One way to do it is with calculus. The distance between any point [tex](x,y)=\left(x,-\dfrac{7x+11}9\right)[/tex] on the line to the origin is given by
[tex]d(x)=\sqrt{x^2+\left(-\dfrac{7x+11}9\right)^2}=\dfrac{\sqrt{130x^2+154x+121}}9[/tex]
Now, both [tex]d(x)[/tex] and [tex]d(x)^2[/tex] attain their respective extrema at the same critical points, so we can work with the latter and apply the derivative test to that.
[tex]d(x)^2=\dfrac{130x^2+154x+121}{81}\implies\dfrac{\mathrm dd(x)^2}{\mathrm dx}=\dfrac{260}{81}x+\dfrac{154}{81}[/tex]
Solving for [tex](d(x)^2)'=0[/tex], you find a critical point of [tex]x=-\dfrac{77}{130}[/tex].
Next, check the concavity of the squared distance to verify that a minimum occurs at this value. If the second derivative is positive, then the critical point is the site of a minimum.
You have
[tex]\dfrac{\mathrm d^2d(x)^2}{\mathrm dx^2}=\dfrac{260}{81}>0[/tex]
so indeed, a minimum occurs at [tex]x=-\dfrac{77}{130}[/tex].
The minimum distance is then
[tex]d\left(-\dfrac{77}{130}\right)=\dfrac{11}{\sqrt{130}}[/tex]
