Equation of Straight Line in Plane/Two-Point Form

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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

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:

$\begin {cases} x = x_1 + t \paren {x_2 - x_1} \\ y = y_1 + t \paren {y_2 - y_1} \end {cases}$


Determinant Form

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 1

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 $\paren {x_2, y_2}$:

\(\ds y_1\) \(=\) \(\ds m x_1 + c\)
\(\ds y_2\) \(=\) \(\ds m x_2 + c\)
\(\ds \leadsto \ \ \) \(\ds c\) \(=\) \(\ds y_1 - m x_1\)
\(\ds \leadsto \ \ \) \(\ds y_2\) \(=\) \(\ds m x_2 + y_1 - m x_1\)
\(\text {(2)}: \quad\) \(\ds \leadsto \ \ \) \(\ds m\) \(=\) \(\ds \dfrac {y_2 - y_1} {x_2 - x_1}\)


\(\ds y_1\) \(=\) \(\ds m x_1 + c\)
\(\ds y_2\) \(=\) \(\ds m x_2 + c\)
\(\ds \leadsto \ \ \) \(\ds m\) \(=\) \(\ds \dfrac {y_2 - c} {x_2}\)
\(\ds \leadsto \ \ \) \(\ds y_1\) \(=\) \(\ds \dfrac {y_2 - c} {x_2} x_1 + c\)
\(\ds \leadsto \ \ \) \(\ds y_1 x_2\) \(=\) \(\ds x_1 y_2 + c \paren {x_2 - x_1}\)
\(\text {(3)}: \quad\) \(\ds \leadsto \ \ \) \(\ds c\) \(=\) \(\ds \dfrac {y_1 x_2 - x_1 y_2} {x_2 - x_1}\)


Substituting for $m$ and $c$ in $(1)$:

\(\ds y\) \(=\) \(\ds m x + c\) which is $(1)$
\(\ds \leadsto \ \ \) \(\ds y\) \(=\) \(\ds \dfrac {y_2 - y_1} {x_2 - x_1} x + \dfrac {y_1 x_2 - x_1 y_2} {x_2 - x_1}\) from $(2)$ and $(3)$
\(\ds \leadsto \ \ \) \(\ds y \paren {x_2 - x_1} + x_1 y_2\) \(=\) \(\ds x \paren {y_2 - y_1} + y_1 x_2\)
\(\ds \leadsto \ \ \) \(\ds y \paren {x_2 - x_1} + x_1 y_2 - y_1 x_1\) \(=\) \(\ds x \paren {y_2 - y_1} + y_1 x_2 - x_1 y_1\) adding $y_1 x_1 = x_1 y_1$ to both sides
\(\ds \leadsto \ \ \) \(\ds y \paren {x_2 - x_1} - y_1 \paren {x_2 - x_1}\) \(=\) \(\ds x \paren {y_2 - y_1} - x_1 \paren {y_2 - y_1}\)
\(\ds \leadsto \ \ \) \(\ds \paren {y - y_1} \paren {x_2 - x_1}\) \(=\) \(\ds \paren {x - x_1} \paren {y_2 - y_1}\)
\(\ds \leadsto \ \ \) \(\ds \dfrac {y - y_1} {x - x_1}\) \(=\) \(\ds \dfrac {y_2 - y_1} {x_2 - x_1}\)
\(\ds \leadsto \ \ \) \(\ds \dfrac {x - x_1} {x_2 - x_1}\) \(=\) \(\ds \dfrac {y - y_1} {y_2 - y_1}\)

$\blacksquare$


Proof 2

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:

\(\ds 0\) \(=\) \(\ds \dfrac 1 2 \size {\paren {\begin {vmatrix} x & y & 1 \\ x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ \end {vmatrix} } }\) Area of Triangle in Determinant Form
\(\ds \) \(=\) \(\ds x_1 y_2 - x_2 y_1 + x_2 y - x y_2 + x y_1 - x_1 y\)
\(\ds \leadsto \ \ \) \(\ds x_2 \paren {y - y_1} - y_2 \paren {x - x_1}\) \(=\) \(\ds y x_1 - x y_1\)
\(\ds \) \(=\) \(\ds x_1 \paren {y - y_1} - y_1 \paren {x - x_1}\)
\(\ds \leadsto \ \ \) \(\ds \paren {x_2 - x_1} \paren {y - y_1}\) \(=\) \(\ds \paren {y_2 - y_1} \paren {x - x_1}\)
\(\ds \leadsto \ \ \) \(\ds \dfrac {x_2 - x_1} {x - x_1}\) \(=\) \(\ds \dfrac {y_2 - y_1} {y - y_1}\)

$\blacksquare$


Proof 3

Straight-line-2-points-form-Proof-3.png


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:

\(\ds \dfrac {P_1 K} {P_1 H}\) \(=\) \(\ds \dfrac {K P} {H P_2}\)
\(\ds \leadsto \ \ \) \(\ds \dfrac {x - x_1} {x_2 - x_1}\) \(=\) \(\ds \dfrac {y - y_1} {y_2 - y_1}\)

$\blacksquare$


Proof 4

Straight-line-2-points-form-Proof-4.png


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:

\(\ds OA + AP\) \(=\) \(\ds OP\)
\(\ds \leadsto \ \ \) \(\ds z_1 + AP\) \(=\) \(\ds z\)
\(\ds \leadsto \ \ \) \(\ds AP\) \(=\) \(\ds z - z_1\)

and:

\(\ds OA + AB\) \(=\) \(\ds OB\)
\(\ds \leadsto \ \ \) \(\ds z_1 + AB\) \(=\) \(\ds z_2\)
\(\ds \leadsto \ \ \) \(\ds AB\) \(=\) \(\ds z_2 - z_1\)


Then:

\(\ds \exists t \in \R: \, \) \(\ds AP\) \(=\) \(\ds t AB\) as $A$, $P$ and $B$ are collinear
\(\ds \leadsto \ \ \) \(\ds z - z_1\) \(=\) \(\ds t \paren {z_2 - z_1}\)
\(\ds \leadsto \ \ \) \(\ds z\) \(=\) \(\ds \paren {1 - t} z_1 + t z_2\)
\(\ds \leadsto \ \ \) \(\ds x - x_1\) \(=\) \(\ds t \paren {x_2 - x_1}\)
\(\ds y - y_1\) \(=\) \(\ds t \paren {y_2 - y_1}\)
\(\ds \leadsto \ \ \) \(\ds \dfrac {x - x_1} {x_2 - x_1}\) \(=\) \(\ds \dfrac {y - y_1} {y_2 - y_1}\)

$\blacksquare$


Examples

$\tuple {1, 3}$ and $\tuple {-2, 4}$

Two-Point Form of Equation of Straight Line in Plane/Examples/(1, 3) and (-2, 4)

$\tuple {2, 1}$ and $\tuple {-6, 7}$

Let $\LL$ be the straight line passing through the points $\tuple {2, 1}$ and $\tuple {-6, 7}$.

Then $\LL$ has the equation:

$\dfrac {x - 2} {-6 - 2} = \dfrac {y - 1} {7 - 1}$

which after rearranging, can be written in the general form:

$4 y = -3 x + 10$


Sources

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