To determine which of the statements is true, let's first rewrite each line in the slope-intercept form \( y = mx + b \), where \( m \) is the slope.
-
Line 1: \( 3y = 4x + 3 \)
Divide by 3:
\[ y = \frac{4}{3}x + 1 \]
(Slope \( m_1 = \frac{4}{3} \)) -
Line 2: \( y = 3x - 4 \)
(Slope \( m_2 = 3 \)) -
Line 3: \( 3x + 4y = 8 \)
Rearranging gives:
\[ 4y = -3x + 8 \]
Divide by 4:
\[ y = -\frac{3}{4}x + 2 \]
(Slope \( m_3 = -\frac{3}{4} \)) -
Line 4: \( 4x + 3y = -6 \)
Rearranging gives:
\[ 3y = -4x - 6 \]
Divide by 3:
\[ y = -\frac{4}{3}x - 2 \]
(Slope \( m_4 = -\frac{4}{3} \))
Now, let's analyze the slopes to check for parallelism and perpendicularity:
-
Lines are parallel if they have the same slope.
- Line 1 \( \left(\frac{4}{3}\right) \) and Line 4 \( \left(-\frac{4}{3}\right) \) have different slopes, so they are not parallel.
- Line 2 \( (3) \) and Line 3 \( \left(-\frac{3}{4}\right) \) have different slopes, so they are not parallel.
-
Lines are perpendicular if the product of their slopes is -1.
-
\( m_2 \left(3\right) \) and \( m_4 \left(-\frac{4}{3}\right) \):
\[ 3 \cdot -\frac{4}{3} = -4 \] (not -1, so not perpendicular) -
\( m_1 \left(\frac{4}{3}\right) \) and \( m_3 \left(-\frac{3}{4}\right) \):
\[ \frac{4}{3} \cdot -\frac{3}{4} = -1 \] (they are perpendicular)
-
So, the only correct statement is:
D. Lines 1 and 3 are perpendicular.