Question 336497
Note: *[Tex \LARGE \left(x_{1}, y_{1}\right)] is the first point *[Tex \LARGE \left(a,b\right)]. So this means that {{{x[1]=a}}} and {{{y[1]=b}}}.

Also, *[Tex \LARGE \left(x_{2}, y_{2}\right)] is the second point *[Tex \LARGE \left(b,a\right)].  So this means that {{{x[2]=b}}} and {{{y[2]=a}}}.



{{{m=(y[2]-y[1])/(x[2]-x[1])}}} Start with the slope formula.



{{{m=(a-b)/(b-a)}}} Plug in {{{y[2]=a}}}, {{{y[1]=b}}}, {{{x[2]=b}}}, and {{{x[1]=a}}}



{{{m=(a-b)/(-1(-b+a))}}} Factor out a -1 from the denominator.



{{{m=-(a-b)/(a-b)}}} Rearrange the terms.



{{{m=-cross((a-b))/cross((a-b))}}} Cancel out the common terms.



{{{m=-1}}} Simplify



So the slope of the line through the points (a,b) and (b,a) is {{{m=-1}}}. This is true regardless of the values of 'a' and 'b' as long as {{{a<>b}}}



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Jim