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

the distance from (x,y) to (4,3) is
√((x4)^2 + (y+3)^2)
= √((x4)^2 + ((2x+7)/4+3)^2)
= √5/4 (4x^220x+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 (x4)
This line intersects the other where
2(x4)3 = (2x+7)/4
x = 37/10
. . .
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Ola, I'm not understanding this question, any assistance?