# Calculus

posted by Alessandra

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

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 !

2. Alessandra

the distance is sqrt(x2-x1)^2+(y2 -y1)^2)?

will that be a form of interval?

3. Alessandra

it sid the distance formula isn't what they're looking for, they're looking for interval

4. Damon

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

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