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December 22, 2014

December 22, 2014

Posted by **Allison** on Wednesday, October 22, 2008 at 9:08pm.

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**Damon**, Wednesday, October 22, 2008 at 9:14pmy = x-4 does not go through the origin

to find the shortest distance from a point to a line find the line perpendicular to the given line and through the point first.

slope of perpendicular line = -1/original slope = -1/1 = -1

so

y = -x is our line (b = 0 because it goes through origin)

Now where does that line hit the original one?

y = x - 4

y = -x

so

-x = x - 4

2 x = 4

x = 2

then y = -2

Now we have two points

(0,0) and (2,-2)

what is the distance between them?

d = sqrt (2^2+(-2)^2) = sqrt 8 = 2 sqrt 2

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**Allison**, Wednesday, October 22, 2008 at 9:21pmok, but if i were dealing with the formula ... *absolute value*Ax+By+C/over -> *square root* A^2 +B^2

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- math -
**Damon**, Wednesday, October 22, 2008 at 9:36pmWell, interesting and limited to finding distance only from origin but if you insist:

y = x - 4

write in form

1 x - 1 y - 4 =0

A = 1

B = -1

C = -4

|A+B+C|=4

sqrt(A^2+B^2) = sqrt 2

4/sqrt2 = (4 sqrt 2)/ 2 = 2 sqrt 2

- math -
**Allison**, Wednesday, October 22, 2008 at 9:44pmits the |A+B+C|=4.. because our P.O.I was (2,-2), therefore A=1(2) + B=-1(-2) and C=-4....that makes 2+2-4.. that's 0

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- math -
**Damon**, Wednesday, October 22, 2008 at 9:59pmNow wait a minute.

If you are trying to find the distance ,d, between points P (Xp, Yp) and Q (Xq ,Yq)

distance in x = (Xq-Xp)

distance in y = (Yq-Yp)

d = hypotenuse = sqrt [(Xq-Xp)^2 +(Yq-Yp)^2 ]

here our two points are

(0,0) and (2,-2)

so

d = sqrt [(2-0)^2 + (-2-0)^2 ]

= sqrt (4+4)

=sqrt(8)

= 2 sqrt 2

- math -
**Damon**, Wednesday, October 22, 2008 at 10:08pmAre you sure your formula is not just |A+B+C|/sqrt(A^2+B^2) ?

It does not make sense to find the (2,-2) point and then go to the trouble of using your formula. I have never seen (or needed) this formula.

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**cassidy**, Thursday, October 23, 2008 at 9:50amhel p

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**Reiny**, Wednesday, October 22, 2008 at 10:25pmGiven the line Ax + By + C = 0 and a point (a,b) not on that line, then the shortest distance from the point to the line is

|Aa + Bb + C|/√(A^2 + B^2)

the point is (0,0) and x-y-4=0

so here distance =|0 + 0 - 4|/√(1+1) = 4/√2 = 2√2 as Damon found for you

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