Question 175560
We can see that the equation {{{y=-(2/3)x-12}}} is in slope intercept form {{{y=mx+b}}} where the slope is {{{m=-2/3}}} and the y-intercept is {{{b=-12}}} note: the y-intercept is the point *[Tex \LARGE \left(0,-12\right)]




Since we're looking at the point (-4,-10), this means that {{{x=-4}}} and {{{y=-10}}}



{{{y=-(2/3)x-12}}} Start with the given equation.



{{{-10=-(2/3)(-4)-12}}} Plug in {{{x=-4}}} and {{{y=-10}}}



{{{-10=8/3-12}}} Multiply 



{{{-10=8/3-36/3}}} Multiply 12 by {{{3/3}}}



{{{-10=-28/3}}} Subtract the fractions.



Since the equation is NOT true, this means that the point (-4,-10) is NOT on the line {{{y=-(2/3)x-12}}}. So the flamingo at the point (-4,-10) is NOT on the water line {{{y=-(2/3)x-12}}}



So you can place the flamingo at the point (-4,-10).