nyuuzyou/wikis huggingface dataset:
Equation of Nephroid

== Theorem ==
Let $H$ be the nephroid generated by the epicycle $C_1$ of radius $b$ rolling without slipping around the outside of a deferent $C_2$ of radius $a = 2 b$.
Let $C_2$ be embedded in a cartesian plane with its center $O$ located at the origin.
Let $P$ be a point on the circumference of $C_1$.
Let $C_1$ be initially positioned so that $P$ is its point of tangency to $C_2$, located at point $A = \tuple {a, 0}$ on the $x$-axis.
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 parametric equation:
:$\begin{cases} x & = 3 b \cos \theta - b \cos 3 \theta \\ y & = 3 b \sin \theta - b \sin 3 \theta \end{cases}$
where $\theta$ is the angle between the $x$-axis and the line joining the origin to the center of $C_1$.
== Proof ==
By definition, a nephroid is an epicycloid with $2$ cusps.
:
By Equation of Epicycloid, the equation of $H$ is given by:
:$\begin{cases} x & = \paren {a + b} \cos \theta - b \map \cos {\paren {\dfrac {a + b} b} \theta} \\ y & = \paren {a + b} \sin \theta - b \map \sin {\paren {\dfrac {a + b} b} \theta} \end{cases}$
When $a = 2 b$ the equation of $H$ is now given by:
:$\begin{cases} x & = 3 b \cos \theta - b \cos 3 \theta \\ y & = 3 b \sin \theta - b \sin 3 \theta \end{cases}$
== Sources ==
* : Chapter $\text {B}.21$: The Cycloid: Problem $14$
Category:Nephroids
Equation of Normal to Circle Centered at Origin

== Theorem ==
Let $\CC$ be a circle whose radius is $r$ and whose center is at the origin of a Cartesian plane.
Let $P = \tuple {x_1, y_1}$ be a point on $\CC$.
Let $\NN$ be a normal to $\CC$ passing through $P$.
Then $\NN$ can be defined by the equation:
:$y_1 x - x_1 y = 0$
== Proof ==
Let $\TT$ be the tangent to $\CC$ passing through $P$.
From Equation of Tangent to Circle Centered at Origin, $\TT$ can be described using the equation:
:$x x_1 + y y_1 = r^2$
expressible as:
:$y - y_1 = -\dfrac {x_1} {y_1} \paren {x - x_1}$
where the slope of $\TT$ is $-\dfrac {x_1} {y_1}$.
By definition, the normal is perpendicular to the tangent.
From Condition for Straight Lines in Plane to be Perpendicular, the slope of $\NN$ is $\dfrac {y_1} {x_1}$
Hence the equation for $\NN$ is:
Hence the result.
== Sources ==
* : Chapter $\text {III}$. The Circle: $3$. The normal
Category:Circles
Category:Normals to Curves
Equation of Ovals of Cassini/Cartesian Form

== Theorem ==
Let $P_1$ and $P_2$ be points in the cartesian plane located at $\tuple {a, 0}$ and $\tuple {-a, 0}$ for some constant $a \ne 0$.
Let $b$ be a real constant.
The Cartesian equation:
:$\paren {x^2 + y^2 + a^2}^2 - 4 a^2 x^2 = b^4$
describes the ovals of Cassini.
:
== Proof ==
The ovals of Cassini are the loci of points $M$ in the plane such that:
:$P_1 M \times P_2 M = b^2$
Let $b$ be chosen.
Let $P = \tuple {x, y}$ be an arbitrary point of $M$.
We have:
Hence the result.
== Also presented as ==
== Also see ==
* Equation of Ovals of Cassini/Polar Form
== Sources ==
* 
* 
* 
Category:Ovals of Cassini
Equation of Ovals of Cassini/Cartesian Form/Also presented as

== Equation of Ovals of Cassini: Cartesian Form: Also presented as ==
Some sources present the equation defining the ovals of Cassini as:
:$\paren {\paren {x^2 + a^2} + y^2} \paren {\paren {x^2 - a^2} + y^2} = b^4$
== Sources ==
* 
* 
* 
Category:Ovals of Cassini
Equation of Ovals of Cassini/Polar Form

