Question 657379
 an equation in slope-intercept form:

{{{y=mx+b}}}

for the line that contains the points ({{{2}}}, {{{1}}}) and ({{{-2}}}, {{{10}}}), will be:

{{{y=mx+b}}}....plug in given values for {{{x}}} and {{{y}}} from the point ({{{2}}}, {{{1}}})

{{{1=m*2+b}}}...........1

{{{y=mx+b}}}....plug in given values for {{{x}}} and {{{y}}} from the point ({{{-2}}}, {{{10}}})

{{{10=m*(-2)+b}}}...........2
--------------------------------solve the system

{{{1=2m+b}}}...........1


{{{10=-2m+b}}}...........2
-------------------------------add 1 and 2

{{{1+10=2m+(-2m)+b+b}}}

{{{11=cross(2m)+cross((-2m))+2b}}}

{{{11=2b}}}......solve for {{{b}}}

{{{11/2=b}}}

{{{highlight(5.5=b)}}}

now we can find {{{m}}}

{{{1=2m+b}}}...........1

{{{1-b=2m}}}

{{{1-5.5=2m}}}

{{{-4.5=2m}}}

{{{-4.5/2=m}}}

{{{highlight(-2.25=m)}}}

so, our equation is:

{{{y=-2.25x+5.5}}}


{{{drawing(600,600,-15,15,-15,15,grid(1),circle(2,1,0.2),circle(-2,10,0.2),graph(600,600,-15,15,-15,15,-2.25x+5.5))}}}