Let (x,y) be the point on the line that is closet to the origine. Since, \(y=5x+3,\) so the point (x,y) on line becomes, \((x,5x+3)\) so now we use the distance dormula and minimize it.

Distance between (0,0) and \((x,5x+3)\) is given by

\(D=\sqrt{(x-0)^2+(5x+3-0)^2}\)

\(=\sqrt{x^2+(25x^2+30x+9)}\)

\(\Rightarrow D=\sqrt{26x^2+30x+9}\)

Now to minimize the distance, we find critical point by setting \(D'=0\). Here using chain rule, derivative is:

\(D'=\frac{1}{2}(26x^2+30x+9)^{-\frac{1}{2}}\cdot\frac{d}{dx}(26x^2+30x+9)\)

\(=\frac{1}{2\sqrt{26x^2+30x+9}}\cdot(26\cdot2x+30\cdot1+0)\)

\(\Rightarrow D'=\frac{52x+30}{2\sqrt{26x^2+30x+9}}\)

Now D'=0

\(\Rightarrow \frac{52x+30}{2\sqrt{26x^2+30x+9}}=0\)

\(\Rightarrow 52x+30=0\)

\(\Rightarrow x=-\frac{30}{52}=-\frac{15}{26}\)