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

3 answers

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!

No which option
I think moved down 5 units

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.