Calculus
posted by ryan .
Find the point on the line 6 x + 5 y  1 =0 which is closest to the point ( 4, 1 )

Let the point be P(x,y) , point (4,1) be A
Since the shortest distance must be where AP is perpendicular to the given line.
slope of given line = 6/5
so slope of AP is  5/6
equation of AP:
y1 = (5/6)(x+4)
times 6
6y  6 = 5x 20
5x + 6y = 14 , #2
6x  5y = 1 , #1, the given line
#1 times 6 >36x  30y = 6
#2 times 5 > 25x + 30y = 70
add them:
61x = 76
x = 76/61
back into #1
6(76/61)  5y = 1
5y = 1 + 456/61
5y = 395/61
y =  79/61
The closest point is (76/61 , 79/61)
check my arithmetic, expected "nicer " numbers. 
just as a check, the distance from point P(h,k) to the line ax+by+c=0 is
ah+bk+c/√(a^2+b^2)
In this case, that is
24+5+1/√(36+25) = 30/√61
Hmmm. What went wrong above?
Oops. Forgot to multiply 6 by 6