calculus

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

• calculus -

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
. . .

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Ola, I'm not understanding this question, any assistance?

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