== Theorem ==
Let $P_1$ and $P_2$ be points in a polar coordinate plane located at $\polar {a, 0}$ and $\polar {a, \pi}$ for some constant $a \ne 0$.
Let $b$ be a real constant.
The polar equation:
:$r^4 + a^4 - 2 a^2 r^2 \cos 2 \theta = b^4$
describes the ovals of Cassini.
:
== Proof ==
The ovals of Cassini are the loci of points $M$ in the plane such that:
:$P_1 M \times P_2 M = b^2$
Let $b$ be chosen.
Let $P = \tuple {x, y}$ be an arbitrary point of $M$.
Let $d_1 = \size {P_1 P}$ and $d_2 = \size {P_2 P}$.
We have:
Hence the result.
== Also see ==
* Equation of Ovals of Cassini/Cartesian Form
== Sources ==
* : $\S 11$: Special Plane Curves: Ovals of Cassini: $11.31$
* : $\S 9$: Special Plane Curves: Ovals of Cassini: $9.31.$
* 
Category:Ovals of Cassini
Equation of Parabola in Reduced Form

== Theorem ==
Let $K$ be a parabola aligned in a cartesian plane in reduced form.
That is:
:$(1)$ its focus is at the point $\tuple {c, 0}$
:$(2)$ its directrix is aligned with the line $x = -c$
for some $c \in \R_{> 0}$.
=== Cartesian Frame ===
=== Parametric Form ===
== Also see ==
* Equation of Ellipse in Reduced Form
* Equation of Hyperbola in Reduced Form
Category:Equation of Parabola in Reduced Form
Category:Parabolas
Equation of Plane/General Equation

== Theorem ==
A plane $P$ is the set of all $\tuple {x, y, z} \in \R^3$, where:
:$\alpha_1 x + \alpha_2 y + \alpha_3 z = \gamma$
where $\alpha_1, \alpha_2, \alpha_3, \gamma \in \R$ are given, and not all of $\alpha_1, \alpha_2, \alpha_3$ are zero.
== Proof ==
== Also presented as ==
== Sources ==
* : Chapter $\text V$: Vector Spaces: $\S 28$: Linear Transformations
* 
Category:General Equation of Plane
Category:Equations of Planes
Equation of Plane/Intercept Form

== Theorem ==
Let $P$ be a plane which intercepts the $x$-axis, $y$-axis and $z$-axis respectively at $\tuple {a, 0, 0}$, $\tuple {0, b, 0}$ and $\tuple {0, 0, c}$, where $a b c \ne 0$.
Then $P$ can be described by the equation:
:$\dfrac x a + \dfrac y b + \dfrac y c = 1$
== Proof ==
== Sources ==
* 
* 
Category:Intercept Form of Equation of Plane
Category:Equations of Planes
Equation of Plane/Normal Form

== Theorem ==
Let $P$ be a plane.
Let $\LL$ be the normal to $P$ through the origin.
Let:
:the length of $\LL$ be $p$
:the direction cosines of $\LL$ be $l$, $m$ and $n$.
Then $P$ can be defined by the equation:
:$l x + m y + n z = p$
== Proof ==
== Sources ==
* 
* 
Category:Normal Form of Equation of Plane
Category:Equations of Planes
Equation of Plane/Vector Form

== Theorem ==
Let $P$ be a plane which passes through a point $C$ whose position vector relative to the origin $O$ is $\mathbf c$.
Let $\mathbf p$ be the vector perpendicular to $P$ from $O$.
Let $\mathbf r$ be the position vector of an arbitrary point on $P$.
Then $P$ can be represented by the equation:
:$\mathbf p \cdot \paren {\mathbf r - \mathbf c} = 0$
where $\cdot$ denotes dot product.
== Proof ==
:
It is seen that $\mathbf r - \mathbf c$ lies entirely within the plane $P$.
As $P$ is perpendicular to $\mathbf p$, it follows that $\mathbf r - \mathbf c$ is perpendicular to $\mathbf p$.
Hence by Dot Product of Perpendicular Vectors:
:$\mathbf p \cdot \paren {\mathbf r - \mathbf c} = 0$
== Also presented as ==
== Sources ==
* : Chapter $\text {II}$: The Products of Vectors: $2$. The Scalar Product
Category:Vector Equation of Plane
Category:Equations of Planes
Equation of Plane Shear

