As you say, we have 3 points, $A(10,10)$, $B(15,20)$, and $C(16,16)$.
The distance formula says that the distance between $A(x_1,y_1)$ and $B(x_2,y_2)$ is $$\sqrt {(x_2-x_1)^2+(y_2-y_1)^2}$$
Here, the distance $AB$ is $5\sqrt5$
Similarly, $AC$ is $6\sqrt2$
And $BC$ is $\sqrt{17}$
Since the perpendicular is the smallest distance, let a point $D(x,y)$ on $AB$ such that $AB\underline{|} CD$, we can form 2 right triangles $\delta ADC$ and $\delta BDC$
We already know that $AB$ is $5\sqrt5$
Let $CD=x$
Applying the Pythagorean theorem:-
$AD=\sqrt{72-x^2}$
And
$BD=\sqrt{17-x^2}$
Since $AB=AD+BD$
Therefore
$$5\sqrt5=\sqrt{72-x^2}+\sqrt{17-x^2}$$
Solving, $x={6\over\sqrt5}$
Therefore, here the smallest distance is $6\over\sqrt5$
The shortest distance between the blue point and the given line is measured along the perpendicular to that line through the blue point. Geometrically, this is a basic compass-and-straightedge construction.
Analytically, there are several ways to go about this. We’ll do it by translating the geometric construction above into analytic terms. Assume for the moment that the given line is neither vertical nor horizontal. Its equation is $y=-x\tan\theta$, where $\theta$ is the rotation angle. The negative sign appears because you’re taking positive angles as clockwise, which is the opposite of the usual sign convention for a right-handed coordinate system. Let the coordinates of the blue point $B$ be $(x_b,y_b)$. Two lines are perpendicular if the product of their slopes is $-1$, therefore the equation of the perpendicular line through $B$ is $y-y_b = (x-x_b) \cot\theta$. Now, solve for the intersection point, giving $$ x=(x_b\cos\theta-y_b\sin\theta) \cos\theta \\ y=(y_b\sin\theta-x_b\cos\theta) \sin\theta.$$ You can then use the formula for the distance between two points to get, after some simplification, $|x_b\sin\theta+y_b\cos\theta|$ for the distance from $B$ to the line.
We can’t use this derivation for a horizontal or vertical line because one of the slopes is infinite, but a slight modification to the equation of the line handles those cases, too. Expand $\tan\theta$ as ${\sin\theta\over\cos\theta}$ in the equation of the line, multiply through by $\cos\theta$ and rearrange to get $x\sin\theta + y\cos\theta = 0$. A similar transformation produces $(x-x_b)\cos\theta-(y-y_b)\sin\theta=0$ as the equation of the perpendicular. The rest of the derivation proceeds the same way as above.
Notice that the distance of $B$ from the line is just the absolute value of what you get from plugging its coordinates into the more general equation of the line. This can be generalized: Given the equation of any line $ax+by+c=0$, not necessarily passing through the origin, the distance of a point $(x_0,y_0)$ from it is given by $${|ax_0+by_0+c| \over \sqrt{a^2+b^2}}.$$ I’ll leave proving this more general formula to you. One way to prove it follows the same steps as the above construction for a line through the origin.
Best Answer
Using the perpendicular distance formula: $d = \frac{ax_1 + by_1 + c}{\sqrt{a^2+b^2}}$ with $x_1 = -1, y_1 = 0$ (the coordinates of the given point) and $a = 1, b = k, c = 0$ (by writing the equation of the line in general form). This gives $d = \frac{1}{\sqrt{1+k^2}}$.
Note that this even works for $k < 0$ (but obviously not for $k = 0$).