Find the slopes of the lines to determine which lines are perpendicular. (1 point) Responses Line C is perpendicular to line D . Line upper C is perpendicular to line upper D . Line A is perpendicular to line C . Line upper A is perpendicular to line upper C . Line A is perpendicular to line D . Line upper A is perpendicular to line upper D . Line B is perpendicular to line C .

Answer 1

To determine if two lines are perpendicular, we need their slopes. Two lines are perpendicular if the product of their slopes is -1 (i.e., one slope is the negative reciprocal of the other).

Here's a breakdown of how to identify perpendicular lines based on their slopes:

1. Calculate the slopes: If you have the equations of the lines, convert them to slope-intercept form (y = mx + b) to find the slopes (m) of each line.

2. Check for perpendicularity: Compare the slopes using the criteria mentioned above. For example, if the slope of line A is \( m_a \) and line C's slope is \( m_c \), check if \( m_a \cdot m_c = -1 \).

If you provide the slopes or equations of lines A, B, C, and D, I can identify which lines are perpendicular based on that information.

Answer 2

A is, y=-5x+3. B is, y=-1/6x-1. C is, y=5x+2. D is, y=1/5-2

Answer 3

To determine which lines are perpendicular, we first need to find the slopes of each line from their equations.

1. Line A: \( y = -5x + 3 \)

   • Slope (m) of Line A: \( m_A = -5 \)

2. Line B: \( y = -\frac{1}{6}x - 1 \)

   • Slope (m) of Line B: \( m_B = -\frac{1}{6} \)

3. Line C: \( y = 5x + 2 \)

   • Slope (m) of Line C: \( m_C = 5 \)

4. Line D: \( y = \frac{1}{5}x - 2 \)

   • Slope (m) of Line D: \( m_D = \frac{1}{5} \)

Next, we check which pairs of lines have slopes that are negative reciprocals of each other:

• Line A and Line C: \[ m_A \cdot m_C = (-5) \cdot (5) = -25 \quad (\text{not perpendicular}) \]

• Line A and Line D: \[ m_A \cdot m_D = (-5) \cdot \left(\frac{1}{5}\right) = -1 \quad (\text{perpendicular}) \]

• Line A and Line B: \[ m_A \cdot m_B = (-5) \cdot \left(-\frac{1}{6}\right) = \frac{5}{6} \quad (\text{not perpendicular}) \]

• Line B and Line C: \[ m_B \cdot m_C = \left(-\frac{1}{6}\right) \cdot (5) = -\frac{5}{6} \quad (\text{not perpendicular}) \]

• Line B and Line D: \[ m_B \cdot m_D = \left(-\frac{1}{6}\right) \cdot \left(\frac{1}{5}\right) = -\frac{1}{30} \quad (\text{not perpendicular}) \]

• Line C and Line D: \[ m_C \cdot m_D = (5) \cdot \left(\frac{1}{5}\right) = 1 \quad (\text{not perpendicular}) \]

From the analysis above, the only pair of lines that are perpendicular is Line A and Line D.