Answer:
Choice D
Step-by-step explanation:
For this one I would find if the point lands on the line.
Choice A:
What we have to do is to plug in -4 for x and 4 for y.
[tex]y=2x+10\\(4)=2(-4)+10\\4=-8+10\\4\ne2[/tex]
The point is not on this line so this cannot be it.
Choice B:
We pug what we know again.
[tex]y=\frac{1}{3}x+1\\(4)=\frac{1}{3}(-4)+1\\(4)=-\frac{4}{3}+1\\4\ne-\frac{1}{3}[/tex]
The point is not on this line so it can't be it.
Choice C:
We pug in what we know again.
[tex]y=3x-9\\(4)=3(-4)-9\\(4)=-12-9\\4\ne-21[/tex]
The point is not on this line so it can't be it.
The next one has to be it, but we'll check it just in case.
Choice D:
We plug in what we know again.
[tex]y= \frac{1}{2}x+6\\(4)=\frac{1}{2}(-4)+6\\(4)=-2+6\\4=4[/tex]
The point is on this line so this is the line.
