Answer:
[tex]\large\boxed{y-3=\dfrac{2}{3}(x-2)}[/tex]
Step-by-step explanation:
[tex]\text{Let}\\\\k:y=m_1x+b_1\\\\l:y=m_2x+b_2\\\\l\ \perp\ k\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}\\\\l\ \parallel\ k\iff m_1=m_2\\\\===============================[/tex]
[tex]\text{The point-slope form of an equation of a line:}\\\\y-y_1=m(x-x_1)\\\\m-slope\\\\(x_1,\ y_1)-point\ on\ a\ line\\\\===============================[/tex]
[tex]\text{We have the equation of a line:}\ y-9=\dfrac{2}{3}(x+7)\to m_1=\dfrac{2}{3}.\\\\\text{A slope of parallel line:}\ m_2=m_1=\dfrac{2}{3}.\\\\\text{Put the value of the slope and the coordinates of the point (2, 3)}\\\text{to the equation of a line in point-slope form:}\\\\y-3=\dfrac{2}{3}(x-2)[/tex]
