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

- anonymous

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

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- anonymous

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

- amistre64

what methods do you have available at your disposal?

- 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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## More answers

- 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

ok how do we do that

- 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

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

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

- amistre64

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

- anonymous

I don't know

- anonymous

we don't know whats m

- amistre64

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

- anonymous

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

- 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

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

- amistre64

and for simplicity, assume a slope of 1 for brevity

- anonymous

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

- 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

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

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

- amistre64

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

- anonymous

4+4=sq.rt 8

- amistre64

so 2 sqrt2

- amistre64

but this assumes a slope; m=1

- 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

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

- amistre64

we can use trig?

- anonymous

we can use anything ..I guess

- 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

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

- amistre64

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

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

- 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

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

- amistre64

and m = -a/b

- anonymous

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

- amistre64

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

- amistre64

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

- anonymous

wait 1 sec ..let me comprehend this

- amistre64

my C is off, foiled by a negative

- anonymous

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

- amistre64

yeah

- anonymous

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

- amistre64

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

- anonymous

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

- 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

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

- anonymous

where did u get the sq.rt of 2 frm

- 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

but b=-1

- amistre64

(-1)^2 = 1

- anonymous

o yea I see.. forget about the sq.

- anonymous

so were finish with this question?

- amistre64

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

- anonymous

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

- 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

is that another equation that we can use to find distance

- amistre64

that is the formula you presented to start with

- 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

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

- amistre64

good luck

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