Answer
$\left(-\frac{6}{5},\frac{3}{5}\right)$
Work Step by Step
The distance $d$ from the origin $(0,0)$ to a point $(x,2x+3)$ on the line is given by
$d$ = $\sqrt {(x-0)^{2}+(2x-3-0)^{2}}$
and the square of the is distance is
$S$ = $d^{2}$ = ${(x)^{2}+(2x-3)^{2}}$
$S'(x)$ = $2x+2(2x-3)(2)$ = $10x+12$
$S'(x)$ = $0$ => $x$ = $-\frac{6}{5}$
$S''(x)$ = $10$
$S''(x)$ $\gt$ $0$
so we know that $S$ has a minimum at $x$ = $-\frac{6}{5}$
Thus the y-value is $2\left(-\frac{6}{5}\right)+3$ = $\frac{3}{5}$ and the point is $\left(-\frac{6}{5},\frac{3}{5}\right)$.
