find the distance from the line y-6=m(x+1) to the point given below

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find the distance from the line y-6=m(x+1) to the point given below

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

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