_0}$
Then the perpendicular distance $d$ from $P$ to $\LL$ is given by:
:$d = \dfrac {\size {a x_0 + b y_0 + c} } {\sqrt {a^2 + b^2} }$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $6$. Distance of a point from a line
* : $\S 10$: Formulas from Plane Analytic Geometry: Distance from Point $\tuple {x_1, y_1}$ to Line $A x + B y + C = 0$: $10.8$
* : $\S 8$: Formulas from Plane Analytic Geometry: Distance from Point $\tuple {x_1, y_1}$ to Line $A x + B y + C = 0$: $8.8.$
* : Appendix $4$: Geometry: equations of lines and planes
* : Appendix $5$: Geometry: equations of lines and planes
Category:Perpendicular Distance from Straight Line in Plane to Point