Question

Find the distance from the origin to the graph of 7x+9y+11=0

Answer 1

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

$d\left(-\dfrac{77}{130}\right)=\dfrac{11}{\sqrt{130}}$