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

Answer 1

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 = (2-7x)/5:

d^2 = (-6-x)^2 + (-2-((2-7x)/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 = (2-7x)/5

intersect in a single point.

25(x+6)^2 + (7x-2)^2 = 25r^2
74x^2 + 272x + 904-24r^2 = 0

If there is a single solution, the discriminant is zero, so

262^2 - 4(74)(904-24r^2) = 0
r^2 = 49375/1776
So, we have
25(x+6)^2 + (7x-2)^2 = 25*49375/1776
7x+5y−2=0
I trust we will come up with the same point. You can verify if you wish.