Question

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

Answer 1

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

Answer 2

what methods do you have available at your disposal?

Answer 3

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

Answer 4

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.

Answer 5

ok how do we do that

Answer 6

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

Answer 7

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 ...

Answer 8

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

Answer 9

i think that formula of yours is from the distance between 2 planes tho

Answer 10

I don't know

Answer 11

we don't know whats m

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

and for simplicity, assume a slope of 1 for brevity

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

4+4=sq.rt 8

Answer 23

so 2 sqrt2

Answer 24

but this assumes a slope; m=1

Answer 25

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

Answer 26

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

Answer 27

we can use trig?

Answer 28

we can use anything ..I guess

Answer 29

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)