Question

A straight line passes through the point T (4,1) and has a gradient of 3/5. Determine the equation of this line. A straight line is drawn through the points A (1,1) and B (5,-2). Determine the equation of the line which passes through D (3,2) and is perpendicular to AB? Write answer in (y=mx+c) form

Answer 1

`T(\stackrel{x_1}{4}~,~\stackrel{y_1}{1})\hspace{10em} \stackrel{slope}{m} ~=~ \cfrac{3}{5} \\\\\\ \begin{array} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{1}=\stackrel{m}{ \cfrac{3}{5}}(x-\stackrel{x_1}{4}) \\\\\\ y-1=\cfrac{3}{5}x-\cfrac{12}{5}\implies y=\cfrac{3}{5}x-\cfrac{12}{5}+1\implies {\Large \begin{array}{llll} y=\cfrac{3}{5}x-\cfrac{7}{5} \end{array}} \\\\[-0.35em] ~\dotfill`

keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of the line AB

`(\stackrel{x_1}{1}~,~\stackrel{y_1}{1})\qquad (\stackrel{x_2}{5}~,~\stackrel{y_2}{-2}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{-2}-\stackrel{y1}{1}}}{\underset{run} {\underset{x_2}{5}-\underset{x_1}{1}}} \implies \cfrac{ -3 }{ 4 } \implies - \cfrac{3 }{ 4 } \\\\[-0.35em] ~\dotfill`

`\stackrel{~\hspace{5em}\textit{perpendicular lines have \underline{negative reciprocal} slopes}~\hspace{5em}} {\stackrel{slope}{\cfrac{-3}{4}} ~\hfill \stackrel{reciprocal}{\cfrac{4}{-3}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{4}{-3}\implies \cfrac{4}{3}}}`

so we're really looking for the equation of a line that has a slope of 4/3 and that it passes through (3 , 2)

`(\stackrel{x_1}{3}~,~\stackrel{y_1}{2})\hspace{10em} \stackrel{slope}{m} ~=~ \cfrac{4}{3} \\\\\\ \begin{array} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{2}=\stackrel{m}{ \cfrac{4}{3}}(x-\stackrel{x_1}{3}) \\\\\\ y-2=\cfrac{4}{3}x-4\implies {\Large \begin{array}{llll} y=\cfrac{4}{3}x-2 \end{array}}`