Anonymous
Anonymous asked in Science & MathematicsMathematics · 7 years ago

# Find the point on the line 2x+3y-3=0 closest to the point (3,5)?

Thank you! I just can't get the correct numbers

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Non-calculus solution using perpendicular lines:

3y = 3 - 2x

y = 1 - 2/3 x

Slope = -2/3

Slope of perpendicular = 3/2

Line with slope 3/2 through 3,5:

y = 3/2 x + b

5 = 3/2 * 3 + b

b = 1/2

Line is y = 3/2 x + 1/2

or

2y = 3x + 1

or 3x - 2y + 1 = 0

Original is

2x + 3y - 3 = 0

Original times 2: 4x + 6y - 6 = 0

New times 3: .... 9x - 6y + 3 = 0

Add: 13x - 3 = 0

13x = 3

x = 3/13

Using the new equation

y = 3/2 * 3/13 + 1/2 = (9+13)/26 = 22/26 = 11/13

(x,y) = 3/13, 11/13

Calculus answer using derivatives to minimize the distance

2x + 3y = 3

3y = 3 - 2x

y = (3 - 2x)/3

General point on line is (x, (3-2x)/3)

Distance to (3,5) squared (no need to bother with square roots:

minimizing the square minimizeds the distance, too)

=

(3-x)^2 + (5 - (3-2x)/3)^2 =

9 - 6x + x^2 + 25 - 10/3 * (3 - 2x) + (9 - 12x + 4x^2) / 9 =

9 - 6x + x^2 + 25 - 10 + 20x/3 + 1 - 12x/9 + 4 x^2 / 9 =

81 - 54x + 9x^2 + 225 - 90 + 60x + 9 - 12x + 4x^2 all over 9

we can also ignore the over 9 part and the constants

leaving

-6x + 13x^2

Derivative = 26x - 6

set to 0

26x = 6

x = 3/13

y = (3 - 2x)/3 = (3 - 6/13) / 3 = 1 - 6/39 = 33/39 = 11/13

(x,y) = 3/13, 11/13