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through: \( (-1,0) \), perp. to \( y=\frac{1}{3} x+2 \)

Ask by Hampton Johnston. in the United States
Jan 24,2025

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Answer

The equation of the line that passes through \((-1, 0)\) and is perpendicular to \( y = \frac{1}{3}x + 2 \) is: \[ y = -3x - 3 \]

Solution

To find the equation of a line that passes through the point \((-1, 0)\) and is perpendicular to the line \( y = \frac{1}{3}x + 2 \), follow these steps: ### 1. Determine the Slope of the Given Line The given line is in the slope-intercept form \( y = mx + b \), where: - \( m \) is the slope - \( b \) is the y-intercept For \( y = \frac{1}{3}x + 2 \): - Slope (\( m_1 \)) = \( \frac{1}{3} \) ### 2. Find the Slope of the Perpendicular Line Perpendicular lines have slopes that are **negative reciprocals** of each other. Therefore, if the slope of one line is \( m \), the slope of the perpendicular line (\( m_2 \)) is: \[ m_2 = -\frac{1}{m_1} \] Calculating: \[ m_2 = -\frac{1}{\frac{1}{3}} = -3 \] ### 3. Use the Point-Slope Form to Find the Equation The point-slope form of a line is: \[ y - y_1 = m(x - x_1) \] where \( (x_1, y_1) \) is a point on the line and \( m \) is the slope. Given point \( (-1, 0) \) and slope \( m_2 = -3 \): \[ y - 0 = -3(x - (-1)) \] \[ y = -3(x + 1) \] \[ y = -3x - 3 \] ### 4. Final Equation The equation of the line that passes through \((-1, 0)\) and is perpendicular to \( y = \frac{1}{3}x + 2 \) is: \[ \boxed{y = -3x - 3} \] ### Graphical Representation To visualize, here's a simple sketch of both lines: 1. **Given Line:** \( y = \frac{1}{3}x + 2 \) (Light Blue) 2. **Perpendicular Line:** \( y = -3x - 3 \) (Orange) ![Graph of the two lines](https://i.imgur.com/yourimagelink.png) *Note: Replace the image link with an actual graph if needed.* ### Verification To ensure correctness, let's verify that the two lines are perpendicular: - Slope of the given line (\( m_1 \)): \( \frac{1}{3} \) - Slope of the perpendicular line (\( m_2 \)): \( -3 \) Check the product: \[ m_1 \times m_2 = \frac{1}{3} \times (-3) = -1 \] Since the product is \(-1\), the lines are indeed perpendicular.

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Beyond the Answer

To find the equation of a line that passes through the point \((-1, 0)\) and is perpendicular to the line given by \(y = \frac{1}{3}x + 2\), first, we need to determine the slope of the original line. The slope \(m\) of \(y = \frac{1}{3}x + 2\) is \(\frac{1}{3}\). Since perpendicular lines have slopes that are negative reciprocals of each other, the slope of the line we’re looking for will be \(-3\) (the negative reciprocal of \(\frac{1}{3}\)). Using the point-slope form of the line equation, \(y - y_1 = m(x - x_1)\), with the point \((-1, 0)\) and our perpendicular slope \(m = -3\): \[ y - 0 = -3(x + 1) \] Simplifying, we get: \[ y = -3x - 3 \] So, the equation of the line through the point \((-1, 0)\) and perpendicular to \(y = \frac{1}{3}x + 2\) is: \[ y = -3x - 3 \]

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