Find the point (x,y) of the line 3x+5y-3 that is the closest to the point (0,0)

Question

The equation to be optimized is the distance between the point (0,0) and a point (x,y) on the line 3x+5y−3=0. This distance d is defined by the following equation: d²=(x−0)²+(y−0)².

Answer 1

There are 2 ways to approach this problem.

First, the perpendicular distance from a point to a line is given by the following formula:

d=i (ax₀ + by₀ +c)/sqrt(a² + b²) where (x₀,y₀) is the point and ax + by + c =0 is the equation of the line.

Derivation of this formula requires a fair bit of algebra,,,you should look it up if you need it.

Second, if you have calculus, you can use the formula given in the problem substituting

y= (3/5)(1-x) into that formula and then setting the derivative of d with respect to x equal to zero to find the minimum of d.

Answer 2

y = - 3/5x + 3/5;

f(x) = d² = (x - 0)² + (-3/5x + 3/5 - 0)² = x² + 9/25x² - 18/25x + 9/25 = 34/25x² - 18/25x + 9/25;

f'(x) = 68/25x - 18/25 = 0; x = 9/34; f''(x) = 68/25 > 0. So, at x = 9/34 f(x) has minimum.

y = -3/5 ·9/34 + 3/5 = 3/5·25/34 = 15/34

Point (9/34, 15/34) is closest to (0, 0)