Question

5) Line I passes through the points (12,7) and (2,-3). Find the equation of line k that passes throu

point (-3,11) and is parallel to line 1.

Answer 1

keeping in mind that parallel lines have exactly the same slope, let's check for the slope of the Line I

`(\stackrel{x_1}{12}~,~\stackrel{y_1}{7})\qquad (\stackrel{x_2}{2}~,~\stackrel{y_2}{-3}) ~\hfill~ \stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{-3}-\stackrel{y1}{7}}}{\underset{\textit{\large run}} {\underset{x_2}{2}-\underset{x_1}{12}}} \implies \cfrac{ -10 }{ -10 } \implies 1`

so we are really looking at the equation of a line whose slope is 1 and it passes through (-3 , 11) for Line K

`(\stackrel{x_1}{-3}~,~\stackrel{y_1}{11})\hspace{10em} \stackrel{slope}{m} ~=~ 1 \\\\\\ \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}{11}=\stackrel{m}{1}(x-\stackrel{x_1}{(-3)}) \implies y -11 = 1 ( x +3) \\\\\\ y-11=x+3\implies {\Large \begin{array}{llll} y=x+14 \end{array}}`