anonymous
  • anonymous
Choose the point-slope form of the equation below that represents the line that passes through the points (-6, 4) and (2, 0).
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
@abb0t
misssunshinexxoxo
  • misssunshinexxoxo
y − 4 = −one half(x + 6)
misssunshinexxoxo
  • misssunshinexxoxo
@TheDragon do you know point slope intercept form?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
The choices are y - 4 = -1/2 (x + 6) y - 4 = 2 (x + 6) y + 6 = -1/2 (x - 4) y + 6 = 2 (x - 4)
anonymous
  • anonymous
thanks, and kind of @misssunshinexxoxo
abb0t
  • abb0t
For this one, you do \(\sf \large \color{blue}{\frac{y_2-y_1}{x_2-x_1}}\) and then plug it into the point-slope equation: So you get \(\large \frac{0-4}{2+6 } = \frac{-4}{8} = \frac{-1}{2} \) so your slope is \(-\frac{1}{2}\) now pick any point, i will pick second one y - 0 = \(\frac{1}{2}\)(x-2) or more simplified y = -\(\frac{1}{2}\)(x-2) which is the same thing as \(\sf \color{red}{y=-\frac{1}{2}x + 1}\)
anonymous
  • anonymous
Thank you @abb0t can you help me on another question?

Looking for something else?

Not the answer you are looking for? Search for more explanations.