We are given the point $(8, 5)$ and a condition that the line must be parallel to the line described by the equation $9x - y = 8$.

Step 1: Identify the slope of the given line

We need to find the slope of the line $9x - y = 8$. First, rewrite the equation in slope-intercept form ($y = mx + b$), where $m$ is the slope:

$9x - y = 8 \quad \text{(Original equation)}$

Solve for $y$:

$y = 9x - 8$

From this, we can see that the slope of the given line is $m = 9$.

Step 2: Use the slope-point form to find the equation of the new line

Since the new line is parallel to the given line, it will have the same slope, $m = 9$. The line also passes through the point $(8, 5)$.

We can use the point-slope form of a line equation:

$y - y_1 = m(x - x_1)$

Substitute $m = 9$, $x_1 = 8$, and $y_1 = 5$:

$y - 5 = 9(x - 8)$

Step 3: Simplify to slope-intercept form

Expand and simplify the equation:

$y - 5 = 9x - 72$ $y = 9x - 72 + 5$ $y = 9x - 67$

Thus, the equation of the line in slope-intercept form is:

$\boxed{y = 9x - 67}$

Step 4: Convert to standard form

The standard form of a line is $Ax + By = C$, where $A$, $B$, and $C$ are integers, and $A$ is positive.

Starting with $y = 9x - 67$, rearrange the terms to get:

$9x - y = 67$

Thus, the equation of the line in standard form is:

$\boxed{9x - y = 67}$

Final Answers:

• (a) Slope-intercept form: $y = 9x - 67$
• (b) Standard form: $9x - y = 67$

Would you like further details or have any questions?

Here are 5 related questions for further practice:

1. Write the equation of a line passing through the point $(2, -3)$ and parallel to the line $4x - y = 7$.
2. Find the equation of a line passing through $(1, 2)$ and perpendicular to the line $y = -3x + 4$.
3. Determine the slope of the line parallel to $5x + 2y = 10$.
4. Find the equation of a line passing through $(4, 1)$ and with a slope of $-2$.
5. Convert the equation $7x + 3y = 9$ into slope-intercept form.

Tip: A line parallel to another has the same slope, while a line perpendicular to it has the negative reciprocal of its slope.