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I'm very new to proofs and I was asked to prove this by assuming I have two lines that pass through the given points and then prove that the assumed lines are actually the same.

Suppose that (a,b) and (c,d) are two distinct points that are real numbered ordered pairs. Use a uniqueness proof to prove that there exists a unique line passing through the two points.

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  • $\begingroup$ You'll need to provide some context. First off, what is your definition of a line? $\endgroup$ – dxiv Sep 16 '16 at 1:22
  • $\begingroup$ Generally the technique is to assume that there are two lines that meet the criteria, then show that they are in fact the same line. $\endgroup$ – Doug M Sep 16 '16 at 1:23
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Hypothesis

There is a unique line that passes through two points $ (a,b) $ and $ (c,d) $.

Definitions

$ \forall x \in X, \exists y \in Y, : f(x) \rightarrow y := (x,y) \in X \times Y $ has a gradient in the form of $ m = \frac{y - y_{1}}{x - x_{1}} $ - $ (1) $

There exists a function in the form of $ y = mx + c $ were $ m $ is the gradient, $ c $ is the y-intercept and $ x, y $ are ordered pairs. - $(2)$

Proposition

Using $ (1) $ and $ (2) $ we find $ y - y_{1} = m(x - x_{1}) $

Therefore $ y - b = mx - md $ and $ y - a = mx - mc $

$ y = m(x-d) + b $ and $ y = m(x-c) + d $

equating the above

$ m(x - a) + b = m(x - c) + d $

$ d - b = m(a - c) $

$ m = \frac{d - b}{c - a} $

$$ \hspace{15cm} \square $$

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