nyuuzyou/wikis huggingface dataset:
Equation of Straight Line in Plane/Point-Slope Form/Parametric Form

== Theorem ==
Let $\LL$ be a straight line embedded in a cartesian plane, given in point-slope form as:
::$y - y_0 = \paren {x - x_0} \tan \psi$
where $\psi$ is the angle between $\LL$ and the $x$-axis.
Then $\LL$ can be expressed by the parametric equations:
:$\begin {cases} x = x_0 + t \cos \psi \\ y = y_0 + t \sin \psi \end {cases}$
== Proof ==
Let $P_0$ be the point $\tuple {x_0, y_0}$.
Let $P$ be an arbitrary point on $\LL$.
Let $t$ be the distance from $P_0$ to $P$ measured as positive when in the positive $x$ direction.
The equation for $P$ is then:
The result follows.
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $4$. Special forms of the equation of a straight line: $(1)$ Gradient forms
Category:Equation of Straight Line in Plane/Point-Slope Form
Equation of Straight Line in Plane/Slope-Intercept Form

== Theorem ==
Let $\LL$ be the straight line in the Cartesian plane such that:
:the slope of $\LL$ is $m$
:the $y$-intercept of $\LL$ is $c$
Then $\LL$ can be described by the equation:
:$y = m x + c$
such that $m$ is the slope of $\LL$ and $c$ is the $y$-intercept.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Also presented as ==
== Also known as ==
== Examples ==
== Sources ==
* : Chapter $\text I$: Functions and their Graphs: $1$. Constants and Variables
* : $\S 10$: Formulas from Plane Analytic Geometry: Equation of Line joining Two Points $\map {P_1} {x_1, y_1}$ and $\map {P_2} {x_2, y_2}$: $10.4$
* 
* 
* 
* 
* : $\S 8$: Formulas from Plane Analytic Geometry: Equation of Line joining Two Points $\map {P_1} {x_1, y_1}$ and $\map {P_2} {x_2, y_2}$: $8.4.$
* Slope-intercept form
* 
Category:Equation of Straight Line in Plane/Slope-Intercept Form
Category:Equations of Straight Lines in Plane
Equation of Straight Line in Plane/Slope-Intercept Form/Also known as

== Slope-Intercept Form of Equation of Straight Line in Plane: Also known as == 
The slope-intercept form of the equation of a straight line in the plane is also known as the gradient-intercept form.
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $4$. Special forms of the equation of a straight line: $(1)$ Gradient forms
Category:Equation of Straight Line in Plane/Slope-Intercept Form
Equation of Straight Line in Plane/Slope-Intercept Form/Also presented as

== Slope-Intercept Form of Equation of Straight Line in Plane: Also presented as == 
The slope-intercept form of the equation of a straight line in the plane can also be seen presented as:
:$y = x \tan \psi + c$
where $\psi$ is the angle that $\LL$ makes with the $x$-axis.
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $4$. Special forms of the equation of a straight line: $(1)$ Gradient forms
Category:Equation of Straight Line in Plane/Slope-Intercept Form
Equation of Straight Line in Plane/Slope-Intercept Form/Examples/Arbitrary Example 1

== Examples of Slope-Intercept Form of Equation of Straight Line in Plane ==
The straight line $\LL$ given by the equation:
:$y = 2 x + 4$
has:
:a slope of $2$, hence the angle between $\LL$ and the $x$-axis is $\arctan 2$
:a $y$-intercept of $\tuple {0, 4}$.
== Sources ==
* 
* 
Category:Equation of Straight Line in Plane/Slope-Intercept Form
Equation of Straight Line in Plane/Slope-Intercept Form/Proof 1

