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January 31, 2015

January 31, 2015

Posted by **ryan** on Saturday, November 23, 2013 at 3:22pm.

- Calculus -
**Reiny**, Saturday, November 23, 2013 at 3:53pmLet 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)

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.

- Calculus -
**Steve**, Saturday, November 23, 2013 at 7:17pmjust 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

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