Question

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

Answer 1

easy way:

d = |(-4)(-5)+(7)(-5)+3|/√(4^2+7^2) = 12/√65
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calculus way:

the distance d can be found using

d^2 = (-5-x)^2 + (-5-y)^2
= (x+5)^2 + (5+(3+4x)/7)^2

d = 1/7 √(65x^2+794x+2669)
dd/dx = (130x+794) / 14√(65x^2+794x+2669)

dd/dx=0 when x = -373/65
so, y = -241/65

The distance between (-5,-5) and (-373/65,-241/65) = 12/√65
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Perpendicular line way:
perp line is
y+5 = -7/4 (x+5)
They intersect at (-373/65,-241/65)
and the distance is 12/√65