calculus help amigos

Find the point on the line 5x+5y+7=0
which is closest to the point (3,−4)

asked by AMELIA
  1. Just use your "closest distance from a given point to a line" formula

    D = | 5(3) + 5(-4) + 7 |/√(5^2 + 5^2))
    = 2/√50
    = 2/5√2
    = 2/5√2 units or √2/5 after rationalizing.

    or , (the long and traditional way):

    you know the slope of the given line is -1
    so the slope of a perpendicular from (3,-4) is +1

    equation of that perpendicular is:
    y+4 = 1(x-3)
    y = x - 7
    sub into the given line to find the intersection:
    5x + 5(x-7) + 7 0
    5x + 5x - 35 + 7 = 0
    10x = 28
    x = 2.8
    y = 2.8-7 = -4.2


    distance between (3,-4) and (2.8, -4.2)
    = √( (3-2.8)^2 + (-4+4.2)^2)
    = √(.04+.04)
    = √.08 = √(8/100)
    = √8/10
    = 2√2/10 = √2/5, same as above

    posted by Reiny

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