math
posted by Allison .
Find the exact value of the shortest distance from the origin to each line. y=x4 is the equation of the line. When i do the problem I keep on getting 0 but that's not the answer!!! please help me.

y = x4 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+24.. 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 = (XqXp)
distance in y = (YqYp)
d = hypotenuse = sqrt [(XqXp)^2 +(YqYp)^2 ]
here our two points are
(0,0) and (2,2)
so
d = sqrt [(20)^2 + (20)^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 xy4=0
so here distance =0 + 0  4/√(1+1) = 4/√2 = 2√2 as Damon found for you