== Theorem ==
== Proof ==
Let $\LL$ be the straight line defined by the general equation:
:$\alpha_1 x + \alpha_2 y = \beta$
We have:
Setting $x = 0$ we obtain:
:$y = \dfrac {\beta} {\alpha_2}$
which is the $y$-intercept.
Differentiating $(1)$ $x$ gives:
:$y' = -\dfrac {\alpha_1} {\alpha_2}$
By definition, this is the slope of $\LL$ and is seen to be constant.
The result follows by setting:
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $4$. Special forms of the equation of a straight line: $(1)$ Gradient forms
Category:Equation of Straight Line in Plane/Slope-Intercept Form
Equation of Straight Line in Plane/Slope-Intercept Form/Proof 2

== Theorem ==
== Proof ==
By definition, the $y$-intercept of $\LL$ is $\tuple {0, c}$.
We calculate the $x$-intercept $X = \tuple {x_0, 0}$ of $\LL$:
Hence from the two-intercept form of the Equation of straight line in the plane, $\LL$ can be described as:
== Sources ==
* : $\text {III}$. Analytical Geometry: The Straight Line: Equation of a Straight Line: Gradient-intercept form
Category:Equation of Straight Line in Plane/Slope-Intercept Form
Equation of Straight Line in Plane/Slope-Intercept Form/Proof 3

== Theorem ==
== Proof ==
:
Let the straight line $\LL$ make an angle $\psi$ with the $x$-axis.
The slope $m$ of $\LL$ is then given by:
:$m = \tan \psi$
Let $B := \tuple {0, c}$ be the $y$-intercept of $\LL$.
Let $P := \tuple {x, y}$ be an arbitrary point on $\LL$.
Let the perpendicular $MP$ to the $x$-axis be constructed from $P$ to intersect the $x$-axis at $M$
Let $BK$ be constructed perpendicular $MP$ from $B$ to intersect $MP$ at $K$.
Then we have:
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $4$. Special forms of the equation of a straight line: $(1)$ Gradient forms
Category:Equation of Straight Line in Plane/Slope-Intercept Form
Equation of Straight Line in Plane/Two-Intercept Form

== Theorem ==
Let $\LL$ be a straight line which intercepts the $x$-axis and $y$-axis respectively at $\tuple {a, 0}$ and $\tuple {0, b}$, where $a b \ne 0$.
Then $\LL$ can be described by the equation:
:$\dfrac x a + \dfrac y b = 1$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Also known as ==
== Examples ==
== Sources ==
* : $\S 10$: Formulas from Plane Analytic Geometry: Equation of Line in terms of $x$ Intercept $a \ne 0$ and $y$ Intercept $b \ne 0$: $10.5$
* 
* 
* : $\S 8$: Formulas from Plane Analytic Geometry: Equation of Line in terms of $x$ Intercept $a \ne 0$ and $y$ Intercept $b \ne 0$: $8.5.$
* Two-intercept form
* 
Category:Equation of Straight Line in Plane/Two-Intercept Form
Category:Equations of Straight Lines in Plane
Equation of Straight Line in Plane/Two-Intercept Form/Also known as

== Equation of Straight Line in Plane: Two-Intercept Form: Also known as == 
The two-intercept form of the Equation of Straight Line in Plane is also known as the intercept form.
== Sources ==
* 
* 
* 
* 
Category:Equation of Straight Line in Plane/Two-Intercept Form
Equation of Straight Line in Plane/Two-Intercept Form/Examples/Arbitrary Example 1

== Example of Two-Intercept Form of Equation of Straight Line in Plane ==
Let $\LL$ be the straight line given in general form as:
:$4 y = 2 x - 8$
Dividing through by $8$ and rearranging, we obtain:
:$\dfrac x 4 - \dfrac y 2 = 1$
So the $x$-intercept is $4$ and the $y$-intercept is $-2$.
== Sources ==
* 
* 
Category:Equation of Straight Line in Plane/Two-Intercept Form
Equation of Straight Line in Plane/Two-Intercept Form/Proof 1

