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

Damon
rewrite as
y = (6/4)x + 1/4
line perpendicular has slope +4/6 = +2/3
y = (2/3) x + b
we want it through 0, 1
1 = (2/3) 0 + b
b = 1
so
y = (2/3) x  1
where does that hit original line?
(2/3) x  1 = (3/2) x + 1/4
4 x  6 = 9 x + 3/2
13 x = 15/2
x = 15/26
then y = 5/13
now find the distance from
(0,1) to (15/26 , 5/13) yuuk
check my arithmetic ! 
Alessandra
the distance is sqrt(x2x1)^2+(y2 y1)^2)?
will that be a form of interval? 
Alessandra
it sid the distance formula isn't what they're looking for, they're looking for interval

Damon
Oh, well I guess I found your point :)
(15/26,5/13)
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