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Normally the equation of a line is written on standard form, slope-intercept form or point-slope form. However, one can equivalently write out the equation of a line on the form $(a_2 - b_1)(y - b_1) = (b_2 - b_1)(x - a_1)$. This form is apparently useful when trying to define all constructible numbers.

Does anyone know an intuitive proof that the equation of a line can be written on this form (i.e. a proof that is not built on 'reverse engineering' the given equation)?

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  • $\begingroup$ What are $a_1,a_2,b_1,b_2$? $\endgroup$ – Aweygan Dec 19 '17 at 14:30
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    $\begingroup$ I think you have a typo. The first factor should probably be $(a_2 - a_1)$. $\endgroup$ – André 3000 Dec 19 '17 at 14:33
  • $\begingroup$ Yes, you've got the wrong formula. $\endgroup$ – Thomas Andrews Dec 19 '17 at 14:39
  • $\begingroup$ @Quasicoherent. Indeed, I've made a mistake when copying the equation. Do you believe that I should correct the errors in my question or leave them be given that AsBk3397 have discussed the errors as part of his answer? $\endgroup$ – K. Claesson Dec 19 '17 at 15:03
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I think $(a_1,b_1)$ and $(a_2, b_2)$ are two points that line includes. But as indicated, it must be $(a_2 - a_1)(y - b_1) = (b_2 - b_1)(x - a_1)$ and actually this expression is not different from the equation of the form $$y - y_1 = m(x-x_1)$$ where $(x_1,y_1)$ is a point on the line and $m$ is the slope. But notice that if we take two distinct points on the line, say $(a_1,b_1)$ and $(a_2, b_2)$, then we can find slope $m = \frac{b_2-b_1}{a_2-a_1}$ when $a_1 \ne a_2$ (Notice that in this case, you have a line perpendicular to the x-axis, which has a slope angle of $90^\circ$). Then if we take one of these points, $(a_1,b_1)$ and write the line equation with slope, we have $$y-b_1 = \frac{b_2-b_1}{a_2-a_1}(x-a_1)$$ and the result follows.

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    $\begingroup$ The key difference between $y=m(x-x_1)$ and the question's formula is that it allows the line $x=2$ as well as the line $y=2$, so it is complete for all lines, not just functions with lines as the graph. Consider the case when $a_1=a_2$ but $b_1\neq b_2$, then your $m$ is not defined. $\endgroup$ – Thomas Andrews Dec 19 '17 at 14:43
  • $\begingroup$ Yes, that's right, I forgot to mention the case where the slope angle is $90^\circ$, thank you for the correction. $\endgroup$ – ArsenBerk Dec 19 '17 at 14:53
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There are at least four ways to write the straight line equation. All are same, nothing is "normal" over other forms.

  • $ y=mx+c$
  • $ \dfrac{b_2-b_1}{a_2-a_1}= \dfrac{y-b_1} {x-a_1} $
  • $x \cos \alpha + y \sin \alpha =p$
  • $ x/a+y/b=1$

The constants of each form are relatable to each other.

In particular, the second form equates slope between two fixed points and between variable points. The line segments have equal slope.

Try writing the second form into the first form so you can recognize slope and intercept with constants from second form.

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