Equation of Straight Line in Plane/Two-Point Form/Proof 1
Jump to navigation
Jump to search
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:
- $\dfrac {x - x_1} {x_2 - x_1} = \dfrac {y - y_1} {y_2 - y_1}$
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 | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds y \paren {x_2 - x_1} - y_1 \paren {x_2 - x_1}\) | \(=\) | \(\ds x \paren {y_2 - y_1} - x_1 \paren {y_2 - y_1}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {y - y_1} \paren {x_2 - x_1}\) | \(=\) | \(\ds \paren {x - x_1} \paren {y_2 - y_1}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac {y - y_1} {x - x_1}\) | \(=\) | \(\ds \dfrac {y_2 - y_1} {x_2 - x_1}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac {x - x_1} {x_2 - x_1}\) | \(=\) | \(\ds \dfrac {y - y_1} {y_2 - y_1}\) |
$\blacksquare$