== Theorem ==
== Proof ==
:
From the General Equation of Straight Line in Plane, $\LL$ can be expressed in the form:
:$(1): \quad \alpha_1 x + \alpha_2 y = \beta$
where $\alpha_1, \alpha_2, \beta \in \R$ are given, and not both $\alpha_1, \alpha_2$ are zero.
Substituting for the two points whose coordinates we know about:
and: 
We know that $\beta \ne 0$ because none of $a, b, \alpha_1, \alpha_2$ are equal to $0$.
Hence:
Category:Equation of Straight Line in Plane/Two-Intercept Form
Equation of Straight Line in Plane/Two-Intercept Form/Proof 2

== Theorem ==
== Proof ==
By definition, $\LL$ passes through $\tuple {a, 0}$ and $\tuple {0, b}$.
From the Two-Point Form of Equation of Straight Line in Plane, $\LL$ can be expressed in the form:
== Sources ==
* : $\text {III}$. Analytical Geometry: The Straight Line: Equation of a Straight Line: Intercept form
Category:Equation of Straight Line in Plane/Two-Intercept Form
Equation of Straight Line in Plane/Two-Intercept Form/Proof 3

== Theorem ==
== Proof ==
:
We have that $\LL$ is passes through the two points $A = \tuple {a, 0}$ and $B = \tuple {0, b}$.
Let $P = \tuple {x, y}$ be an arbitrary point on $\LL$.
We have:
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $4$. Special forms of the equation of a straight line: $(2)$ Intercept form
Category:Equation of Straight Line in Plane/Two-Intercept Form
Equation of Straight Line in Plane/Two-Point Form

== Theorem ==
Let $P_1 := \tuple {x_1, y_1}$ and $P_2 := \tuple {x_2, y_2}$ be points in a cartesian plane.
Let $\LL$ be the straight line passing through $P_1$ and $P_2$.
Then $\LL$ can be described by the equation:
:$\dfrac {y - y_1} {x - x_1} = \dfrac {y_2 - y_1} {x_2 - x_1}$
or:
:$\dfrac {x - x_1} {x_2 - x_1} = \dfrac {y - y_1} {y_2 - y_1}$
This is known as the two-point form.
=== Parametric Form ===
=== Determinant Form ===
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Proof 4 ==
== Examples ==
== Sources ==
* : $\S 10$: Formulas from Plane Analytic Geometry: Equation of Line joining Two Points $\map {P_1} {x_1, y_1}$ and $\map {P_2} {x_2, y_2}$: $10.3$
* 
* 
* 
* 
* : $\S 8$: Formulas from Plane Analytic Geometry: Equation of Line joining Two Points $\map {P_1} {x_1, y_1}$ and $\map {P_2} {x_2, y_2}$: $8.3.$
* Two-point form
* 
Category:Two-Point Form of Equation of Straight Line in Plane
Category:Equations of Straight Lines in Plane
Equation of Straight Line in Plane/Two-Point Form/Determinant Form

== Theorem ==
Let $\LL$ be a straight line embedded in a Cartesian plane, given in two-point form as:
:$\dfrac {x - x_1} {x_2 - x_1} = \dfrac {y - y_1} {y_2 - y_1}$
Then $\LL$ can be expressed in the form:
:$\begin {vmatrix} x & y & 1 \\ x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \end {vmatrix} = 0$
== Proof ==
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $4$. Special forms of the equation of a straight line: $(3)$ Line through two points
Category:Two-Point Form of Equation of Straight Line in Plane
Equation of Straight Line in Plane/Two-Point Form/Parametric Form

== Theorem ==
Let $\LL$ be a straight line embedded in a cartesian plane, given in two-point form as:
::$\dfrac {x - x_1} {x_2 - x_1} = \dfrac {y - y_1} {y_2 - y_1}$
Then $\LL$ can be expressed by the parametric equations:
:$\begin {cases} x = x_1 + t \paren {x_2 - x_1} \\ y = y_1 + t \paren {y_2 - y_1} \end {cases}$
== Proof ==
Let $P = \tuple {x, y}$ be an arbitrary point on $\LL$.
Let $t = \dfrac {x - x_1} {x_2 - x_1} = \dfrac {y - y_1} {y_2 - y_1}$.
We then have:
The result follows.
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $4$. Special forms of the equation of a straight line: $(3)$ Line through two points
Category:Two-Point Form of Equation of Straight Line in Plane
Equation of Straight Line in Plane/Two-Point Form/Proof 1

