Question

Line B is a straight line crosses at (0,0)
Line C is a straight line but below line B crosses at (0,-5)

Line moved down 5units
Decreased by factor of 5

Answer 1

No which option
I think moved down 5 units

Answer 2

To analyze the situation with lines B and C, let's start by defining what we have:

1. **Line B**: This line crosses at the origin (0,0). Since it's a straight line with a y-intercept of 0, we can express Line B's equation as:
\[
y = mx
\]
where \( m \) is the slope of the line.

2. **Line C**: This line is parallel to Line B but shifted downwards. Since it crosses the y-axis at (0, -5), we can express its equation as:
\[
y = mx - 5
\]

Now let's discuss the operations you've mentioned:

- **Moved down 5 units**: This operation suggests that we are further translating Line C downwards. The new y-intercept of Line C after translating down 5 units would be:
\[
-5 - 5 = -10
\]
Therefore, the equation for the new Line C (let's call it Line C') would be:
\[
y = mx - 10
\]

- **Decreased by factor of 5**: This operation implies that we are scaling the line vertically by a factor of \( \frac{1}{5} \). To scale Line C' by a factor of \( \frac{1}{5} \), we can multiply the entire equation of Line C' by \( \frac{1}{5} \):
\[
y = \frac{1}{5}(mx - 10)
\]
This simplifies to:
\[
y = \frac{m}{5}x - 2
\]

In summary, after moving Line C down by 5 units and then scaling it by a factor of 5, the new equation of Line C becomes:
\[
y = \frac{m}{5}x - 2
\]

If you need anything else, such as specific values for \( m \) or any additional properties of these lines, please let me know!

Answer 3

I apologize for any confusion in my previous message. Let’s clarify what happens when we move a line down by 5 units and the operations that follow.

1. **Line B** remains unchanged as:
\[
y = mx \quad \text{(crosses at (0,0))}
\]

2. **Line C** initially described as crossing at (0, -5):
\[
y = mx - 5
\]

### Moved Down 5 Units
If we move **Line C** down by 5 units, the new equation of Line C becomes:
\[
y = mx - 10 \quad \text{(crosses at (0, -10))}
\]

### Decreased by Factor of 5
If we then decrease this new Line C by a factor of 5, we scale the entire equation:
\[
y = \frac{1}{5}(mx - 10)
\]
Which simplifies to:
\[
y = \frac{m}{5}x - 2
\]

### Summary
Thus, after moving Line C down by 5 units and then decreasing it by a factor of 5, the final equation of Line C would be:
\[
y = \frac{m}{5}x - 2
\]

If the option you're considering is only moving the line down by 5 units, then the new equation of Line C would simply be:
\[
y = mx - 10
\]

I hope this effectively addresses your question! If you have specific options you’d like me to evaluate, please provide them.