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Find the exact value of the shortest distance from the origin to each line. y=x-4 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.

  • math -

    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

  • math -

    ok, but if i were dealing with the formula ... *absolute value*Ax+By+C/over -> *square root* A^2 +B^2

  • math -

    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

  • math -

    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

  • math -

    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

  • math -

    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.

  • math -

    hel p

  • math -

    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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