== Theorem ==
== Proof ==
From the slope-intercept form of the equation of the straight line:
:$(1): \quad y = m x + c$
which is to be satisfied by both $\tuple {x_1, y_1}$ and $\tuple {x_2, y_2}$.
We express $m$ and $c$ in terms of $\paren {x_1, y_1}$ and $\paren {x_2, y_2}$:
Substituting for $m$ and $c$ in $(1)$:
Category:Two-Point Form of Equation of Straight Line in Plane
Equation of Straight Line in Plane/Two-Point Form/Proof 2

== Theorem ==
== Proof ==
Let $\tuple {x, y}$ be an arbitrary point on the straight line through $\tuple {x_1, y_1}$ and $\tuple {x_2, y_2}$.
The area of the triangle formed by $\tuple {x, y}$, $\tuple {x_1, y_1}$ and $\tuple {x_2, y_2}$ is equal to $0$.
Hence from Area of Triangle in Determinant Form:
:$\AA = \dfrac 1 2 \size {\paren {\begin {vmatrix} x & y & 1 \\ x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ \end {vmatrix} } }$
Hence:
== Sources ==
* : $\text {III}$. Analytical Geometry: The Straight Line: Equation of a Straight Line
Category:Two-Point Form of Equation of Straight Line in Plane
Equation of Straight Line in Plane/Two-Point Form/Proof 3

== Theorem ==
== Proof ==
:
Let $P = \tuple {x, y}$ be an arbitrary point on the straight line through $P_1 = \tuple {x_1, y_1}$ and $P_2 = \tuple {x_2, y_2}$.
Construct the straight line $P_1 H K$ perpendicular to the $x$-axis.
We have that $\triangle P_1 H P_2$ and $\triangle P_1 K P$ are similar.
Hence:
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $4$. Special forms of the equation of a straight line: $(3)$ Line through two points
Category:Two-Point Form of Equation of Straight Line in Plane
Equation of Straight Line in Plane/Two-Point Form/Proof 4

== Theorem ==
== Proof ==
:
Let $z_1 = x_1 + i y_1$ and $z_2 = x_2 + i y_2$ be the position vectors of the points $A$ and $B$ embedded in the complex plane.
Let $z = x + i y$ be the position vector of an arbitrary point $P$ on the straight line $AB$.
From the diagram:
and:
Then:
== Sources ==
* : $1$: Solved Problems: Graphical Representations of Complex Numbers. Vectors: $11$
Category:Two-Point Form of Equation of Straight Line in Plane
Equation of Straight Line in Space/Parametric Form

== Theorem ==
Let $\LL$ be a straight line embedded in a cartesian $3$-space passing through the point $\tuple {x_1, y_1, z_1}$ and with direction cosines $l$, $m$ and $n$.
$\LL$ is expressed in parametric form by the set of equations:
:$\begin {cases} x & = & x_1 + l d \\ y & = & y_1 + m d \\ z & = & z_1 + n d \end {cases}$
where $d \in \R$ is the distance of the variable point $\tuple {x, y, z}$ from $\tuple {x_1, y_1, z_1}$
== Proof ==
== Sources ==
* 
* 
* 
* 
Category:Equation of Straight Line in Space/Parametric Form
Category:Equations of Straight Lines in Space
Equation of Straight Line in Space/Symmetric Form

== Theorem ==
Let $\LL$ be a straight line embedded in a cartesian $3$-space passing through the point $\tuple {x_1, y_1, z_1}$.
$\LL$ is expressed in symmetric form by the equation:
:$\dfrac {x - x_1} l = \dfrac {y - y_1} m = \dfrac {z - z_1} n$
where $l, m, n \in \R$ are the direction ratios of $\LL$.
== Proof ==
== Also known as ==
== Sources ==
* 
* 
* 
* 
Category:Equation of Straight Line in Space/Symmetric Form
Category:Equations of Straight Lines in Space
Equation of Straight Line in Space/Two-Point Form