== Theorem ==
Let $P$ be a plane shear whose fixed line is the $x$-axis of a Cartesian plane.
Then $P$ can be described by the equation:
:$\tuple {x, y} \mapsto \tuple {x + y \tan \theta, y}$
where $\theta$ is the angle through which a straight line perpendicular to the $x$-axis is moved by $P$.
== Proof ==
== Sources ==
* 
* 
Category:Plane Shears
Equation of Plane Shear/Mistake

== Source Work ==
:
:shear 2.
== Mistake ==
:The shear of the plane with invariant line $x = 0$ and angle $\theta$ maps the point with Cartesian coordinates $\tuple {x, y}$ to $\tuple {x + y \tan \theta, y}$.
== Correction ==
The invariant line for such a shear is actually the $x$-axis, that is, the line $y = 0$.
== Sources ==
* 
* 
Category:Mistakes/The Penguin Dictionary of Mathematics
Equation of Plane Wave/Direction Cosine Form

== Theorem ==
Let $\phi$ be a plane wave propagated with velocity $c$.
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 $\phi$ can be expressed as:
:$\map \phi {x, y, z, t} = \map f {l x + m y + n z - c t}$
== Proof ==
By Equation of Wavefront of Plane Wave, the equation of the wavefront of $\phi$ is given by:
:$l x + m y + n z = K$
Hence it is clear that:
:$\map \phi {x, y, z, t} = \map f {l x + m y + n z - c t}$
is a function which fulfils all the requirements to be a plane wave.
Hence $\phi$ as defined represents a plane wave propagated with velocity $c$ in the direction given.
== Sources ==
* : Chapter $\text {I}$: The Equation of Wave Motion: $\S 4$: $(13)$
Category:Equation of Plane Wave
Equation of Plane Wave is Particular Solution of Wave Equation/Direction Cosine Form

== Theorem ==
Let $\phi$ be a plane wave propagated with velocity $c$ in a Cartesian $3$-space.
Let $\phi$ be expressed as:
:$\map \phi {x, y, z, t} = \map f {l x + m y + n z - c t}$
where $l$, $m$ and $n$ are the direction cosines of the normal to $P$.
Then $\phi$ satisfies the wave equation.
== Proof ==
The wave equation is expressible as:
:$\dfrac 1 {c^2} \dfrac {\partial^2 \phi} {\partial t^2} = \dfrac {\partial^2 \phi} {\partial x^2} + \dfrac {\partial^2 \phi} {\partial y^2} + \dfrac {\partial^2 \phi} {\partial z^2}$
== Sources ==
* : Chapter $\text {I}$: The Equation of Wave Motion: $\S 5$
Category:Equation of Plane Wave is Particular Solution of Wave Equation
Equation of Polar of Point with respect to Circle

== Theorem ==
Let $\CC$ be a circle whose radius is $r$ and whose center is at the origin of a Cartesian plane.
Let $P = \tuple {x_0, y_0}$ be an arbitrary point in the Cartesian plane.
The polar of $P$ with respect to $\CC$ is the straight line whose equation is given by:
:$x x_0 + y y_0 = r^2$
== Proof ==
== Sources ==
* : Chapter $\text {III}$. The Circle: $4$. Pole and polar
Category:Polars of Points
Category:Circles
Equation of Polar of Point with respect to Ellipse

