Least Distance

If any point coincides, then it is the closest.

Checking if there is a real number solution for the two equations $x^2 - 2x +1 = 0$ $(x-1)^2=0$ When x is 1, y is 1. Point (1,1) is satisfying both equations.

Let the point on the parabola ${y}={x}^{2}$ be $({x},{x}^{2} )$ .

Let the point on the straight line $y=2x-1$ be $({x},2{x}-1)$ .

Using the distance formula, we get the distance between the two points as:

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

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

$=x^{2}-2x+1$

To get the minimum distance, we use the concept of minima-maxima.

that is, $\dfrac{\text{d}}{\text{d}x} (x^{2}-2x+1) = 0$

$\implies 2x-2=0$

$\implies x=1$

Thus the point on the parabola is $\boxed {(1,1) }$ .