What is the equation of the line that is perpendicular to the given line and passes through point (2,5)?

Answer:[tex]\large\boxed{y=-\dfrac{1}{3}x+1}[/tex]Step-by-step explanation:[tex]\text{The slope-intercept form of an equation of a line:}\\\\y=mx+b\\\\m-slope\\b-y-intercept\to(0,\ b)[/tex][tex]\text{Let}\ k:y=m_1x+b_1\ \text{and}\ l:y=m_2x+b_2,\ \text{then}\\\\l\ \parallel\ k\iff m_1=m_2\\\\l\ \perp\ k\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}\\=======================\\\\\text{We must find the slope of the given line.}\\\\\text{The formula of a slope:}\\\\m=\dfrac{y_2-y_1}{x_2-x_1}\\\\\text{Substitute the coordinates of the given points from the graph (0, 1) and (-3, 2)}\\\\m_1=\dfrac{2-1}{-3-0}=\dfrac{1}{-3}=-\dfrac{1}{3}[/tex][tex]\text{We have an equation:}\\\\y=-\dfrac{1}{3}x+b\\\\\text{From the coordinates of the point (0, 1) we have the y-intercept:}\ b=1.\\\\\text{Therefore we have the equation:}\\\\y=-\dfrac{1}{3}x+1[/tex]