Prove that the point-slope form of a linear equation does not depend on the point

Question

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?

Answer 1

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 $$(y_1−y_2)x+(x_2−x_1)y=x_2y_1−x_1y_2$$