How Far Away is Your Point

The shortest distance between a point and a line is between the point and the intersection of the original line and a perpendicular line, which is going through the point.

By rearranging the equation of the original line, we get:

$y = 2x + 1$

The equation of the perpendicular line, going through the point (2 , 1):

$y = -0.5x + 2$

By solving the simultaneous equations, we will have the coordinates of the point of intersection:

(0.4 , 1.8)

Finally, the distance between these two points:

$\sqrt {(2 - 0.4)^2 + (1 - 1.8)^2} = \sqrt {1.6^2 + (- 0.8)^2 } = \boxed { \frac {4}{5} \sqrt {5} }$

The equation of the line is

$y-1=2x$

The equation of a circle of radius $r$ centered at $\left(1,2\right)$ is

${ \left( y-1 \right) }^{ 2 }={ r }^{ 2 }-{ \left( x-2 \right) }^{ 2 }$

We solve for $x$ to find the intersection points in terms of $r$ , getting

$x=\dfrac { 1 }{ 5 } (2\pm \sqrt { 5{ r }^{ 2 }-16 } )$

For this circle to be tangent to the line, the radical needs to vanish, which is true if

$r=\dfrac{4\sqrt{5}}{5}$

Edit: Another approach:

At $x=2$ , the point on the line is $(2,5)$ .
At $y=1$ , the point on the line is $(0,1)$ .
The distance between the two points is $2\sqrt{5}$ .
The area of the right triangle formed by the $3$ points is $4$ .
The altitude of the triangle from the hypotenuse is $\dfrac{4\sqrt{5}}{5}$ , which, of course, is the minimum distance we seek.

$\sin{\alpha}=\dfrac{2}{\sqrt{5}}=\dfrac{h}{2}$

$h=\dfrac{4}{\sqrt{5}}=\dfrac{4\sqrt{5}}{5}$

Solution 2 : the shortest distance between $f(x)=2x+1$ and $(2,1)$ is equal to the shortest distance between $f(x)=2x+1$ and $g(x)=2x-3$

the vertical lenght between $f(x)$ and $g(x)$ is $|f^{-1}(x)-g^{-1}(x)|=2$ and the horizontal lenght between $f(x)$ and $g(x)$ is $|f(x)-g(x)|=4$ now make a right triangle with those lenghts and its hypotenuse will be $2\sqrt{5}$ . let $h$ be hypotenuse's altitude(it's also the shortest distance between graphs) then:
$2\sqrt{5}h=4\cdot 2\Rightarrow h=\dfrac{4\sqrt{5}}{5}$

It can be solved using Lagrange multipliers .

$\begin{cases} r^2 = (x-2)^2 + (y-1)^2 \\ 2x - y +1 = 0 \end{cases}$

Solving for the smallest $r$ with $2x-y+1=0$ as constraint.

$\begin{aligned} F(x,y,\lambda) & = (x-2)^2 + (y-1)^2- \lambda(2x - y +1) \end{aligned}$

\(\begin{array}{} \dfrac {\partial}{\partial x} F(x,y,\lambda) = 2(x-2) - 2 \lambda = 0 & \implies \lambda = x-2 & ...(1) \\ \dfrac {\partial}{\partial y} F(x,y,\lambda) = 2(y-1) + \lambda = 0 & \implies \lambda = 2-2y & ...(2) \\ \dfrac {\partial}{\partial \lambda} F(x,y,\lambda) = 2x - y + 1 = 0 & \implies 2x - y + 1 = 0 & ...(3) \end{array} \)

$\begin{aligned} (1), \ (2): \quad \lambda & = x-2 = 2-2y \\ \implies x = 4-2y \end{aligned}$

$\begin{aligned} (3): \quad 2x - y + 1 & = 0 \\ 8 - 4y - y + 1 & = 0 \\ \implies y & = \frac 95 \quad \implies x & = \frac 25 \end{aligned}$

$\begin{aligned} \implies r^2 & = (x-2)^2 + (y-1)^2 \\ & = \left(\frac 25 - 2 \right)^2 + \left(\frac 95 - 1 \right)^2 \\ & = \frac {80}{25} \\ \implies r & = \boxed{\dfrac {4\sqrt 5}5} \end{aligned}$

If equation is ax+by+c=0 and given point is(m,n). Then shortest distance is Mod[(am+bn+c)]/root of a^2+b^2 On solving...... 2(2)+1-1/root of 2^2+1................ 4/root of 5..........on rationalisation we get 4×root 5/5

In the spirit of diversity, here's a calculus solution.