Question

which of the following gives an equation of a line that passes through the point (6 over 5, -19 over 5) and is parallel to the line that passes through th organ point (-2,-12)

Answer 1

One equation for this would be

$y = \frac{41}{16} x-\frac{55}{8}$

We start by finding the slope between the two points:

$m=\frac{y_2-y_1}{x_2-x_1}=\frac{-12-\frac{-19}{5}}{-2-\frac{6}{5}} \\ \\=(-12+\frac{19}{5}) \div (-2-\frac{6}{5}) \\ \\=(\frac{-60}{5}+\frac{19}{5}) \div (\frac{-10}{5}-\frac{6}{5}) \\ \\=\frac{-41}{5} \div \frac{-16}{5}=\frac{-41}{5} \times \frac{-5}{16}=\frac{41}{16}$

A line parallel to this one will have the same slope.  We will use point-slope form to write our equation:

$y-y_1=m(x-x_1) \\ \\y-\frac{-19}{5}=\frac{41}{16}(x-\frac{6}{5}) \\ \\y+\frac{19}{5}=\frac{41}{16}x- \frac{41}{16} \times \frac{6}{5} \\ \\y+\frac{19}{5}=\frac{41}{16}x-\frac{246}{80} \\ \\y+\frac{304}{80}=\frac{41}{16}x-\frac{246}{80} \\ \\y=\frac{41}{16}x-\frac{246}{80}-\frac{304}{80} \\ \\y=\frac{41}{16}x-\frac{550}{80} \\ \\y=\frac{41}{16}x-\frac{55}{8}$