Derivation/Representation of Equation of Lines

Lines are best represented by equations in Cartesian Coordinate System. As with any geometic object in any coordinate system, equations of Lines can be represented in Coordinate form, Parametric form, Position Vector form and Non-Position Vector form.

Equation of Lines in 2D

Equations in Coordinate Form

The following are Coordinate Form representation/derivation of equations of lines in 2D:

Standard Form or Implicit Form: The equation of the Line in 2D in Standard Form or Implicit Form is given as:

\(Ax + By + C=0\)

Any equation of line derived from any given inputs is generally brought into the Standard Form for final representation.
Following are certain important observations abount scalar equations of Lines in 2D:
1. It is represented by a single Linear Equation in either 1 or 2 variables. The variables are denoted by the coordinates \(x\) and/or \(y\)
2. The co-efficients of \(x\) and \(y\) (given as \(A\) and \(B\) repectively in the above equation) cannot both be simultaneously zero. Though either one can be. If the co-efficient of \(x\) is 0 (i.e. \(A=0\)) then the line is parallel to or contains X axis and is perpendicular to Y axis. If the co-efficient of \(y\) is 0 (i.e. \(B=0\)) then the line is parallel to or contains Y axis and is perpendicular to X axis.
3. \(C\) is the constant of the equation. If \(C=0\), the line passes through origin.
2 Point Form: The equation of the line that passes through two points \((x_1, y_1)\) and \((x_2, y_2)\) can be derived as:

\(\frac{y-y_1}{y_1-y_2}=\frac{x-x_1}{x_1-x_2}\)

\(\Rightarrow (x_1-x_2)(y-y_1)=(y_1-y_2)(x-x_1)\)

The above equation can be brought to Standard Form of line equation by solving and rearranging the terms.

Please note the following:
1. Any of the coordinate points can be subtracted from \(x\) (Only that corresponding coordinate point must be subtracted from \(y\).). Similar consistency must be maintained while subtracting the coordinate points.
2. The difference between Y coordinates of two points \((y_1-y_2)\)(or the displacement of line along Y axis between these two points) gives the Y Direction Number (henceforth refered as \(y_d\)). The difference between corresponding X coordinates of two points \((x_1-x_2)\) (or the displacement of line along X axis between these two points) gives the X Direction Number (henceforth refered as \(x_d\)). The ratio between \(x_d\) and \(y_d\), known as Direction Ratio of the line, is constant between any two points in a line. The value of \(\frac{y_d}{x_d}\) is known as the Slope of the line.
3. The value of either \(x_d\) or \(y_d\) may be 0 (but not both). If \(x_d=0\) the Slope is undefined and the line is perpendicular to X axis. If \(y_d=0\) the Slope is 0 and the line is perpendicular to Y axis. However, In such cases also the cross multiplication is done to arrive at the equation of line.
Determinant Form: The Determinant equation of the line that passes through two points \((x_1, y_1)\) and \((x_2, y_2)\) is given as:

\(\begin{vmatrix} x & y & 1\\ x_1 & y_1 & 1 \\ x_2& y_2 & 1\end{vmatrix}=0\) OR \(\begin{vmatrix} (x - x_1) & (y - y_1)\\ (x_1 - x_2) & (y_1 - y_2)\end{vmatrix}=0\)

The above equation can be brought to Standard Form of line equation by solving the determinant and rearranging the terms.
Just like 2 Point Form of equation, any of the coordinate points can be subtracted from \(x\) (Only that corresponding coordinate point must be subtracted from \(y\).). Similar consistency must be maintained while subtracting the coordinate points.
Point Direction Ratio Form: The equation of the line that passes through a point \((x_1, y_1)\) and has direction ratio \((x_d, y_d)\) can be derived as:

\(\frac{y-y_1}{y_d}=\frac{x-x_1}{x_d}\)

\(\Rightarrow (x_d)(y-y_1)=(y_d)(x-x_1)\)

The above equation can be brought to Standard Form of line equation by solving and rearranging the terms.
Point Slope Form: The equation of the line that passes through a point \((x_1, y_1)\) and has Slope m can be derived as:

\(\frac{y-y_1}{x-x_1}=m\)

\(\Rightarrow (y-y_1)=m(x-x_1)\)

The above equation can be brought to Standard Form of line equation by solving and rearranging the terms.
Normal Form: The equation of the line that passes through a point \((x_1, y_1)\) and has direction ratio of the line perpendicular to it as \((x_n, y_n)\) (also known as normal to the line) can be derived as:

\(x_n(x-x_1) + y_n(y-y_1) =0 \)

