calculus

Find the point on the line 2x+4y+7=0 which is closest to the point (4,−3).

  1. 2
asked by Emily
  1. the distance from (x,y) to (4,-3) is

    √((x-4)^2 + (y+3)^2)
    = √((x-4)^2 + ((2x+7)/4+3)^2)
    = √5/4 (4x^2-20x+61)
    This has a minimum at x = 37/10

    So, the point is (37/10,-18/5)

    Or, you can consider the perpendicular line through (4,-3). It is
    y+3 = 2 (x-4)
    This line intersects the other where

    2(x-4)-3 = -(2x+7)/4
    x = 37/10
    . . .

    posted by Steve

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