32. Which of these lines is perpendicular to the line 5y = -6x + 2? А. бу — -5х +3 В. бу — 5х + 3 %3D С. Бу — -6х + 8 D. 5y = 6x + 8

Elementary Geometry For College Students, 7e
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ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
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Question 32

**Question 32: Which of these lines is perpendicular to the line \(5y = -6x + 2\)?**

*Options:*

- **A. \(6y = -5x + 3\)**

- **B. \(6y = 5x + 3\)**

- **C. \(5y = -6x + 8\)** (Selected)

- **D. \(5y = 6x + 8\)**

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**Explanation:**

To determine which line is perpendicular to the given line \(5y = -6x + 2\), we need to convert it to the slope-intercept form \(y = mx + b\), where \(m\) is the slope.

For the line \(5y = -6x + 2\):
\[ y = -\frac{6}{5}x + \frac{2}{5} \]
The slope (\(m\)) of this line is \(-\frac{6}{5}\).

Two lines are perpendicular if the product of their slopes (\(m_1 \cdot m_2\)) is \(-1\).

Let's analyze the options.

- **Option A**: \(6y = -5x + 3\)
\[ y = -\frac{5}{6}x + \frac{1}{2} \]
The slope is \(-\frac{5}{6}\). \(-\frac{6}{5} \cdot -\frac{5}{6} = 1 \neq -1\), so it is not perpendicular.

- **Option B**: \(6y = 5x + 3\)
\[ y = \frac{5}{6}x + \frac{1}{2} \]
The slope is \(\frac{5}{6}\). \(-\frac{6}{5} \cdot \frac{5}{6} = -1\), so this line is perpendicular.

- **Option C**: \(5y = -6x + 8\)
\[ y = -\frac{6}{5}x + \frac{8}{5} \]
The slope is \(-\frac{6}{5}\), which is the same slope as the given line, so it is not perpendicular. 

- **Option D**: \(5y = 6x
Transcribed Image Text:**Question 32: Which of these lines is perpendicular to the line \(5y = -6x + 2\)?** *Options:* - **A. \(6y = -5x + 3\)** - **B. \(6y = 5x + 3\)** - **C. \(5y = -6x + 8\)** (Selected) - **D. \(5y = 6x + 8\)** --- **Explanation:** To determine which line is perpendicular to the given line \(5y = -6x + 2\), we need to convert it to the slope-intercept form \(y = mx + b\), where \(m\) is the slope. For the line \(5y = -6x + 2\): \[ y = -\frac{6}{5}x + \frac{2}{5} \] The slope (\(m\)) of this line is \(-\frac{6}{5}\). Two lines are perpendicular if the product of their slopes (\(m_1 \cdot m_2\)) is \(-1\). Let's analyze the options. - **Option A**: \(6y = -5x + 3\) \[ y = -\frac{5}{6}x + \frac{1}{2} \] The slope is \(-\frac{5}{6}\). \(-\frac{6}{5} \cdot -\frac{5}{6} = 1 \neq -1\), so it is not perpendicular. - **Option B**: \(6y = 5x + 3\) \[ y = \frac{5}{6}x + \frac{1}{2} \] The slope is \(\frac{5}{6}\). \(-\frac{6}{5} \cdot \frac{5}{6} = -1\), so this line is perpendicular. - **Option C**: \(5y = -6x + 8\) \[ y = -\frac{6}{5}x + \frac{8}{5} \] The slope is \(-\frac{6}{5}\), which is the same slope as the given line, so it is not perpendicular. - **Option D**: \(5y = 6x
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