\(\Rightarrow x_nx + y_ny -x_nx_1 - y_ny_1 =0 \)

Simplyfying the above equation we get an equation of the Standard Form or Implicit Form of line

\(Ax + By + C = 0\)

In the above equation

\(A = x_n\)
\(B = y_n\)
\(C = -x_nx_1 - y_ny_1\)

The relation between direction ratio of line \((x_d,y_d)\) and direction ratio of the normal to the line \((x_n,y_n)\) is as follows:

\(x_n=y_d \hspace{.5cm} y_n=-x_d\) OR \(x_n=-y_d \hspace{.5cm} y_n=x_d\)

Please notice that the X and the Y direction numbers get interchanged and one of them (and not both) gets a negative sign.
Unit Normal Distance Form: The Standard Form of line equation is given as:

\(Ax + By + C=0\)

The Unit Normal Distance Form of Line is given by dividing both sides by \(\sqrt{A^2 + B^2}\) and rearranging as following:

\(\frac{A}{\sqrt{A^2 + B^2}}x + \frac{B}{\sqrt{A^2 + B^2}}y =\frac{-C}{\sqrt{A^2 + B^2}}\)

In this form of equation, the term \(\frac{-C}{\sqrt{A^2 + B^2}}\) represents the signed distance of line from the origin (0,0).
Intercept Form: The Standard Form of line equation is given as:

\(Ax + By + C=0\)
\(\Rightarrow Ax + By=-C\)

Dividing by -C on both sides we get

\(\frac{x}{\frac{-C}{A}} + \frac{y}{\frac{-C}{B}} = 1\)

In the above equation the value \(\frac{-C}{A}\) is known as X Intercept (the \(x\) coordinate of the point where the line meets the X axis) and the value \(\frac{-C}{B}\) is known as Y Intercept (the \(y\) coordinate of the point where the line meets the Y axis). Hence the equation of the line having intercepts \(c_x\) and \(c_y\) on X and Y axis respectively is given as:

\(\frac{x}{c_x} + \frac{y}{c_y} = 1 \)

Please note that if the Slope of Line is positive then one of either the X Intercept is negative or the Y Intercept is negative. If the Slope of Line is negative then either both the X Intercept and the Y Intercept are negative or both the X Intercept and the Y Intercept are positive.
Slope Intercept Forms or Explicit Forms: The Standard Form of line equation is given as:

\(Ax + By + C=0\)

The following 2 Forms of equations can be obtained by rearranging the equation

\(y=\frac{-Ax}{B} - \frac{C}{B}\)
\(\Rightarrow y=\frac{-A}{B}x + \frac{-C}{B}\)

OR

\(x=\frac{-By}{A} - \frac{C}{A}\)
\(\Rightarrow x=\frac{y}{\frac{-A}{B}} + \frac{-C}{A}\)

In the above equations \(\frac{-A}{B}\) gives the Slope of the line (generally denoted by m). The value \(\frac{-C}{A}\) is known as X Intercept (denoted by \(c_x\)). The value \(\frac{-C}{B}\) is known as Y Intercept (denoted by \(c_y\)).
Hence, The equation of the line having a Slope m, Y Intercept as \(c_y\) and X Intercept as \(c_x\) can be given as:

\(y = mx + c_y\)

OR

\(x = \frac{y}{m} + c_x\)

Please note that the above equations can be brought back to Standard Form of line equation just by rearranging the terms.
Polar Form: The Polar Form Equation of the line that is situated at distance d from origin, perpendicular to it and and the perpendicular making an angle A with the X axis is given as:

\(x\hspace{.2cm}cos(A) + y\hspace{.2cm}sin (A) = d\)

Derivation:

Since the Line is at distance d from the origin and the perpendicular makes an angle A with positive direction of X axis, the projection of origin on the line is \((d \cos(A),d \sin(A))\).
Also the direction ratio of the normal to the line is same i.e. = \((d \cos(A), d \sin(A))\).
Hence, the direction ratio of the Line is = \((-d \sin(A), d \cos(A))\).
Therefore, the Slope of the Line is = \(\frac{-d \cos(A)}{d \sin(A)}=\frac{-\cos(A)}{\sin(A)}\).
Using the Point Slope Form of equation of the Line we have

\(\frac{y-d \sin(A)}{x-d \cos(A)}= \frac{-cos(A)}{\sin(A)}\)
\(\Rightarrow \sin(A)(y-d \sin(A))= -\cos(A) (x-d \cos(A))\)
\(\Rightarrow x\hspace{.2cm}cos(A) + y\hspace{.2cm}sin (A) = d\)

Please note that the above equation can be brought to Standard Form of line equation by solving and rearranging the terms.
Symmetric Form: The Symmetric Form Equation of the line passing through a point \((x_1,y_1)\) making an angle A with the positive direction of X axis is given as:

\(\frac{x-x_1}{cos(A)}= \frac{y-y_1}{sin(A)}\)

Please note that the above equation can be brought to Standard Form of line equation by solving and rearranging the terms.