== Theorem ==
Let $\KK$ be an ellipse embedded in a Cartesian plane in reduced form with the equation:
:$\dfrac {x^2} {a^2} + \dfrac {y^2} {b^2} = 1$
Let $P = \tuple {x_0, y_0}$ be an arbitrary point in the Cartesian plane.
The polar of $P$ with respect to $\KK$ is the straight line whose equation is given by:
:$\dfrac {x x_0} {a^2} + \dfrac {y y_0} {b^2} = 1$
== Proof ==
== Also see ==
* Definition:Chord of Contact on Ellipse: when $P$ is specifically outside $\KK$
== Sources ==
* : Chapter $\text {IV}$. The Ellipse: $3$.
Category:Polars of Points
Category:Ellipses
Equation of Rectangular Hyperbola in Reduced Form

== Theorem ==
Let $\KK$ be a rectangular hyperbola whose transverse axis and conjugate axis are of length $2 a$.
Let $\KK$ be aligned in a cartesian plane in reduced form.
$\KK$ can be expressed by the equation:
:$x^2 - y^2 = a^2$
== Proof ==
From Equation of Hyperbola in Reduced Form in Cartesian Frame, a hyperbola can be expressed by the equation:
:$\dfrac {x^2} {a^2} - \dfrac {y^2} {b^2} = 1$
For a rectangular hyperbola:
:$a = b$
Hence $\KK$ can be expressed by the equation:
:$\dfrac {x^2} {a^2} - \dfrac {y^2} {a^2} = 1$
Multiplying both sides by $a^2$
:$x^2 - y^2 = a^2$
== Sources ==
* 
* 
Category:Rectangular Hyperbolas
Equation of Rectangular Hyperbola in Standard Form/Parametric Form

== Theorem ==
Let $\KK$ be a rectangular hyperbola in standard form.
$\KK$ can be expressed in parametric form as:
:$\begin {cases} x = c t \\ y = \dfrac c t \end {cases}$
== Proof ==
Let the point $\tuple {x, y}$ satisfy the equations:
Then:
== Sources ==
* 
* 
Category:Equation of Hyperbola in Reduced Form
Equation of Sphere/Rectangular Coordinates

== Theorem ==
The equation of a sphere with radius $R$ and center $\tuple {a, b, c}$ expressed in Cartesian coordinates is:
:$\paren {x - a}^2 + \paren {y - b}^2 + \paren {z - c}^2 = R^2$
=== Corollary ===
== Proof ==
Let the point $\tuple {x, y, z}$ satisfy the equation:
:$(1): \quad \paren {x - a}^2 + \paren {y - b}^2 + \paren {z - c}^2 = R^2$
By the Distance Formula in 3 Dimensions, the distance between this $\tuple {x, y, z}$ and $\tuple {a, b, c}$ is:
:$\sqrt {\paren {x - a}^2 + \paren {y - b}^2 + \paren {z - c}^2}$
But from equation $(1)$, this quantity equals $R$.
Therefore the distance between points satisfying the equation and the center is constant and equal to the radius.
Thus $\tuple {x, y, z}$ lies on the surface of a sphere with radius $R$ and center $\tuple {a, b, c}$.
Now suppose that $\tuple {x, y, z}$ does not satisfy the equation:
:$\paren {x - a}^2 + \paren {y - b}^2 + \paren {z - c}^2 = R^2$
Then by the same reasoning as above, the distance between $\tuple {x, y, z}$ and $\tuple {a, b, c}$ does not equal $R$.
Therefore $\tuple {x, y, z}$ does not lie on the surface of a sphere with radius $R$ and center $\tuple {a, b, c}$.
Hence it follows that the points satisfying $(1)$ are exactly those points which are the sphere in question.
== Sources ==
* : $12.20$: Formulas from Solid Analyic Geometry: Equation of Sphere in Rectangular Coordinates
* 
* 
* 
Category:Spheres
Category:Examples of Surfaces
Equation of Sphere/Rectangular Coordinates/Corollary