== Theorem ==
Let $\LL$ be a straight line embedded in a cartesian $3$-space passing through the points $\tuple {x_1, y_1, z_1}$ and $\tuple {x_2, y_2, z_2}$.
$\LL$ is expressed in two-point form by the equation:
:$\dfrac {x - x_1} {x_2 - x_1} = \dfrac {y - y_1} {y_2 - y_1} = \dfrac {z - z_1} {z_2 - z_1}$
== Proof ==
== Sources ==
* 
* 
Category:Equation of Straight Line in Space/Two-Point Form
Category:Equations of Straight Lines in Space
Equation of Straight Line in Space/Vector Form

== Theorem ==
Let $\mathbf a$ and $\mathbf b$ denote the position vectors of two points in space
Let $L$ be a straight line in space passing through $\mathbf a$ which is parallel to $\mathbf b$.
Let $\mathbf r$ be the position vector of an arbitrary point on $L$.
Then:
:$\mathbf r = \mathbf a + t \mathbf b$
for some real number $t$, which may be positive or negative, or even $0$ if $\mathbf r = \mathbf a$.
== Proof ==
:
Let $a$ and $b$ be points as given, with their position vectors $\mathbf a$ and $\mathbf b$ respectively.
Let $P$ be an arbitrary point on the straight line $L$ passing through $\mathbf a$ which is parallel to $\mathbf b$.
By the parallel postulate, $L$ exists and is unique.
Let $\mathbf r$ be the position vector of $P$.
Let $\mathbf r = \mathbf a + \mathbf x$ for some $\mathbf x$.
Then we have:
:$\mathbf x = \mathbf r - \mathbf a$
As $\mathbf a$ and $\mathbf r$ are both on $L$, it follows that $\mathbf x$ is parallel to $\mathbf b$.
That is:
:$\mathbf x = t \mathbf b$
for some real number $t$.
Hence the result.
== Also presented as ==
== Sources ==
* : Introduction: Vector Notation and Formulae
* : Chapter $\text I$: Definitions. Elements of Vector Algebra: $3$. Addition and Subtraction of Vectors
* : Chapter $\text I$: Vector Algebra: $\S 9$: $(17)$
Category:Equation of Straight Line in Space/Vector Form
Category:Equations of Straight Lines in Space
Category:Vectors
Equation of Straight Line in Space/Vector Form/Also presented as

== Vector Equation of Straight Line in Space: Also presented as ==
Let $a$ and $b$ be points in cartesian $3$-space with position vectors $\mathbf a$ and $\mathbf b$ respectively.
The vector form of the equation of a straight line in space can be presented in the forms:
:$\mathbf r = \mathbf a + t \paren {\mathbf b - \mathbf a}$
or:
:$\paren {\mathbf r - \mathbf a} \times \paren {\mathbf b - \mathbf a} = \mathbf 0$
which is the same straight line.
== Proof ==
We have that $\mathbf r = \mathbf a + t \paren {\mathbf b - \mathbf a}$ is exactly the same form as $\mathbf r = \mathbf a + t \mathbf b$ but substituting $\mathbf b - \mathbf a$ for $\mathbf b$.
From the exposition of the proof, $\mathbf r = \mathbf a + t \mathbf b$ is all the points in the straight line passing through $a$ parallel to $\mathbf b$.
Hence $\mathbf r = \mathbf a + t \paren {\mathbf b - \mathbf a}$ is all the points in the straight line passing through $a$ parallel to $\mathbf b - \mathbf a$.
But $\mathbf b - \mathbf a$ is the vector that takes you:
:from $a$, whose position vector is $\mathbf a$
:to $b$, whose position vector is $\mathbf b$.
Hence $b$ is itself on the straight line passing through $\mathbf a$ parallel to $\mathbf b - \mathbf a$.
Hence $\mathbf r = \mathbf a + t \paren {\mathbf b - \mathbf a}$ is the straight line passing through $a$ and $b$.
We note that, by Cross Product of Parallel Vectors:
:$\paren {\mathbf r - \mathbf a} \times \paren {\mathbf b - \mathbf a} = \mathbf 0$
defines the set of points in $\R^3$ passing through $\mathbf a$ parallel to $\mathbf b - \mathbf a$.
That is, from the above, the set of points in the straight line passing through $a$ and $b$.
== Sources ==
* 
* 
Category:Equation of Straight Line in Space/Vector Form
Equation of Straight Line through Intersection of Two Straight Lines

