Question 655658
If {{{f(2)=4}}},{{{f(4)=2}}}, {{{f(3)=3}}}, then the graph of the function contains the point ({{{2}}},{{{4}}}),({{{4}}},{{{2}}}), and ({{{3}}},{{{3}}}). 

Use the two point form of an equation of a straight line and the points ({{{2}}},{{{4}}}),({{{4}}},{{{2}}}): 

{{{f(x)-y1=((y1-y2)/(x1-x2))(x-x1)}}} where (x1,y1)and (x2,y2) are the coordinates of the given points. 

{{{f(x)-4=((4-2)/(2-4))(x-2)}}}


{{{f(x)-4=(2/-2)(x-2)}}}


{{{f(x)-4=-1(x-2)}}}

{{{f(x)-4=-x+2}}}

{{{f(x)=-x+2+4}}}

{{{highlight(f(x)=-x+6)}}}


{{{ graph( 600, 600, -10, 10, -10, 10, -x+6) }}}


here is a proof that all three of given points lie on this line


*[invoke To_determine_if_3_points_lie_in_a_line 2, 4, 4, 2, 3, 3]