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I am stucked on the following challenge: "If the line determined by two distinct points $(x_1, y_1)$ and $(x_2, y_2)$ is not vertical, and therefore has slope $(y_2-y_1)/(x_2-x_1)$, show that the point-slope form of its equation is the same regardless of which point is used as the given point." Okay, we can separate $(x_0, y_0)$ from the form to get: $$y(x_2-x_1)-x(y_2-y_1) = y_0(x_2-x_1)-x_0(y_2-y_1)$$ But how exclude this point $(x_0,y_0)$ and leave only $x, y, x_1, y_1, x_2, y_2$ in the equation? UPDATE: There is a solution for this challenge: $$(y_1-y_2)x+(x_2-x_1)y=x_2y_1-x_1y_2$$ From the answer I found that $$y_2(x-x_1)-y_1(x-x_2)=y(x_2-x_1)$$ ... but why this is true?

Eugene Shpak
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    I think they mean "whether you use $$(x_1,y_1)$ of $(x_2,y_2)$ as the given point." – saulspatz Mar 24 '21 at 14:22
  • Nope, because there is a solution for this challenge without x_0 and y_0. I updated the question. – Eugene Shpak Mar 24 '21 at 16:50
  • If you are responding to my comment, that's what I meant, but I see that I messed up the MathJax. I don't follow your update. Why is what true? – saulspatz Mar 24 '21 at 16:55
  • You are right, thanks, finally I got it. The solution was so simple -_-. I posted the answer. P.S. The last equation in the question was formed, by extracting the common part of the reworked point-slope equation and the answer. So it can be called cheating – Eugene Shpak Mar 24 '21 at 21:31
  • No, it called learning, not cheating. This is very common. Once you solve a problem, and see what the answer is, you also see an easier way to get it. – saulspatz Mar 24 '21 at 21:46

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Thanks to saulspatz, the solution is to simply show, that whether we are using $(x_1, y_1)$ or $(x_2, y_2)$ as the given point, the equation does not change. So both equations: $$y-y_1=m(x-x_1)$$ $$y-y_2=m(x-x_2)$$ reduce to the $$(y_1−y_2)x+(x_2−x_1)y=x_2y_1−x_1y_2$$

Eugene Shpak
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