== Theorem ==
The equation of a sphere with radius $R$ whose center is at the origin expressed in Cartesian coordinates is:
:$x^2 + y^2 + z^2 = R^2$
== Proof ==
From Equation of Sphere in Rectangular Coordinates, the equation of a sphere with radius $R$ and center $\tuple {a, b, c}$ expressed in Cartesian coordinates is:
:$\paren {x - a}^2 + \paren {y - b}^2 + \paren {z - c}^2 = R^2$
Setting $a = b = c = 0$ yields the result.
== Sources ==
* : Chapter $6$: Curves and Coordinates: Descartes
Category:Spheres
Category:Examples of Surfaces
Equation of Straight Line Tangent to Circle

== Theorem ==
Let $\tuple {a, b}$ be the center of a circle $\CC$ whose radius is $r$.
Let $P_n = \tuple {x_n, y_n}$ be any point on $\CC$.
The equation of a non-vertical tangent line $\TT$ to $\CC$ is given by:
:$y - y_n = \dfrac {a - x_n} {y_n - b} \paren {x - x_n}$
The equations of the vertical tangent lines to $\CC$ are:
:$x = r - a$ for $P = \tuple {r - a, b}$
:$x = a - r$ for $P = \tuple {a - r, b}$
== Proof ==
=== Non-Vertical Tangent Lines ===
From Equation of Circle, $\CC$ can be described on the $x y$-plane in the form:
:$\paren {x - a}^2 + \paren {y - b}^2 = r^2$
where $P = \tuple {a, b}$ is the center of the circle and $r$ is the radius. 
We use the definition of the derivative as the gradient of the tangent line $\TT$.
Taking the derivative $x$ of both sides of the equation we get:
This is the slope at any point on $\CC$.
From the slope-intercept form of a line, the equation of such a line is:
:$y - y_n = m \paren {x - x_n}$
given any point $\paren {x_n, y_n}$ and the gradient $m$.
For $\TT$:
:$m = \valueat {\dfrac {\d y} {\d x} } {x \mathop = x_n, y \mathop = y_n}$
Thus the equation of $\TT$ is:
:$y - y_n = \dfrac {a - x_n} {y_n - b} \paren {x - x_n}$
=== Vertical Tangent Lines ===
== Sources ==
* 
* 
Category:Analytic Geometry
Category:Equations of Straight Lines in Plane
Category:Tangents to Circles
Category:Circles
Equation of Straight Line in Plane

== Theorem ==
=== General Equation ===
=== Slope-Intercept Form ===
=== Two-Intercept Form ===
=== Normal Form ===
=== Two-Point Form ===
=== Point-Slope Form ===
=== Homogeneous Cartesian Coordinates ===
== Also see ==
* Equation of Straight Line in Space
Category:Equations of Straight Lines in Plane
Category:Straight Lines
Category:Analytic Geometry
Category:Linear Algebra
Equation of Straight Line in Plane/General Equation

== Theorem ==
A straight line $\LL$ is the set of all $\tuple {x, y} \in \R^2$, where:
:$\alpha_1 x + \alpha_2 y = \beta$
where $\alpha_1, \alpha_2, \beta \in \R$ are given, and not both $\alpha_1, \alpha_2$ are zero.
== Proof ==
Let $y = \map f x$ be the equation of a straight line $\LL$.
From Line in Plane is Straight iff Slope is Constant, $\LL$ has constant slope.
Thus the derivative of $y$ $x$ will be of the form:
:$y' = c$
Thus:
where $K$ is arbitrary.
Taking the equation:
:$\alpha_1 x + \alpha_2 y = \beta$
it can be seen that this can be expressed as:
:$y = -\dfrac {\alpha_1} {\alpha_2} x + \dfrac {\beta} {\alpha_2}$
thus demonstrating that $\alpha_1 x + \alpha_2 y = \beta$ is of the form $y = c x + K$ for some $c, K \in \R$.
== Also presented as ==
== Also known as ==
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $3$.
* : Chapter $1$: Introduction to Coordinate Geometry: $(1.2)$
* : Chapter $\text V$: Vector Spaces: $\S 28$: Linear Transformations
* : $\S 10$: Formulas from Plane Analytic Geometry: General Equation of Line: $10.7$
* 
* 
* : Chapter $6$: Curves and Coordinates: Cartesian coordinates
* : $\S 8$: Formulas from Plane Analytic Geometry: General Equation of Line: $8.7.$
* 
Category:Equation of Straight Line in Plane/General Equation
Category:Equations of Straight Lines in Plane
Equation of Straight Line in Plane/General Equation/Also presented as

