# Caluclus

Find the point on the line 7x+1y−5=0 which is closest to the point (−2,−6).

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1. 7x+1y−5=0
y=-7x+5
so the distance minimum will be on a line thru the point (-2,6) and has a slope of 1/7

y=1/7 x+b
-2=1/7 (-2)+b
b=-2+2/7

now look at the lines to find the intersection.
y=-7x+5
y=x/7 -2+2/7
set them equal
-7x+5=x/7-2+2/7
solve for x, then go back and solve for y. That (x,y) is the point on the original line which is closest to (-2,-6)

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2. The distance from (-2,-6) to the line 7x+y-5=0 is

|7(-2)+1(-6)-5|/√(7^2+1^2) = 25/√50 = 5/√2

so, we need to find (x,y) on the line such that

(x+6)^2 + (y+2)^2 = 25/2
(x+6)^2 + (11-7x)^2 = 25/2
x = 3/2
so, y=5-7x = -11/2
So, (3/2,-11/2) is the closest point to (-2,-6)

check: the slope of the line joining those two points is
(1/2)/(7/2) = 1/7
as shown above

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