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,-\frac{7x+11}9\right)` on the line to the origin is given by


`d(x)=\sqrt{x^2+\left(-\frac{7x+11}9\right)^2}=\frac{√(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=(130x^2+154x+121)/(81)\implies(\mathrm dd(x)^2)/(\mathrm dx)=(260)/(81)x+(154)/(81)`

Solving for
`(d(x)^2)'=0`, you find a critical point of
`x=-(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


`(\mathrm d^2d(x)^2)/(\mathrm dx^2)=(260)/(81)>0`

so indeed, a minimum occurs at
`x=-(77)/(130)`.

The minimum distance is then


`d\left(-(77)/(130)\right)=(11)/(√(130))`

Answer 2

Answer: A, roughly 0.96

Explanation:

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