Question

On a coordinate plane, a line goes through (0, negative 1) and (3, 1). A point is at (negative 3, 0).

What is the equation of the line that is parallel to the given line and has an x-intercept of –3?

Answer 1

keeping in mind that parallel lines have exactly the same slope, let's check for the slope of the line above. To get the slope of any straight line, we simply need two points off of it, and we have three, let's use two of those

`(\stackrel{x_1}{0}~,~\stackrel{y_1}{-1})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{1}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{1}-\stackrel{y1}{(-1)}}}{\underset{\textit{\large run}} {\underset{x_2}{3}-\underset{x_1}{0}}} \implies \cfrac{ 1 +1 }{ 3 } \implies \cfrac{ 2 }{ 3 } \implies \cfrac{2}{3}`

any line parallel to that will then have the exact same slope, so we're really looking for the equation of line whose slope is 2/3 and it passes through the x-intercept at -3.

`\stackrel{ x-intercept }{(\stackrel{x_1}{-3}~,~\stackrel{y_1}{0})}\hspace{10em} \stackrel{slope}{m} ~=~ \cfrac{2}{3} \\\\\\ \begin{array}c \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}{0}=\stackrel{m}{\cfrac{2}{3}}(x-\stackrel{x_1}{(-3)}) \\\\\\ y -0 = \cfrac{2}{3} ( x +3) \implies {\Large \begin{array}{llll} y=\cfrac{2}{3}x+2 \end{array}}`