Posted by Allison on Wednesday, October 22, 2008 at 9:08pm.
y = 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
ok, but if i were dealing with the formula ... *absolute value*Ax+By+C/over -> *square root* A^2 +B^2
Well, 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
its 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
Now 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
Are 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.
hel p
Given 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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