## Emra Group Title Find The equestion in slope intercept form for the line contaning the two points. (1,1 ) and ( 5,4 ) 2 years ago 2 years ago

1. Mrs.Math101 Group Title

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2. Mrs.Math101 Group Title

you know how to use the equation?

3. Callum29 Group Title

that gives the gradient of the line

4. Emra Group Title

Do You Start With Y1 Or Y2

5. Callum29 Group Title

call x_1= 1, y_1 =1, x_2=5, y_2 = 4

6. Mrs.Math101 Group Title

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7. Callum29 Group Title

the gradient, m is 3/4

8. Callum29 Group Title

now you want the equation in the form: y=mx + c

9. Mrs.Math101 Group Title

y=mx+B

10. MohamedDada Group Title

the answer is y = 3/5x -1

11. MohamedDada Group Title

i did all my work the answer doesnt show up

12. Mrs.Math101 Group Title

do you get emra?

13. MohamedDada Group Title

no i didnt get emra she's hard to catch o.O

14. Callum29 Group Title

you know m = 3/4. Take the first coordinate pair (1,1). Set y=1 and x=1. Then: $1=3/4 + c$ $c=1/4$ You can check this value by using the other coordinate pair (5,4) $4=15/4 + c$ $c=(16-15)/4 = 1/4$ Therefore the equation of your line is $y=\frac{3}{4}x + \frac{1}{4}$

15. Callum29 Group Title

it's not y=3/5x - 1 though...

16. MohamedDada Group Title

thats how we learned it

17. MohamedDada Group Title

18. MohamedDada Group Title

i see my mistake

19. MohamedDada Group Title

yeah its y = 3/4 x + 1/4

20. Callum29 Group Title

you forgot to minus the 1 from the 5 didn't you?

21. MohamedDada Group Title

no i wrote the slope wrong -.-

22. Emra Group Title

imm confused alot know

23. Emra Group Title

5-1 is 4 right

24. Callum29 Group Title

yeah ^^ i remember when I was doing all that geometry was fun, I went to classes drunk for the lols

25. MohamedDada Group Title

lol

26. Callum29 Group Title

5-1 is 4, bravo, bravo

27. Emra Group Title

heheheheh

28. Emra Group Title

How would you do this one (8,6 ( and (-3,6)

29. Callum29 Group Title

y_2 - y_1 = 6 - 6 = 0 x_2 - x_1 = -3 - 8 = -11

30. Emra Group Title

0 -- 11

31. Emra Group Title

then what do i do

32. Callum29 Group Title

it's zero yes, what does that mean about the line?

33. Callum29 Group Title

hint: m=0

34. Emra Group Title

?/// thats what i dont get ohh m=0 know we have to see what b=

35. Emra Group Title

??

36. Callum29 Group Title

go back to the general equation of a line: $y=mx+b$ in this case, m=0 so it' just $y=b$ This tells you that the line is a straight line forever, parallel to the x axis and intersecting the y axis at 6. Now look at your two pairs of coordinates. y is 6 in both (it better be)! So the equation of your line is......?

37. Emra Group Title

6 ?

38. Emra Group Title

6 -- 6 = 0 ??

39. Callum29 Group Title

yes $y=6$ is the equation of your line, it's a very boring graph if you ask me

40. Emra Group Title

so what the equestions y=0x

41. Callum29 Group Title

yes 0x=0 for any x

42. Callum29 Group Title

so it's not in the equation

43. Emra Group Title

so whats the equestion then is it y=ox

44. Callum29 Group Title

remember x is a variable, and is allowed to take all values on the real line, but it obeys the law 0x=0

45. Emra Group Title

kk is that the equestion

46. Emra Group Title

k how would you do this one ( 100,-300 ) and (101,-299 )

47. Callum29 Group Title

do it how you think...

48. Emra Group Title

-1 ____ -1 = 1 ?

49. Emra Group Title

H-E-L-P

50. Callum29 Group Title

y_2-y_1 = -299 - (-300) = -299 + 300 = 1 x_2 - x_1 = 101 - 100 = 1 so it is 1/1 = 1. Your way worked too which is a good thing!

51. Callum29 Group Title

so yes, you are right

52. Emra Group Title

so what would the equestion bee y=-1x+ ??

53. Callum29 Group Title

use y=x+b and one of your coordinate pairs to solve the equation for b. (Gradient is 1 not -1 by the way).