Write the equation of a line that is perpendicular to y=7/5x+6 and that passes through the point (2,-6)

Given that,The equation of line is y=7/5x+ 6 and that passes through the point (2,-6).To find,The equation of line that is perpendicular to the given line.Solution,The given line is :y=7/5x+ 6 The slope of this line = 7/5For two perpendicular lines, the product of slopes of two lines is :[tex]m_1m_2=-1\\\\m_2=\dfrac{-1}{7/5}\\\\=\dfrac{-5}{7}[/tex]Equation will be :y=-5x/7+ bNow finding the value of b. As it passes through (2,-6). The equation of line will be :[tex]-6=\dfrac{-5}{7}(2)+b\\\\ -6=\dfrac{-10}{7}+b\\\\b=-6+\dfrac{10}{7}\\\\b=\dfrac{-32}{7}[/tex]So, the required equation of line is :y=-5x/7+ (-32/7)[tex]y=\dfrac{-5x}{7}-\dfrac{-32}{7}[/tex]