Find point on line 3x+y=5 that is closest to the point (-5,2)

Question

Find point on line 3x+y=5 that is closest to the point (-5,2)

(x,y) = (____)

Answer 1

This is a problem easily done by geometry. The shortest distance is the perpendicular line. The perpendicular line has a slope that is the negative reciprocal of the slope of 3x + y = 5 (which is -3) so it must be 1/3. It goes through the point (-5, 2) so the equation (using point-slope) is y - 2 = 1/3(x + 5) which converts to y = 1/3x + 11/3. Now, find the intersection point by substituting this into the original to get 3x + (1/3x + 11/3) = 5 gives x = 0.4. Plugging that in to either equation gives y = 3.8 so the point closest is (0.4, 3.8).

However, to make this hard, you could use Calculus. The distance from the line 3x + y = 5 to (-5, 2) can be put into an equation using the distance formula d = sqrt[(x₂ - x₁)² + (y₂ - y₁)²] where (x₁, y₁) is (-5, 2) and (x₂, y₂) is some point on 3x + y = 5. Since 3x + y = 5, we can say y = -3x + 5 so the point (x₂, y₂) can be written as (x, -3x + 5). Plugging the 2 points into the distance formula, we get:

d = sqrt[(x₂ - x₁)² + (y₂ - y₁)²]

d = sqrt[(x - -5)² + (-3x + 5 - 2)²]

d = sqrt[(x + 5)² + (-3x + 3)²]

d = sqrt(x² + 10x + 25 + 9x² -18x + 9)

d = sqrt(10x² - 8x + 34)

Now, minimize this by taking the derivative (requires the chain rule) and setting it equal to zero.

d' = (20x - 8)/[2sqrt(10x² - 8x + 34)]

(20x - 8)/[2sqrt(10x² - 8x + 34)] = 0 Note this only = 0 when the numerator = 0 so:

20x - 8 = 0

20x = 8

x = 8/20 = 0.4

Plugging that into the 3x + y = 5 gives y = 3.8

So the point on 3x + y = 5 closest to (-5, 2) is (0.4, 3.8) [But like I said, it's way easier to solve using geometry]