How do you write an equation of a line going through #(1/8, −1)# and parallel to the line #8x − 9y = 6#?

#y= 8/9x -1 1/9#

Explanation:

Parallel lines have the same slope. So we can find the required slope from the equation of the given line.
Change it into the form #y = color(blue)(m)x+c#

#9y =8x-6#

#y = color(blue)(8/9)x-2/3" "rarr# this gives us # color(blue)(m = 8/9)#

You can use the formula: #y-y_1 = m(x-x_1)# because we have a point #(1/8, -1)# and the slope #color(blue)(m = 8/9)#

#y-y_1 = color(blue)(m)(x-x_1)#

#y -(-1) = color(blue)(8/9)(x-1/8)#

#y+1 = 8/9x -1/9" "larr [cancel8/9 xx1/cancel8 =1/9]#

#y= 8/9x -1 1/9" "larr [-1 -1/9 = -1 1/9]#

As we expected, the slope of the new line is #8/9#