Question 1176519


given:

({{{x}}},{{{-4}}}) and ({{{2}}},{{{8}}})

{{{m=-3}}}

{{{m=(y[2]-y[1])/(x[2]-x[1])}}}


{{{-3=(8-(-4))/(2-x)}}}

{{{-3=(8+4)/(2-x)}}}

{{{-3=12/(2-x)}}}

{{{-3(2-x)=12}}}

{{{(2-x)=12/-3}}}

{{{2-x=-4}}}

{{{2+4=x}}}

{{{x=6}}}

then ({{{x}}},{{{-4}}}) is ({{{6}}},{{{-4}}})