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:
y-1 = (-5/6)(x+4)
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
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
In this case, that is
|24+5+1|/√(36+25) = 30/√61
Hmmm. What went wrong above?
Oops. Forgot to multiply -6 by 6
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