== General Equation of Straight Line in Plane: Also presented as == 
Some sources give the general equation of the straight line in the plane as:
:$a x + b y + c = 0$
where $a, b, c \in \R$ are given, and not both $a, b$ are zero.
Its equivalence to the given form can be seen by equating $a = \alpha_1, b = \alpha_2, c = -\beta$.
Similarly equivalent forms include:
:$a x = b y + c$
and so on.
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $3$.
* : Chapter $1$: Introduction to Coordinate Geometry: $(1.2)$
* : $\S 10$: Formulas from Plane Analytic Geometry: General Equation of Line: $10.7$
* 
* 
* : Chapter $6$: Curves and Coordinates: Cartesian coordinates
* : $\S 8$: Formulas from Plane Analytic Geometry: General Equation of Line: $8.7.$
* 
Category:Equation of Straight Line in Plane/General Equation
Equation of Straight Line in Plane/Homogeneous Cartesian Coordinates

== Theorem ==
A straight line $\LL$ is the set of all points $P$ in $\R^2$, where $P$ is described in homogeneous Cartesian coordinates as:
:$l X + m Y + n Z = 0$
where $l, m, n \in \R$ are given, and not both $l$ and $m$ are zero.
== Proof ==
Let $P = \tuple {X, Y, Z}$ be a point on $L$ defined in homogeneous Cartesian coordinates.
Then by definition:
where $P = \tuple {x, y}$ described in conventional Cartesian coordinates.
Hence:
Hence the result.
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $9$. Parallel lines. Points at infinity
Category:Equations of Straight Lines in Plane
Category:Homogeneous Cartesian Coordinates
Equation of Straight Line in Plane/Normal Form

== Theorem ==
Let $\LL$ be a straight line such that:
:the perpendicular distance from $\LL$ to the origin is $p$
:the angle made between that perpendicular and the $x$-axis is $\alpha$.
Then $\LL$ can be defined by the equation:
:$x \cos \alpha + y \sin \alpha = p$
=== Polar Form ===
== Proof 1 ==
== Proof 2 ==
== Also known as ==
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $4$. Special forms of the equation of a straight line: $(5)$ Normal or canonical form
* : $\S 10$: Formulas from Plane Analytic Geometry: Normal Form for Equation of Line: $10.6$
* : $\S 8$: Formulas from Plane Analytic Geometry: Normal Form for Equation of Line: $8.6.$
Category:Equation of Straight Line in Plane/Normal Form
Category:Equations of Straight Lines in Plane
Equation of Straight Line in Plane/Normal Form/Also known as

== Equation of Straight Line in Plane: Normal Form: Also known as == 
The normal form of a straight line in the plane is also known as the canonical form.
Some sources refer to it as the $\tuple {p, \alpha}$ form, after the symbols conventionally used to define it.
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $4$. Special forms of the equation of a straight line: $(5)$ Normal or canonical form
Category:Equation of Straight Line in Plane/Normal Form
Equation of Straight Line in Plane/Normal Form/Polar Form

