The equation is:
[tex]y=-3x[/tex]
so is in the form y = mx + c
this means that C = 0 so the intersection when x = 0 y = 0 so the first point is (0,0)
we also know that the slope is negative, so only points in the II and IV cuadrant can be part of the line so we evaluate them:
For point A
[tex]\begin{gathered} 9=-3(-3) \\ 9=9 \end{gathered}[/tex]
point A is part of the line (-3,9)
For point D
[tex]\begin{gathered} -6=-3(2) \\ -6=-2 \end{gathered}[/tex]
So D is part of the line
So the solution is: PointsIf you don’t need further explanation on this question, we can end the session. I’d really appreciate you letting me know how I did by rating our session after you exit. Thanks and have a great day! A, B, D
