To derive the equation of a line in the form \( y = mx \) that goes through the origin and is shown to have points (5, y) and (10, y), we first need to determine the slope \( m \) using the two points on the line.
Given the points \( (5, y_1) \) and \( (10, y_2) \), we can infer that because the relationship is proportional, the y-values also maintain a constant ratio.
Assuming that the values of \( y_1 \) and \( y_2 \) correspond proportionally to \( 5 \) and \( 10 \), we can represent them as:
- For \( x = 5 \): \( y_1 = k \cdot 5 \)
- For \( x = 10 \): \( y_2 = k \cdot 10 \)
Now we need to find the slope \( m \):
The slope \( m \) can be calculated using the two points:
\[ m = \frac{y_2 - y_1}{10 - 5} \]
Substituting for \( y_1 \) and \( y_2 \):
\[ m = \frac{k \cdot 10 - k \cdot 5}{10 - 5} = \frac{5k}{5} = k \]
Thus, the slope is directly equivalent to the constant \( k \).
Since the equation is in the form \( y = mx \), and we know that the slope \( m \) equals \( k \):
\[ y = kx \]
To express the relationship clearly, if we choose specific values for \( k \) based on the relationship you want to depict, the final equation remains in the form \( y = kx \), showing that \( y \) is proportional to \( x \).
If a specific value of \( k \) is not provided, we can represent the equation generally as:
\[ y = \frac{y_1}{5} x \]
For example, if \( y_1 = 10 \) at \( x = 5 \), then \( k = \frac{10}{5} = 2 \), and the equation would be:
\[ y = 2x \]
This is one possible equation based on an example value of \( k \). Depending on the actual points on the graph or a specific value for \( y_1 \) or \( y_2 \), you could adjust \( k \) accordingly. If you state what \( y_1 \) and \( y_2 \) correspond to, you will finalize the equation.