Fang02 2 years ago Can someone help me with this please? I don't know how to do this. Write the equation for the horizontal line that contains point G(-8, 8).

1. tcarroll010

Horizontal lines are of the form y=n where you may have been given a point in the form (m, n). You have such a point G.

2. piscez.in

keep in mind that a horizontal line is parallel to x axis, and slope is the value of tan of the angle between the x axis and the line

3. Fang02

@piscez.in: What do you mean by the slope is the value of tan of the angle between the x-axis and the line? Specifically, what do you mean by the value of tan? @tcarroll010: Okay, just a moment please.

4. Fang02

So it would be something like y = 8 - (-8)?

5. piscez.in

@fang02 tan value as in tangent value, like sine cosine, tan etc.

6. Fang02

Hmm..I do not believe I have learned about the tan value then.

7. tcarroll010

He means that the equation of a line in slope-intercept form, in general is y = mx + b and that here we have "0" slope, so the mx term disappears. So, we have y=b, or in my notation, y=n. You have point (-8, 8) which is the same distance away from the x-axis as the "b" intercept, (0, 8), so you can simplify and use the "n".

8. piscez.in

@Fang02 oh im sorry hun, well then forget it. But try to roughly draw the line on a paper. Find the y intercept. And you must already be knowing that the slope of horizontal lines( lines parallel to x axis) is always 1

9. tcarroll010

Not y = 8 - (-8). Instead, y=n where n = 8

10. Fang02

So, the answer is pretty much y = 8 then? No need to be sorry pisez.in.

11. tcarroll010

Yes! You got it. Good job! It's just y=8.

12. Fang02

Okay, thanks a lot! :) Would you mind helping me with another problem real quick?

13. tcarroll010

sure!

14. Fang02

Write an equation in point-slope form of the line through point J(4, 1) with slope -4.

15. Fang02

Point-slope form is y2 - y1 over x2 over x1 So, I think it would be: 1 - 4 over something

16. tcarroll010

point-slope form is going to be:\[y - y _{1} = m(x - x _{1})\] Here, y is y and x is x as variables. x1 and y1 are a SPECIFIC point in the form (x1, y1)

17. tcarroll010

(x1, y1) is the point you are given that is (4, 1), so x1 is 4 and y1 is 1.

18. Fang02

I think that (x1, y1) would be (-4, 1) right?

19. Fang02

Why does the 4 turn positive?

20. tcarroll010

(x1, y1) is the point you are given that is (4, 1), so x1 is 4 and y1 is 1. It's the slope that is -4.

21. Fang02

I understand.

22. tcarroll010

So now, just substitute.

23. Fang02

y - y1 = m(x - x1) y - 1 = -4 (4 - 4) ?

24. tcarroll010

Not quite there yet. What happened to "x"? "x" stays as "x" just like "y" stays as "y". You only substitute for y1, x1, and m.

25. Fang02

Oh okay. So, y = 1 = -4(x - 4)

26. tcarroll010

You still have a small typo, but you are extremely close now.

27. Fang02

Would I put x1 in parenthesis to show that it is positive and is just being subtracted? y = 1 = -4(x - (4))

28. tcarroll010

You don't have to do that, you can just leave it as y - 1 = -4(x - 4). Notice how after the "y" I changed that from "=" to "-"

29. Fang02

Oh okay, I understand now.

30. tcarroll010

Very nice working with you and you will get there. Just keep up the good work!

31. Fang02

Thanks! You too! :)