Equations in Parametric Form

The following are Parametric Form representation of equations of lines in 2D:

Point Direction Ratio Form: The Parametric Equation of a Line passing through point \((x_1,y_1)\) having a direction ratio \((x_d,y_d)\) can be found by equating the Point Direction Ratio Form of Equation to a constant \(t\) as follows:

\(\frac{x-x_1}{x_d}=\frac{y-y_1}{y_d}=t\)

\(\Rightarrow x= x_1 + tx_d ,\hspace{.5cm}y= y_1 + ty_d\)

\(t\) can be any real value.
Explicit Form: The Parametric Equation of a Line can also be given from the Slope Intercept Forms / Explicit Forms of Equations as follows:

\(x=t ,\hspace{.5cm} y=mt + c_y\)    (When \(y\) is given as a function of \(x\))

\(y=t ,\hspace{.5cm} x=\frac{t}{m} + c_x\)    (When \(x\) is given as a function of \(y\))

\(x= C ,\hspace{.5cm}y=t\)   (Line perpendicular to \(x\) axis)

\(y= C ,\hspace{.5cm}x=t \)   (Line perpendicular to \(y\) axis)

\(t\) can be any real value. \(C\) is constant. \(m\) is the slope of the Line. \(c_x\) and \(c_y\) are the Intercepts on X and Y axis respectively

Equations in Vector Form

The following are Vector Form representation of equations of lines in 2D:

Position Vector Form: The Position Vector Form of equation of Line can be found by using the value of \(x\) and \(y\) from the Parametric Equations as follows:

\(\vec{P}= (x_1 + tx_d)\hat{\textbf{i}} + (y_1 + ty_d)\hat{\textbf{j}}\)   (For a Line passing through point \((x_1,y_1)\) having a direction ratio \((x_d,y_d)\))

\(\vec{P}= t\hat{\textbf{i}} + (mt + c_y)\hat{\textbf{j}}\)   (When \(y\) is given as a function of \(x\))

\(\vec{P}= (\frac{t}{m} + c_x)\hat{\textbf{i}} + t\hat{\textbf{j}}\)   (When \(x\) is given as a function of \(y\))

\(\vec{P}= C\hat{\textbf{i}} + t\hat{\textbf{j}}\)   (Line perpendicular to \(x\) axis)

\(\vec{P}= t\hat{\textbf{i}} + C\hat{\textbf{j}}\)   (Line perpendicular to \(y\) axis)
Normal Form: The Normal Form of vector equation of Line in 2D is given as:

\((\vec{r}-\vec{a})\cdot\vec{n}=0\) ...(1)
\(\Rightarrow \vec{r}\cdot\vec{n}-\vec{a}\cdot\vec{n}=0\)
\(\Rightarrow \vec{r}\cdot\vec{n}=\vec{a}\cdot\vec{n}\)
\(\Rightarrow \vec{r}\cdot\vec{n}=C\) ...(2)

The equations (1) & (2) above are the Normal Form of vector equations of Line in 2D. In above equations

\(\vec{r}\) is the position vector of any arbitrary point on the Line (for equation (1) & (2)) or a vector from any point Outside the Line to a point On the Line (for equation (2) only).
\(\vec{a}\) is the position vector of a given point on the Line.
\(\vec{n}\) is a vector normal to the Line.
\(C\) is a Constant of Line equation if \(\vec{r}\) is the position vector of any arbitrary point on the Line. Otherwise its an equation specific constant.

Please note that the above equations correspond to the Normal Form of Scalar Equation of Line in 2D.
Unit Normal Distance Form: The Unit Normal Distance Form of vector equation of a Line in 2D is given as:

\(\vec{r}\cdot\hat{n}=D\)

In above equation

\(\vec{r}\) is either the position vector of any arbitrary point On the Line or a vector from any point Outside the Line to a point On the Line.
\(\hat{n}\) is a unit normal vector to the Line.
\(D\) is signed distance of the Line from the origin if \(\vec{r}\) is the position vector of any arbitrary point on the Line. Otherwise its signed distance of point Outside the Line to the Line.

