How to find the point on the line (3x+5y-3) that is closest to point (0,0).

Answer 1

Hi Julie,

This is an example of a minimization problem

First. Let us come up with an equation to describe the distance between the origin (0,0) and any given point on the line 3x+5y=3. Our goal is going to be to eventually minimize this equation

by the Pythagorean theorem, this distance, D, is:

D = sqrt (x²+y²) where x and y are coordinates that pass through this line

notice that the line equation can be rearranged to give:

y = (3/5) - (3x/5)

so plugging this y into our distance equation, we get:

D = sqrt (x²+((3/5)-(3x/5))²)

Now, we minimize this equation. To do this, we can start by taking the derivative with respect to x (you can work this part on your own following the rules of differentiation, let me know if you have difficulty)

d(D)/dx = (34x-9) / 25 sqrt(x²+((3/5)-(3x/25))²)

Next, we set this derivative equal to zero. This is because we are looking for a minimum point of the distance formula, where the derivative of the distance formula (d(D)/dx) will be zero due to the distance reaching its smallest value.

d(D)/dx = (34x-9) / 25 sqrt(x²+((3/5)-(3x/25))²) = 0

Then, you can solve this equation for x. This value of x will be the x coordinate that minimizes the distance formula. All that is left to do from here is to solve for what y coordinate corresponds to this x coordinate along the line, which can be solved by plugging in this x into the line equation 3x + 5y = 3 and solving for y. These x and y values will represent the point on the line that is closest to the origin point (0,0).

Answer 2

You actually do not need calculus to solve this problem. Find the equation of the line that is perpendicular to the given line, passing through (0,0) and find the point of intersection of the two lines.

Since the given line has a slope of -3/5, the perpendicular will have a slope of 5/3.

So the two equations are y = (-3/5)x + 3/5 and y = (5/3)x.

You can solve this system with:

(5/3)x = (-3/5)x + 3/5.

Since this question was listed under "Calculus" the intent likely was to solve as a minimization problem, but you can use the above to check your solution.

Answer 3

The distance to (0,0) is √(x² + y²) in general but r = √(x² + ((3 – 3x)/5)²) = √(36x²/25 – 18x/25 + 36/25) along the line. Clearly there is no maximum r so the minimum r must be at the root of 0 = dr/dx = (72x – 18)/50r that is, x = 1/4.