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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..\[d=\frac{ Ax+By+C }{ \pm \sqrt{A^2+B^2} }\]

ok how do we do that

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) ??

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)?

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

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

we can use trig?

we can use anything ..I guess

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

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

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

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)

is that another equation that we can use to find distance

that is the formula you presented to start with

good luck