Please note that the above equations correspond to the Unit Normal Distance Form of Scalar Equation of Line in 2D.
Point Direction Ratio Form: The Point Direction Ratio Form of vector equation of a Line passing through a point having position vector \(\vec{a}\) and having a direction vector \(\vec{b}\) is given as:

\((\vec{r}-\vec{a})\times\vec{b}=0\)

In this equation \(\vec{r}\) is position vector of any point on Line.
2 Point Form: The 2 Point Form of vector equation of a Line passing through 2 points having position vectors \(\vec{a}\) and \(\vec{b}\) is given as:

\((\vec{r}-\vec{a})\times(\vec{b}-\vec{a})=0\)

In this equation \(\vec{r}\) is position vector of any point on Line. Please note that any position vector may be subtracted from \(\vec{r}\).

Equation of Lines in 3D

Equations in Coordinate Form

Point Direction Ratio Form:The Point Direction Ratio Form of scalar equation of a Line passing through point \((x_1,y_1,z_1)\) having a direction ratio \((x_d,y_d,z_d)\) is given as following:

\(\frac{x-x_1}{x_d}=\frac{y-y_1}{y_d}=\frac{z-z_1}{z_d}\)

Please note the following this equation is similar to the Point Slope Form of line equation in 2D.
Three Plane Form: The Three Plane Form equation can be formed by resolving Point Direction Ratio Form into 3 equations (1 each for \(x\) and \(y\), \(y\) and \(z\) and \(z\) and \(x\)) as given in the following:

\(Ax + By = C\)
\(Dy + Ez = F\)
\(Gz + Hx = I\)

These 3 equations represent the equation of 3 planes at the intersection of which the line lies.
Explicit Form: The Point Direction Ratio Form or Three Plane Form of equations can also be resolved into sets of 2 explicit equations of planes each as given in the following:

\(y = Ax + B , \hspace{.5cm} z = Cx + D\)     (When \(y\) and \(z\) are given as a function of \(x\))

\(z = Ay + B , \hspace{.5cm} x = Cy + D\)     (When \(z\) and \(x\) are given as a function of \(y\))

\(x = Az + B , \hspace{.5cm} y = Cz + D\)     (When \(x\) and \(y\) are given as a function of \(z\))

\(x = C , \hspace{.5cm} z = Ay + D\)    (Line perpendicular to \(x\) axis, \(z\) given as a function of \(y\))

\(x = C , \hspace{.5cm} y = Az + D\)    (Line perpendicular to \(x\) axis, \(y\) given as a function of \(z\))

\(y = C , \hspace{.5cm} x = Az + D\)    (Line perpendicular to \(y\) axis, \(x\) given as a function of \(z\))

\(y = C , \hspace{.5cm} z = Ax + D\)    (Line perpendicular to \(y\) axis, \(z\) given as a function of \(x\))

\(z = C , \hspace{.5cm} y = Ax + D\)    (Line perpendicular to \(z\) axis, \(y\) given as a function of \(x\))

\(z = C , \hspace{.5cm} x = Ay + D\)    (Line perpendicular to \(z\) axis, \(x\) given as a function of \(y\))

Each of the above set of plane equation is of Coordinate Representation of Line in 3D.

Equations in Parametric Form

The following are Parametric Form representation/derivation of equations of lines in 3D:

Point Direction Ratio Form: The Parametric Equation of a Line passing through point \((x_1,y_1,z_1)\) having a direction ratio \((x_d,y_d,z_d)\) can be found by equating the Point Direction Ratio Form of Equation to a constant \(t\) as follows:

\(\frac{x-x_1}{x_d}=\frac{y-y_1}{y_d}=\frac{z-z_1}{z_d}=t\)

\(\Rightarrow x= x_1 + tx_d ,\hspace{.5cm}y= y_1 + ty_d ,\hspace{.5cm}z= z_1 + tz_d\)

\(t\) can be any real value.
Explicit Form: The Parametric Equation of a Line can also be given on the basis of Explicit Equations as follows:

\(x=t ,\hspace{.5cm} y=At + B ,\hspace{.5cm} z=Ct + D\)    (When \(y\) and \(z\) are given as a function of \(x\))

\(y=t ,\hspace{.5cm} z=At + B ,\hspace{.5cm} x=Ct + D\)    (When \(z\) and \(x\) are given as a function of \(y\))

\(z=t ,\hspace{.5cm} x=At + B ,\hspace{.5cm} y=Ct + D\)    (When \(x\) and \(y\) are given as a function of \(z\))

\(x=C ,\hspace{.5cm} y=t ,\hspace{.5cm} z=At + B\)    (Line perpendicular to \(x\) axis, \(z\) given as a function of \(y\))

