## anonymous one year ago find the distance from the line y-6=m(x+1) to the point given below

1. anonymous

$(\sqrt{(1+m^2)},6+\sqrt{(1+m^2)} )$$(1,4)$and $(-2,1)$these are the three given points

2. amistre64

what methods do you have available at your disposal?

3. anonymous

I seeing this formula but not sure what to do..$d=\frac{ Ax+By+C }{ \pm \sqrt{A^2+B^2} }$

4. amistre64

i cant say i know what that refers to either ... but i have an idea on how to find a solution. we need to know the distance formula, how to determine a perpendicular slope, formulate a line equation from a point and a slope, and determine where 2 lines cross at.

5. anonymous

ok how do we do that

6. amistre64

another idea involves making a circle at a point, seeing where the line crosses it, and taking the midpoint, then distancing the results

7. amistre64

well, given a slope of a line: m -1/m creates a perpendicular slope ... for a given point (a,b), the form of the perp line is: -1/m(x-a) +b finding where the two lines meet ... is equating them -1/m (x-a) + b = m(x+1)+6 -x/m +a/m + b = mx +m +6 a/m + b -(m +6) = mx+x/m a/m + b -(m +6) = x(m+1/m) (a/m + b -(m +6))/(m+1/m) = x knowing x, we find y by substitution etc ...

8. amistre64

but if you can figure out that formula, and it is one for your course ... go for it :)

9. amistre64

|dw:1443919404316:dw| i think that formula of yours is from the distance between 2 planes tho

10. anonymous

I don't know

11. anonymous

we don't know whats m

12. amistre64

m is just a general value ... the solution will be general as well

13. anonymous

so y=(a/m + b -(m +6))/(m+1/m) ??

14. amistre64

given y = m(x+1)+6 and x = (a/m + b -(m +6))/(m+1/m) , i hope i did that right :) y = mx + m + 6 mx = (a + bm -m^2 +6m)/(m+1/m) = -m(m^2- (6+b)m -a)/(m^2+1) y = -m(m^2- (6+b)m -a)/(m^2+1) + m + 6 might be prudent to find the values for x and y before trying to process a general formula, it is starting to look messy

15. amistre64

lets use the (1,4) for (a,b)

16. amistre64

and for simplicity, assume a slope of 1 for brevity

17. anonymous

ok so we can use any poits for (a,b)?

18. amistre64

x = (a/m + b -(m +6))/(m+1/m) x = (1/1 + 4 -(1 +6))/(1+1/1) x = (-2)/(2) = -1 y = m(-1+1)+6 = 6 well, (a,b) should be a stated point, since that is the one we are trying to define a line with and cross it

19. amistre64

http://www.wolframalpha.com/input/?i=y%3D1%28x%2B1%29%2B6%2C+y%3D-1%28x-1%29%2B4 the good news maybe that our formulas for finding the crossing point are good

20. anonymous

I think ill understand this ...if I wasn't this sleepy

21. amistre64

what is the distance from (1,4) to (-1,6) ?

22. anonymous

4+4=sq.rt 8

23. amistre64

so 2 sqrt2

24. amistre64

but this assumes a slope; m=1

25. amistre64

trying to think of a vector way to approach it, or translation of the line so that we are finding the distance from the origin

26. anonymous

ok if it work....but this is geometry though

27. amistre64

we can use trig?

28. anonymous

we can use anything ..I guess

29. amistre64

well, y-6 = m(x+1) can be dropped 6 units to pass thru the origin as y = m(x+1), since we drop the line, we drop the point by 6 as well (1,4-6) = (1,-2) |dw:1443922039651:dw|

30. amistre64

$\vec n=(a,b-6)$ $\vec \mu=(1,m )$ $cos(\alpha)=\frac{a+m(b-6)}{\sqrt{a^2+(b-6)^2}~\sqrt{1+m^2} }$ $\alpha=cos^{-1}\left(\frac{a+m(b-6)}{\sqrt{a^2+(b-6)^2}~\sqrt{1+m^2} }\right)$ $d=|n|sin(\alpha)$ $d=\sqrt{a^2+(b-6)^2}~sin(\alpha)$

31. amistre64

lets test this thought out ... m=1, a=1,b=4 sqrt(1+4)sin(cos^(-1)((1+1(-2))/(sqrt(5)sqrt(2)))) well something went aloof, we didnt get 2 sqrt2 inthe simplification

32. anonymous

ok I'm lost... I'm sleepy ...but I think m shd be -1

33. amistre64

m is the slope of the stated line ... we can simplify the process if we make it a concrete value like 1 or -1 but overall the distance solution required has to keep m general

34. amistre64

http://www.mathwords.com/d/distance_point_to_line.htm or this defines the parts of your formula for us

35. amistre64

and m = -a/b

36. anonymous

yea that same formula I have here ...but idk

37. amistre64

y - 6 = mx +m (m)x + (-1)y = (m+6) A = m, B=-1, C=m+6

38. amistre64

so we know ABandC, and stated point values ...

39. anonymous

wait 1 sec ..let me comprehend this

40. amistre64

my C is off, foiled by a negative

41. anonymous

y-6=mx+m -6-m=mx-y -(6+m)=mx-y

42. amistre64

yeah

43. anonymous

ok so mx-y+(6+m)=0 a=m b=-1 c=(6+m) ok I see

44. amistre64

ugh, im tired too, im used to C being on the other side of =

45. anonymous

no c=6+m remember we bring it back over the equal sign so it become positive

46. amistre64

yeah :) so my faux pas actually worked in my favor the first time so the point (1,4), assuming m=1 should be |1-4+7|/sqrt(2) = 4/sqrt(2) = 2sqrt(2)

47. amistre64

and that is what we worked so hard to determine the first go around lol

48. anonymous

where did u get the sq.rt of 2 frm

49. amistre64

sqrt(a^2+b^2) sqrt(m^2+1) , assumed m=1 sqrt(2) as a divisor 1/sqrt(2) .... rationalize the denominator (just multiply it all by sqrt(2)/sqrt(2) ) sqrt(2)/2 well, 4/2 = 2 sooo that reduces to 2s qrt(2)

50. anonymous

but b=-1

51. amistre64

(-1)^2 = 1

52. anonymous

o yea I see.. forget about the sq.

53. anonymous

so were finish with this question?

54. amistre64

all there is to it is plugging in our known values and leaving the rest to the imagination

55. anonymous

ok but what about the other point (-2,1)

56. amistre64

$\frac{|mx_0-y_0+m+6|}{\sqrt{m^2+1}}$ $\frac{|-2m-1+m+6|}{\sqrt{m^2+1}}$ $\frac{|-m+5|}{\sqrt{m^2+1}}$

57. anonymous

is that another equation that we can use to find distance

58. amistre64

59. amistre64

all i did was determine what A B and C were, and then its just a matter of plugging in the point values

60. anonymous

ok I will have to look this through back when I'm fully awake to understand everything ..thanks very much

61. amistre64

good luck