## Emra 3 years ago Find The equestion in slope intercept form for the line contaning the two points. (1,1 ) and ( 5,4 )

1. Mrs.Math101

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2. Mrs.Math101

you know how to use the equation?

3. Callum29

that gives the gradient of the line

4. Emra

Do You Start With Y1 Or Y2

5. Callum29

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

6. Mrs.Math101

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7. Callum29

the gradient, m is 3/4

8. Callum29

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

9. Mrs.Math101

y=mx+B

the answer is y = 3/5x -1

i did all my work the answer doesnt show up

12. Mrs.Math101

do you get emra?

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

14. Callum29

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

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

thats how we learned it

i see my mistake

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

20. Callum29

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

no i wrote the slope wrong -.-

22. Emra

imm confused alot know

23. Emra

5-1 is 4 right

24. Callum29

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

lol

26. Callum29

5-1 is 4, bravo, bravo

27. Emra

heheheheh

28. Emra

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

29. Callum29

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

30. Emra

0 -- 11

31. Emra

then what do i do

32. Callum29

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

33. Callum29

hint: m=0

34. Emra

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

35. Emra

??

36. Callum29

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

6 ?

38. Emra

6 -- 6 = 0 ??

39. Callum29

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

40. Emra

so what the equestions y=0x

41. Callum29

yes 0x=0 for any x

42. Callum29

so it's not in the equation

43. Emra

so whats the equestion then is it y=ox

44. Callum29

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

kk is that the equestion

46. Emra

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

47. Callum29

do it how you think...

48. Emra

-1 ____ -1 = 1 ?

49. Emra

H-E-L-P

50. Callum29

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

so yes, you are right

52. Emra

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

53. Callum29

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