Question

Find an equation of the line that passes through (2, -2) and parallel to the line passing through (4, 5) and (6, 4).

Answer 1

Line passing through (4, 5) and (6, 4):

`m = (4 - 5)/(6 - 4) = - (1)/(2)`

Line passing through (2, -2) and with slope -1/2:

-2 = (-1/2)(2) + b

-2 = -1 + b, so b = -1

y = (-1/2)x - 1

-2y = x + 2

-x - 2y = 2

x + 2y = -2

Answer 2

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

`(\stackrel{x_1}{4}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{6}~,~\stackrel{y_2}{4}) ~\hfill~ \stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{4}-\stackrel{y1}{5}}}{\underset{\textit{\large run}} {\underset{x_2}{6}-\underset{x_1}{4}}} \implies \cfrac{ -1 }{ 2 } \implies - \cfrac{1}{2}`

so we're really looking for the equation of a line whose slope is -1/2 and it passes through (2 , -2)

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