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To find the equation of a line passing through the point \(\left(\frac{1}{8}, -1\right)\) and parallel to the line \(8x - 9y = 8\), follow these steps:

### Step 1: Determine the Slope of the Given Line
Since the line we want to find is parallel to \(8x - 9y = 8\), it will have the same slope as this line.

1. Rewrite \(8x - 9y = 8\) in slope-intercept form \(y = mx + b\).
   \[
   -9y = -8x + 8
   \]
   \[
   y = \frac{8}{9}x - \frac{8}{9}
   \]
   Here, the slope \(m\) of the given line is \(\frac{8}{9}\).

### Step 2: Use the Point-Slope Form
Since the desired line is parallel, it also has a slope of \(\frac{8}{9}\). We use the point-slope form of a line equation:
\[
y - y_1 = m(x - x_1)
\]
Substitute \(m = \frac{8}{9}\) and the point \(\left(\frac{1}{8}, -1\right)\):
\[
y - (-1) = \frac{8}{9}\left(x - \frac{1}{8}\right)
\]
\[
y + 1 = \frac{8}{9}x - \frac{8}{72}
\]
\[
y = \frac{8}{9}x - \frac{1}{9} - 1
\]


Find a linear equation whose graph is the straight line through (1/8, -1) and parallel to the line 8x - 9y = 8.

Learn how to find the equation of a line passing through the point (1/8, -1) and parallel to the line 8x - 9y = 8. This problem involves algebraic concepts like linear equations, slopes, and the point-slope form. Suitable for Grade 10 students.

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