== Theorem ==
Let $\LL$ be a straight line such that:
:the perpendicular distance from $\LL$ to the origin is $p$
:the angle made between that perpendicular and the $x$-axis is $\alpha$.
Let $\LL$ be defined in normal form:
:$x \cos \alpha + y \sin \alpha = p$
Then $\LL$ can be presented in polar coordinates as:
:$r \map \cos {\theta - \alpha} = p$
== Proof ==
Let $O$ be the origin of the Cartesian plane and the pole of the corresponding polar frame.
Let $OX$ denote the polar axis, coincident with the $x$-axis.
Let $P$ be an arbitrary point on $\LL$, expressed in polar coordinates as $\polar {r, \theta}$.
Let $N$ be the point on $\LL$ where the normal to $\LL$ intersects $\LL$.
We have that $OP$ is at an angle $\theta$ to $OX$ and is of length $r$.
We have that $ON$ is at an angle $\alpha$ to $OX$ and is of length $p$.
Hence $\angle NOP = \theta = \alpha$.
We also have that $\angle ONP$ is a right angle.
Thus:
:$p = r \map \cos {\theta - \alpha}$
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $4$. Special forms of the equation of a straight line: $(4)$ Polar equation
Category:Equations of Straight Lines in Plane
Equation of Straight Line in Plane/Normal Form/Proof 1

== Theorem ==
== Proof ==
:
Let $A$ be the $x$-intercept of $\LL$.
Let $B$ be the $y$-intercept of $\LL$.
Let $A = \tuple {a, 0}$ and $B = \tuple {0, b}$.
From the Equation of Straight Line in Plane: Two-Intercept Form, $\LL$ can be expressed in the form:
:$(1): \quad \dfrac x a + \dfrac y a = 1$
Then:
Substituting for $a$ and $b$ in $(1)$:
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $4$. Special forms of the equation of a straight line: $(5)$ Normal or canonical form
Category:Equation of Straight Line in Plane/Normal Form
Equation of Straight Line in Plane/Normal Form/Proof 2

== Theorem ==
== Proof ==
:
Let $P = \tuple {x, y}$ be an arbitrary point on $\LL$.
Let $O$ be the origin of the Cartesian plane in which $\LL$ is embedded.
Let $PQ$ be the perpendicular dropped from $P$ to the $x$-axis.
Let $QS$ be the perpendicular dropped from $Q$ to the line $ON$.
Let $PR$ be the perpendicular dropped from $P$ to the line $QS$.
By definition of cosine:
:$OS = OQ \cos \alpha$
By definition of sine:
:$PR = PQ \sin \alpha$
Then:
== Sources ==
* : $\text {III}$. Analytical Geometry: The Straight Line: Equation of a Straight Line: $\tuple {p, \alpha}$ form
Category:Equation of Straight Line in Plane/Normal Form
Equation of Straight Line in Plane/Point-Slope Form

== Theorem ==
Let $\LL$ be a straight line embedded in a cartesian plane, given in slope-intercept form as:
:$y = m x + c$
where $m$ is the slope of $\LL$.
Let $\LL$ pass through the point $\tuple {x_0, y_0}$.
Then $\LL$ can be expressed by the equation:
:$y - y_0 = m \paren {x - x_0}$
=== Parametric Form ===
== Proof ==
As $\tuple {x_0, y_0}$ is on $\LL$, it follows that:
Substituting back into the equation for $\LL$:
== Also presented as ==
== Examples ==
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $2$.
* 
* 
* 
* 
* Point-slope form
* 
Category:Equation of Straight Line in Plane/Point-Slope Form
Category:Equations of Straight Lines in Plane
Equation of Straight Line in Plane/Point-Slope Form/Also presented as

== Point-Slope Form of Equation of Straight Line in Plane: Also presented as == 
The point-slope form of the equation of a straight line $\LL$ in the plane can also be seen presented as:
:$y - y_0 = \paren {x - x_0} \tan \psi$
where $\psi$ is the angle that $\LL$ makes with the $x$-axis.
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $2$.
Category:Equation of Straight Line in Plane/Slope-Intercept Form
Equation of Straight Line in Plane/Point-Slope Form/Examples/Arbitrary Example 1

== Example of Point-Slope Form of Equation of Straight Line in Plane ==
Let $\LL$ be the straight line with slope $2$ passing through the point $\tuple {5, 4}$.
The equation of $\LL$ is then:
:$y - 4 = 2 \paren {x - 5}$
from which, after rearranging, we obtain:
:$y = 2 x - 6$
== Sources ==
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Category:Equation of Straight Line in Plane/Point-Slope Form