\(x=C ,\hspace{.5cm} y=At + B ,\hspace{.5cm} z=t\)    (Line perpendicular to \(x\) axis, \(y\) given as a function of \(z\))

\(y=C ,\hspace{.5cm} z=t ,\hspace{.5cm} x=At + B\)    (Line perpendicular to \(y\) axis, \(x\) given as a function of \(z\))

\(y=C ,\hspace{.5cm} z=At + B ,\hspace{.5cm} x=t\)    (Line perpendicular to \(y\) axis, \(z\) given as a function of \(x\))

\(z=C ,\hspace{.5cm} x=t ,\hspace{.5cm} y=At + B\)    (Line perpendicular to \(z\) axis, \(y\) given as a function of \(x\))

\(z=C ,\hspace{.5cm} x=At + B ,\hspace{.5cm} y=t\)    (Line perpendicular to \(z\) axis, \(x\) given as a function of \(y\))

\(t\) can be any real value. \(C\) is a constant.

Equations in Vector Form

Position Vector Form: The following are Position Vector form of equation of Line in 3D that can be found by using the value of \(x\), \(y\) and \(z\) from the Parametric Equations:

\(\vec{P}= (x_1 + tx_d)\hat{\textbf{i}} + (y_1 + ty_d)\hat{\textbf{j}} + (z_1 + tz_d)\hat{\textbf{k}}\)    (For a Line passing through point \((x_1,y_1,z_1)\) having a direction ratio \((x_d,y_d,z_d)\))

\(\vec{P}= t\hat{\textbf{i}} + (At + B)\hat{\textbf{j}} + (Ct + D)\hat{\textbf{k}}\)   (When \(y\) and \(z\) are given as a function of \(x\))

\(\vec{P}= (Ct + D)\hat{\textbf{i}} + t\hat{\textbf{j}} + (At + B)\hat{\textbf{k}}\)   (When \(z\) and \(x\) are given as a function of \(y\))

\(\vec{P}= (At + B)\hat{\textbf{i}} + (Ct + D)\hat{\textbf{j}} + t\hat{\textbf{k}} \)   (When \(x\) and \(y\) are given as a function of \(z\))

\(\vec{P}= C\hat{\textbf{i}} + t\hat{\textbf{j}} + (At + B)\hat{\textbf{k}}\)   (Line perpendicular to \(x\) axis, \(z\) given as a function of \(y\))

\(\vec{P}= C\hat{\textbf{i}} + (At + B)\hat{\textbf{j}} + t\hat{\textbf{k}}\)   (Line perpendicular to \(x\) axis, \(y\) given as a function of \(z\))

\(\vec{P}= (At + B)\hat{\textbf{i}} + C\hat{\textbf{j}} + t\hat{\textbf{k}}\)   (Line perpendicular to \(y\) axis, \(x\) given as a function of \(z\))

\(\vec{P}= t\hat{\textbf{i}} + C\hat{\textbf{j}} + (At + B)\hat{\textbf{k}}\)   (Line perpendicular to \(y\) axis, \(z\) given as a function of \(x\))

\(\vec{P}= t\hat{\textbf{i}} + (At + B)\hat{\textbf{j}} + C\hat{\textbf{k}}\)   (Line perpendicular to \(z\) axis, \(y\) given as a function of \(x\))

\(\vec{P}= (At + B)\hat{\textbf{i}} + t\hat{\textbf{j}} + C\hat{\textbf{k}}\)   (Line perpendicular to \(z\) axis, \(x\) given as a function of \(y\))
Point Direction Ratio Form: The Point Direction Ratio Form of vector equation of a Line passing through a point having position vector \(\vec{a}\) and having a direction vector \(\vec{b}\) is given as:

\((\vec{r}-\vec{a})\times\vec{b}=0\)

In this equation \(\vec{r}\) is position vector of any point on Line.
2 Point Form: The 2 Point Form of vector equation of a Line passing through 2 points having position vectors \(\vec{a}\) and \(\vec{b}\) is given as:

\((\vec{r}-\vec{a})\times(\vec{b}-\vec{a})=0\)

In this equation \(\vec{r}\) is position vector of any point on Line. Please note that any position vector may be subtracted from \(\vec{r}\).

Related Topics and Calculators

Introduction to Lines, Finding Points on Line/Intercepts of Line, Types of Lines in 2D, Types of Lines in 3D, Condition for Collinearity of 3 Points, Angular Slope of a Line in 2D, Angular Normal of a Line in 2D, Angle Between 2 Lines, Relation Between 2 Lines, Condition for Concurrency of Lines, Family of Lines in 2D

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