Question
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Are the lines given by the equations \( -4 x+y=5 \) and \( -x+4 y=12 \) parallel, perpendicular, or neither? Why?

Ask by Schultz Li. in the United States
Jan 14,2025

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Tutor-Verified Answer

Answer

The lines are neither parallel nor perpendicular because their slopes are different and not negative reciprocals.

Solution

To determine whether the lines given by the equations \( -4x + y = 5 \) and \( -x + 4y = 12 \) are parallel, perpendicular, or neither, we can analyze their slopes. 1. **Find the slope of the first line (\( -4x + y = 5 \)):** \[ y = 4x + 5 \] - **Slope (\( m_1 \))**: 4 2. **Find the slope of the second line (\( -x + 4y = 12 \)):** \[ 4y = x + 12 \quad \Rightarrow \quad y = \frac{1}{4}x + 3 \] - **Slope (\( m_2 \))**: \( \frac{1}{4} \) 3. **Compare the slopes:** - **Parallel Lines:** Slopes are equal. - Here, \( m_1 = 4 \) and \( m_2 = \frac{1}{4} \). Since \( 4 \neq \frac{1}{4} \), the lines are **not parallel**. - **Perpendicular Lines:** Slopes are negative reciprocals (i.e., \( m_1 \times m_2 = -1 \)). - \( 4 \times \frac{1}{4} = 1 \), which is **not equal** to \(-1\). Therefore, the lines are **not perpendicular**. **Conclusion:** The two lines are **neither parallel nor perpendicular**. **Answer:** They are neither parallel nor perpendicular because their slopes are different and not negative reciprocals.

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The Deep Dive

To determine the relationship between the two lines given by the equations \( -4x + y = 5 \) and \( -x + 4y = 12 \), we’ll start by rewriting these equations in slope-intercept form (\(y = mx + b\)), where \(m\) is the slope. For the first equation \( -4x + y = 5 \): 1. Rearrange to solve for \(y\): \[ y = 4x + 5 \] The slope (\(m_1\)) is \(4\). For the second equation \( -x + 4y = 12 \): 1. Rearrange to solve for \(y\): \[ 4y = x + 12 \] \[ y = \frac{1}{4}x + 3 \] The slope (\(m_2\)) is \(\frac{1}{4}\). Now comparing the slopes: - The slopes \(m_1 = 4\) and \(m_2 = \frac{1}{4}\). - Two lines are perpendicular when the product of their slopes equals \(-1\): \[ m_1 \times m_2 = 4 \times \frac{1}{4} = 1 \] Since the product is \(1\) (not \(-1\)), the lines are neither parallel nor perpendicular.

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