Question 1179372


find the value of {{{y}}} such that the points are collinear:

({{{-6}}}, {{{-4}}}), ({{{0}}},{{{ y}}}), ({{{-3}}}, {{{-3}}})


the points are collinear if they lie on same line

{{{y=mx+b}}}

use given points to calculate a slope {{{m}}}

({{{-6}}}, {{{-4}}}) and ({{{-3}}}, {{{-3}}})

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

{{{m=(-3-(-4))/(-3-(-6))}}}

{{{m=(-3+4)/(-3+6)}}}

{{{m=1/3}}}

{{{y=(1/3)x+b}}}...........use one point to calculate {{{b}}} 


{{{-3=(1/3)(-3)+b}}}

{{{-3=-1+b}}}

{{{-3+1=b}}}

{{{b=-2}}}

your equation is:


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

now we can find missing coordinate of the point ({{{0}}},{{{ y}}})


{{{y=(1/3)*0-2}}}

{{{y=-2}}}


the point is ({{{0}}},{{{ -2}}})



{{{ drawing( 600, 600, -10, 10, -10, 10,
circle(-3,-3,.12), locate(-3,-3,p(-3,-3)),
circle(0,-2,.12), locate(0,-2,p(0,-2)),
circle(-6,-4,.12), locate(-6,-4,p(-6,-4)),
graph( 600, 600, -10, 10, -10, 10, (1/3)x-2)) }}}