anonymous
  • anonymous
find the distance from the line y-6=m(x+1) to the point given below
Algebra
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[(\sqrt{(1+m^2)},6+\sqrt{(1+m^2)} )\]\[(1,4)\]and \[(-2,1)\]these are the three given points
amistre64
  • amistre64
what methods do you have available at your disposal?
anonymous
  • anonymous
I seeing this formula but not sure what to do..\[d=\frac{ Ax+By+C }{ \pm \sqrt{A^2+B^2} }\]

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amistre64
  • 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.
anonymous
  • anonymous
ok how do we do that
amistre64
  • amistre64
another idea involves making a circle at a point, seeing where the line crosses it, and taking the midpoint, then distancing the results
amistre64
  • 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 ...
amistre64
  • amistre64
but if you can figure out that formula, and it is one for your course ... go for it :)
amistre64
  • amistre64
|dw:1443919404316:dw| i think that formula of yours is from the distance between 2 planes tho
anonymous
  • anonymous
I don't know
anonymous
  • anonymous
we don't know whats m
amistre64
  • amistre64
m is just a general value ... the solution will be general as well
anonymous
  • anonymous
so y=(a/m + b -(m +6))/(m+1/m) ??
amistre64
  • 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
amistre64
  • amistre64
lets use the (1,4) for (a,b)
amistre64
  • amistre64
and for simplicity, assume a slope of 1 for brevity
anonymous
  • anonymous
ok so we can use any poits for (a,b)?
amistre64
  • 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
amistre64
  • 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
anonymous
  • anonymous
I think ill understand this ...if I wasn't this sleepy
amistre64
  • amistre64
what is the distance from (1,4) to (-1,6) ?
anonymous
  • anonymous
4+4=sq.rt 8
amistre64
  • amistre64
so 2 sqrt2
amistre64
  • amistre64
but this assumes a slope; m=1
amistre64
  • 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
anonymous
  • anonymous
ok if it work....but this is geometry though
amistre64
  • amistre64
we can use trig?
anonymous
  • anonymous
we can use anything ..I guess
amistre64
  • 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|
amistre64
  • amistre64
\[\vec n=(a,b-6)\] \[\vec \mu=(1,m )\] \[cos(\alpha)=\frac{a+m(b-6)}{\sqrt{a^2+(b-6)^2}~\sqrt{1+m^2} }\] \[\alpha=cos^{-1}\left(\frac{a+m(b-6)}{\sqrt{a^2+(b-6)^2}~\sqrt{1+m^2} }\right) \] \[d=|n|sin(\alpha)\] \[d=\sqrt{a^2+(b-6)^2}~sin(\alpha)\]
amistre64
  • 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
anonymous
  • anonymous
ok I'm lost... I'm sleepy ...but I think m shd be -1
amistre64
  • 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
amistre64
  • amistre64
http://www.mathwords.com/d/distance_point_to_line.htm or this defines the parts of your formula for us
amistre64
  • amistre64
and m = -a/b
anonymous
  • anonymous
yea that same formula I have here ...but idk
amistre64
  • amistre64
y - 6 = mx +m (m)x + (-1)y = (m+6) A = m, B=-1, C=m+6
amistre64
  • amistre64
so we know ABandC, and stated point values ...
anonymous
  • anonymous
wait 1 sec ..let me comprehend this
amistre64
  • amistre64
my C is off, foiled by a negative
anonymous
  • anonymous
y-6=mx+m -6-m=mx-y -(6+m)=mx-y
amistre64
  • amistre64
yeah
anonymous
  • anonymous
ok so mx-y+(6+m)=0 a=m b=-1 c=(6+m) ok I see
amistre64
  • amistre64
ugh, im tired too, im used to C being on the other side of =
anonymous
  • anonymous
no c=6+m remember we bring it back over the equal sign so it become positive
amistre64
  • 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)
amistre64
  • amistre64
and that is what we worked so hard to determine the first go around lol
anonymous
  • anonymous
where did u get the sq.rt of 2 frm
amistre64
  • 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)
anonymous
  • anonymous
but b=-1
amistre64
  • amistre64
(-1)^2 = 1
anonymous
  • anonymous
o yea I see.. forget about the sq.
anonymous
  • anonymous
so were finish with this question?
amistre64
  • amistre64
all there is to it is plugging in our known values and leaving the rest to the imagination
anonymous
  • anonymous
ok but what about the other point (-2,1)
amistre64
  • amistre64
\[\frac{|mx_0-y_0+m+6|}{\sqrt{m^2+1}}\] \[\frac{|-2m-1+m+6|}{\sqrt{m^2+1}}\] \[\frac{|-m+5|}{\sqrt{m^2+1}}\]
anonymous
  • anonymous
is that another equation that we can use to find distance
amistre64
  • amistre64
that is the formula you presented to start with
amistre64
  • amistre64
all i did was determine what A B and C were, and then its just a matter of plugging in the point values
anonymous
  • anonymous
ok I will have to look this through back when I'm fully awake to understand everything ..thanks very much
amistre64
  • amistre64
good luck

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