Answer:
[tex]\large\boxed{y=\dfrac{1}{3}x+2\dfrac{1}{3}}[/tex]
Step-by-step explanation:
[tex]\text{Let}\ k:y=_1x+b_1\ \text{and}\ l:y=m_2x+b_2.\\\\l\ \perp\ k\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}\\============================\\\\\text{We have}\ y=-3x+1\to m_1=-3.\\\\\text{Therefore}\ m_2=-\dfrac{1}{-3}=\dfrac{1}{3}.\\\\\text{The equation of the searched line:}\ y=\dfrac{1}{3}x+b.\\\\\text{The line passes through }(2,\ 3).[/tex]
[tex]\text{Put the coordinates of the point to the equation.}\ x=2,\ y=3:\\\\3=\dfrac{1}{3}(2)+b\\\\3=\dfrac{2}{3}+b\qquad\text{subtract}\ \dfrac{2}{3}\ \text{from both sides}\\\\b=2\dfrac{1}{3}[/tex]
