Let's re-evaluate the problem. Given the slope of the line passing through the points (7, 5) and (21, 15) is $\frac{5}{7}$, we can determine another point for a line with a slope that is one-third of $\frac{5}{7}$ passing through the origin.

The slope of the new line is: $\text{New slope} = \frac{1}{3} \times \frac{5}{7} = \frac{5}{21}$

This new line passes through the origin (0, 0). We can use this slope to find another point on the line. The point-slope form of the line equation is: $y = \frac{5}{21}x$

To find another point, let's choose $x = 21$: $y = \frac{5}{21} \times 21 = 5$

Therefore, the point is (21, 5).

If (21, 5) is not an option, let's choose another $x$ value to find another valid point.

Choosing $x = 21$: $y = \frac{5}{21} \times 21 = 5$

Choosing $x = 42$: $y = \frac{5}{21} \times 42 = 10$

So, the point is (42, 10).

Let's list these options in the dropdown menus again:

• The slope of the line passing through the points (7, 5) and (21, 15) is $\frac{5}{7}$.
• Another line with a slope that is one-third that of the slope you just calculated passes through the origin and the point (42, 10).

Would you like more details or have any questions? Here are 5 questions you might consider asking next:

1. How do you find the y-intercept of a line given its slope and a point on the line?
2. What is the difference between slope-intercept form and point-slope form?
3. How can you verify if a point lies on a given line?
4. How do you find the perpendicular bisector of a line segment?
5. What are some real-life applications of linear equations and slopes?

Tip: Always check multiple possible points when verifying answers for questions involving linear equations and slopes.