Question 1126760

the equation of the line:

{{{y=mx+b}}} where {{{m}}} is a slope and {{{b}}} is y-intercept

use given points to ind a slope:  ({{{0}}}, {{{-2}}}) and ({{{6}}}, {{{0}}})

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

{{{m=(0-(-2))/(6-0)}}}

{{{m=2/6}}}

{{{m=1/3}}}

since {{{b}}} is y-intercept, and we know that y-intercept is at  ({{{0}}}, {{{y}}}), from given point  ({{{0}}}, {{{-2}}}) we see that {{{b=-2}}}

so, your equation is

{{{y=(1/3)x-2}}} 


{{{drawing( 600, 600, -10, 10, -10, 10,
circle(0,-2,.12),circle(6,0,.12),
locate(0,-2,p(0,-2)),locate(6,0.6,p(6,0)),
 graph( 600, 600, -10, 10, -10, 10, (1/3)x-2)) }}}