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

Question

How do I solve this? I got down to the distance formula and the seconded distance formula and the derivative of that, but I think I messed it up.

Answer 1

Did you use this as your Distance² ?

(x-1)² + ((2/3)x + (5/3))²

= (13/9)x² +(2/9) x + (34/9)

So derivative = 26/9 x + 2/9
Critical value: x = -1/13 and this is, in fact, a minimum.

So the point is (-1/13, 8/13).

Answer 2

If you drew a line between the point (1,-1) and the point on the line it's closest to, you'd get a line perpendicular to the original line. The original line is -4x + 6y - 4 = 0, which we can rewrite as y = (2/3)x + 2/3

So the perpendicular line has a slope of -3/2. The line with that slope that contains (1,-1) is:
y = mx + b
-1 = (-3/2)1 + b
-1 + 3/2 = b, so b = 1/2
y = (-3/2)x + 1/2

Find out where this line intersects the original line, and we've found the point closest to (1,-1):
y = (2/3)x + 2/3
y = (-3/2)x + 1/2
(2/3)x + 2/3 = (-3/2)x + 1/2
6(2/3)x + 6*2/3 = 6*(-3/2)x + 6*1/2
4x + 4 = -9x + 3
1 = -13x
x = -1/13, so y is (2/3)(-1/13) + 2/3 = (2/3)(-1/13 + 1) = (2/3)(12/13) = (2)(4/13) = 8/13

So the point is (-1/13, 8/13)

If you really wanted to use the distance formula, call the point on the line (x1, y1). Since y = (2/3)x + 2/3, this is (x1, (2/3)x1 + 2/3). So the distance formula gives
√[ (x1 - 1)² + ((2/3)x1 + 2/3 - (-1))² ] =
√[ (x1 - 1)² + ((2/3)x1 + 5/3)² ] =
√[ x1² - 2x1 + 1 + (4/9)x1² + (20/9)x1 + 25/9 ] =
√[ (13/9)x1² + (2/9)x1 + 34/9 ] =
√[ 13x1² + 2x1 + 34 ] / 3 =

Taking the derivative of this and setting it equal to zero gives:
(1/3)(1/2)[ 13x1² + 2x1 + 34 ]^(-1/2) * (26x1 + 2) = 0
(26x1 + 2) / √[ 13x1² + 2x1 + 34 ] = 0
(26x1 + 2) = 0
x1 = -1/13, so we get the same coodinates as before

Answer 3

Find the slope m of the line.

Drop a perpendicular line from (1,-1) to the original line. The perpendicular line will have a slope of -1/m.

With slope -1/m and point (1,-1), find the equation of the perpendicular line.

The intersection of the two lines is the point you're looking for. Since it lies on both lines, it should satisfy both equations so just solve for x and y (two unknowns two equations) to get the answer.

Answer 4

Your missing the forest because of the trees.
Rewrite your line as y = 2/3 x + 2/3.
The closest this line comes to (1,-1) is given by a line that passes through (1,-1) and is perpendicular to the given line. This line has slope of -3/2. You can find the equation of that line, and find where it intercepts the given line.

Answer 5

- 6x + 4y + a million = 0 4y = 6x - a million y = (3/2)x - a million/4 y = - (2/3)x - 4/3 (3/2)x + (2/3)x = a million/4 - 4/3 18x + 8x = 3 - 16 26x = - 13 x = - 13/26 = - a million/2 y = - 3/4 - a million/4 = - a million y = a million/3 - 4/3 = - a million (x,y) = ( - a million/2, - a million)

Answer 6

-4x + 6y - 4 =0
y = (2/3) + (2/3)x

D = sqrt( [y - (-1)]^2 + [x - (1)]^2 )
=sqrt( [(2/3) + (2/3)x - (-1)]^2 + [x - (1)]^2 )
=sqrt( [(5/3) + (2/3)x )]^2 + [x - 1]^2 )

=sqrt( [(25/9) + (20/9)x + (4/9)x^2 ] + [x^2 -2x + 1])
=sqrt( (34/9) + (2/9)x + (13/9)x^2 )

Minimize D

D' = [ (2/9) + (26/9)x ] / [2*sqrt( (34/9) + (2/9)x + (13/9)x^2 )]
D' = 0 ---> [ (2/9) + (26/9)x ] = 0
x = (-1/13)

y = y = (2/3) + (2/3)(-1/13) = (8/13)