Answer
The point is $(4,3)$.
Work Step by Step
Let's first look at the line $6x+2y=1$ and rearrange it so that it's in the form $y=mx+b$.
$6x+2y=1$
$2y=1-6x$
$y=\frac{1}{2}-3x$
$y=-3x+\frac{1}{2}$
We see that the slope of this line is $-3$.
We now need to find the slope of a line that would be perpendicular to this line by finding the opposite reciprocal of $-3$, which is $\frac{1}{3}$. This means that we want to find the $x$ value that gives us $y'(x)=\frac{1}{3}$.
$y=\sqrt {1+2x}=(1+2x)^{\frac{1}{2}}$
We'll be using the power rule. To refresh your memory:
$y=u^n$
$y'=n'u^{n-1}$
So
$y'=\frac{1}{2}(1+2x)^{-\frac{1}{2}}*2$
$y'=\frac{1}{\sqrt{1+2x}}=\frac{1}{3}$
$\sqrt{1+2x}=3$
$1+2x=9$
$2x=8$
$x=4$
And the point is $(4,3)$.
