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

the shortest distance will be along the line perpendicular, which passes through (6,2)
The given line has slope 7/5, so the perpendicular has slope 5/7
y+2 = 5/7 (x+6)
The two lines intersect at (33/37,61/37)

Or, you can figure the distance to (6,2) knowing that y = (27x)/5:
d^2 = (6x)^2 + (2((27x)/5))^2)
= 1/25 (74x^2 + 132x + 1044)
dd/dx = (37x+33)/√(nonzero junk)
(x,y) = (33/37,61/37)

Or, you can find where the circle at (6,2) is tangent to the line. That will be where
(x+6)^2 + (y+2)^2 = r^2 and
y = (27x)/5
intersect in a single point.
25(x+6)^2 + (7x2)^2 = 25r^2
74x^2 + 272x + 90424r^2 = 0
If there is a single solution, the discriminant is zero, so
262^2  4(74)(90424r^2) = 0
r^2 = 49375/1776
So, we have
25(x+6)^2 + (7x2)^2 = 25*49375/1776
7x+5y−2=0
I trust we will come up with the same point. You can verify if you wish.
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