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:

$\dfrac {x - x_1} {x_2 - x_1} = \dfrac {y - y_1} {y_2 - y_1}$


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:

$\begin {vmatrix} x & y & 1 \\ x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \end {vmatrix}$


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 $\paren {x_2, y_2}$:

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


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


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
\(\ds \leadsto \ \ \) \(\ds y \paren {x_2 - x_1} - y_1 \paren {x_2 - x_1}\) \(=\) \(\ds x \paren {y_2 - y_1} - x_1 \paren {y_2 - y_1}\)
\(\ds \leadsto \ \ \) \(\ds \paren {y - y_1} \paren {x_2 - x_1}\) \(=\) \(\ds \paren {x - x_1} \paren {y_2 - y_1}\)
\(\ds \leadsto \ \ \) \(\ds \dfrac {y - y_1} {x - x_1}\) \(=\) \(\ds \dfrac {y_2 - y_1} {x_2 - x_1}\)
\(\ds \leadsto \ \ \) \(\ds \dfrac {x - x_1} {x_2 - x_1}\) \(=\) \(\ds \dfrac {y - y_1} {y_2 - y_1}\)

$\blacksquare$


Sources

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