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