Equation of Straight Line in Plane/Two-Point Form/Proof 2

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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:

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


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} } } = 0$

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

$\blacksquare$


Sources

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