== Theorem ==
Let $\LL_1$ and $\LL_2$ be straight lines embedded in a cartesian plane $\CC$, expressed using the general equations:
Let $\LL_3$ be a third straight line embedded in $\CC$, passing through the point of intersection of $\LL_1$ and $\LL_2$.
$\LL_3$ can be expressed using the general equation:
:$(1): \quad \paren {l_1 x + m_1 y + n_1} - k \paren {l_2 x + m_2 y + n_2} = 0$
== Proof ==
Let $P = \tuple {x, y}$ be the point of intersection of $\LL_1$ and $\LL_2$.
We have that:
and so equation $(1)$:
:$\paren {l_1 x + m_1 y + n_1} - k \paren {l_2 x + m_2 y + n_2} = 0$
is satisfied by the point $P$.
Then:
Each of $l_1 - k l_2$, $m_1 - k m_2$ and $n_1 - k n_2$ is a real number.
Hence $(1)$ is the equation of a straight line.
== Also presented as ==
This result can also be conveniently presented as follows:
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $10$. Equation of a straight line through the intersection of two given lines
Category:Equations of Straight Lines in Plane
Category:Equation of Straight Line through Intersection of Two Straight Lines
Equation of Straight Line through Intersection of Two Straight Lines/Also presented as

== Theorem ==
Let $u = l_1 x + m_1 y + n_1$.
Let $v = l_2 x + m_2 y + n_2$.
Let $\LL_1$ be defined by the equation $u = 0$.
Let $\LL_2$ be defined by the equation $v = 0$.
Then the equation of the straight line passing through the point of intersection of $\LL_1$ and $\LL_2$ can be written as:
:$u - k v = 0$
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $10$. Equation of a straight line through the intersection of two given lines
Category:Equation of Straight Line through Intersection of Two Straight Lines
Equation of Tangent Plane to Surface

== Theorem ==
Let $\SS$ be a surface embedded in a Cartesian $3$-space.
Let $\TT$ be a tangent plane to $\SS$ at the point $P = \tuple {x_0, y_0, z_0}$.
Then $\TT$ can be expressed using the equation:
:$l \paren {x - x_0} + m \paren {y - y_0} + n \paren {z - z_0} = 0$
where $l$, $m$ and $n$ are the direction cosines of the normal to $S$ at $P$.
== Proof ==
== Sources ==
* 
* 
Category:Tangent Planes
Equation of Tangent 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 $\TT$ be a tangent to $\CC$ passing through $P$.
Then $\TT$ can be defined by the equation:
:$x x_1 + y y_1 = r^2$
== Proof 1 ==
== Proof 2 ==
Category:Equation of Tangent to Circle Centered at Origin
Category:Circles
Category:Tangents
Equation of Tangent to Circle Centered at Origin/Proof 1

== Theorem ==
== Proof ==
From Equation of Straight Line Tangent to Circle we have that for a general circle of radius $r$ and center $\tuple {a, b}$:
:$y - y_1 = \dfrac {a - x_1} {y_1 - b} \paren {x - x_1}$
is the equation of a tangent $\TT$ to $\CC$ passing through $\tuple {x_1, y_1}$.
Setting the center to $\tuple {0, 0}$:
Category:Equation of Tangent to Circle Centered at Origin
Equation of Tangent to Circle Centered at Origin/Proof 2

