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Write an equation for a line perpendicular to \( y=3 x-5 \) and passing through the point \( (3,-2) \) \( y=\square \)

Ask by Burton Frank. in the United States
Jan 27,2025

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Answer

The equation of the line perpendicular to \( y = 3x - 5 \) and passing through the point \( (3, -2) \) is \( y = -\frac{1}{3}x - 1 \).

Solution

To find the equation of a line that is perpendicular to \( y = 3x - 5 \) and passes through the point \( (3, -2) \), follow these steps: 1. **Determine the slope of the given line:** The equation \( y = 3x - 5 \) is in the slope-intercept form \( y = mx + b \), where \( m \) is the slope. \[ \text{Slope of the given line, } m = 3 \] 2. **Find the slope of the perpendicular line:** Perpendicular lines have slopes that are negative reciprocals of each other. Therefore, the slope \( m_{\perp} \) of the perpendicular line is: \[ m_{\perp} = -\frac{1}{m} = -\frac{1}{3} \] 3. **Use the point-slope form to write the equation of the perpendicular line:** The point-slope form is: \[ y - y_1 = m_{\perp}(x - x_1) \] Plugging in the slope \( m_{\perp} = -\frac{1}{3} \) and the point \( (3, -2) \): \[ y - (-2) = -\frac{1}{3}(x - 3) \] Simplifying: \[ y + 2 = -\frac{1}{3}x + 1 \] \[ y = -\frac{1}{3}x + 1 - 2 \] \[ y = -\frac{1}{3}x - 1 \] **Final Equation:** \[ y = -\frac{1}{3}x - 1 \]

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Bonus Knowledge

To find a line perpendicular to \( y = 3x - 5 \), we first need to determine its slope. The slope of this line is 3. The slope of a line perpendicular to it will be the negative reciprocal, which is \( -\frac{1}{3} \). Now, we use the point-slope form of the equation of a line: \[ y - y_1 = m(x - x_1) \] Plugging in the slope \(-\frac{1}{3}\) and the point \( (3, -2) \): \[ y - (-2) = -\frac{1}{3}(x - 3) \] Simplifying: \[ y + 2 = -\frac{1}{3}x + 1 \] Therefore, the equation of the line can be rewritten as: \[ y = -\frac{1}{3}x - 1 \]

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