Question

Find the point of the line y=x nearest to the point (4,1)

Answer 1

Any point in the line y=x is (x,x), where x is a real number.

Distance between (x,x) and (4,1) equals:

$d(x)=\sqrt{(x-4)^2+(x-1)^2}$

$d(x)=\sqrt{x^2-8x+16+x^2-2x+1}$

$d(x)=\sqrt{2x^2-10x+17}$

We need to find the minimum of d(x). Square root function is increasing, so finding the minimum of d(x) is equivalent to finding the minimum of square root's argument.

$f(x)=2x^2-10x+17$

Find the extremum point:

$f'(x)=0$

$4x-10=0$

$x=2.5$

To determine if this point is a minimum we calculate the second derivative at this point.

$f''(x)=4$

$f''(2.5)=4>0$

The second derivative is greater than zero, so x=2.5 is minimum point of f(x) and d(x).

So, point (2.5, 2.5) is the nearest point in the line y=x to the point (4, 1).