== Theorem ==
== Proof ==
From the slope-intercept form of a line, the equation of a line passing through $P$ is:
:$y - y_1 = \mu \paren {x - x_1}$
If this line passes through another point $\tuple {x_2, y_2}$ on $\CC$, the slope of the line is given by:
:$\mu = \dfrac {y_2 - y_1} {x_2 - x_1}$
Because $P$ and $Q$ both lie on $\CC$, we have:
As $Q$ approaches $P$, we have that $y_2 \to y_1$ and $x_2 \to x_1$.
The limit of the slope is therefore:
:$-\dfrac {2 x_1} {2 y_1} = -\dfrac {x_1} {y_1}$
The equation of the tangent $\TT$ to $\CC$ passing through $\tuple {x_1, y_1}$ is therefore:
== Sources ==
* : Chapter $\text {III}$. The Circle: $2$. The tangent at a given point
Category:Equation of Tangent to Circle Centered at Origin
Equation of Tangent to Ellipse in Reduced Form

== Theorem ==
Let $E$ 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 $P = \tuple {x_1, y_1}$ be a point on $E$.
The tangent to $E$ at $P$ is given by the equation:
:$\dfrac {x x_1} {a^2} + \dfrac {y y_1} {b^2} = 1$
== Proof ==
From the slope-intercept form of a line, the equation of a line passing through $P$ is:
:$y - y_1 = \mu \paren {x - x_1}$
If this line passes through another point $\tuple {x_2, y_2}$ on $E$, the slope of the line is given by:
:$\mu = \dfrac {y_2 - y_1} {x_2 - x_1}$
Because $P$ and $Q$ both lie on $E$, we have:
As $Q$ approaches $P$, we have that $y_2 \to y_1$ and $x_2 \to x_1$.
The limit of the slope is therefore:
:$-\dfrac {2 b^2 x_1} {2 a^2 y_1} = -\dfrac {b^2 x_1} {a^2 y_1}$
The equation of the tangent $\TT$ to $\CC$ passing through $\tuple {x_1, y_1}$ is therefore:
== Sources ==
* : Chapter $\text {IV}$. The Ellipse: $3$.
Category:Ellipses
Category:Tangents
Equation of Tangents to Circle from Point

== Theorem ==
Let $\CC$ be a circle embedded in the Cartesian plane of radius $r$ with its center located at the origin.
Let $P = \tuple {x_0, y_0}$ be a point in the plane of $\CC$ which is outside $\CC$.
The tangents to $\CC$ which pass through $P$ can be described using the equation:
:$\paren {x x_0 + y y_0 - r^2}^2 = \paren {x^2 + y^2 - r^2} \paren { {x_0}^2 + {y_0}^2 - r^2}$
== Proof ==
From Equation of Circle center Origin, $\CC$ can be described as:
:$x^2 + y^2 = r^2$
Let $\LL$ be an arbitrary straight line through $P$ which intersects $\CC$ at $U$ and $V$.
Let $Q = \tuple{x, y}$ be an arbitrary point on $\LL$.
Let $k$ be the position-ratio of one of the points $U$ and $V$ with respect to $P$ and $Q$.
By Joachimsthal's equation:
:$(1): \quad k^2 \paren {x^2 + y^2 - r^2} + 2 k \paren {x x_0 + y y_0 - r^2} + \paren { {x_0}^2 + {y_0}^2 - r^2} = 0$
which is a quadratic in $k$.
When $\LL$ is tangent to $\CC$, the points $U$ and $V$ coincide.
Hence $(1)$ has equal roots.
From Solution to Quadratic Equation, it follows that the discriminant of $(1)$ is zero.
That is:
:$\paren {x x_0 + y y_0 - r^2}^2 - \paren {x^2 + y^2 - r^2} \paren { {x_0}^2 + {y_0}^2 - r^2} = 0$
== Sources ==
* : Chapter $\text {III}$. The Circle: $12$. Equation of the tangents to a given circle from a given point
Category:Circles
Category:Tangents to Circles
Equation of Tractrix/X-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 $y$-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.
:
=== Cartesian Form ===
=== Parametric Form ===
Category:Equations of Tractrices