To solve this problem, let's break it down into two parts.

Part (a): Finding the slope of the sales line and the equation of the line

1. Identify the points:
   We have two points on the line based on the information given:

   • In the fourth year: $(x_1, y_1) = (4, 45000)$
   • In the sixth year: $(x_2, y_2) = (6, 65000)$

2. Calculate the slope $m$:
   The slope of a line is calculated using the formula: $m = \frac{y_2 - y_1}{x_2 - x_1}$ Substituting the values: $m = \frac{65000 - 45000}{6 - 4} = \frac{20000}{2} = 10000$ So, the slope $m$ is $10000$.

3. Find the equation of the line:
   The general form of the equation of a line is: $y = mx + b$ We can use one of the points to solve for $b$ (the y-intercept). Using the point $(4, 45000)$ and substituting $m = 10000$: $45000 = 10000(4) + b$ $45000 = 40000 + b$ $b = 45000 - 40000 = 5000$ Therefore, the equation of the sales line is: $y = 10000x + 5000$

Part (b): Finding the number of years until sales surpass $120,000

We need to find $x$ when $y = 120000$. Using the equation we derived:

$120000 = 10000x + 5000$

Subtract 5000 from both sides:

$115000 = 10000x$

Divide both sides by 10000:

$x = \frac{115000}{10000} = 11.5$

So, the sales will surpass $120,000 after approximately 11.5 years. Since we're dealing with years, we round up to the next whole number, so it will take 12 years for the sales to surpass $120,000.

Summary:

• The equation of the sales line is $y = 10000x + 5000$.
• The sales will surpass $120,000 in the 12th year of operation.

Do you want more details or have any questions? Here are some related questions:

1. How would the equation change if the sales in the sixth year were $75,000?
2. What would be the sales in the 10th year according to this model?
3. How would you interpret the y-intercept in this context?
4. What does the slope of the line represent in terms of the company's sales?
5. How would you graph this line on a coordinate plane?

Tip: Always check if the data can reasonably be modeled by a linear equation before applying linear models.