To determine which lines are perpendicular, we first need to find the slope of each line. Two lines are perpendicular if the product of their slopes is -1, meaning that one slope is the negative reciprocal of the other.
-
From the equation \(4x - 3y = -6\):
- Rearranging this to slope-intercept form (y = mx + b):
\[
-3y = -4x - 6 \implies y = \frac{4}{3}x + 2
\]
- Slope \(m_1 = \frac{4}{3}\).
- Rearranging this to slope-intercept form (y = mx + b):
\[
-3y = -4x - 6 \implies y = \frac{4}{3}x + 2
\]
-
From the equation \(5x - 3y = -3\):
- Rearranging to slope-intercept form:
\[
-3y = -5x - 3 \implies y = \frac{5}{3}x + 1
\]
- Slope \(m_2 = \frac{5}{3}\).
- Rearranging to slope-intercept form:
\[
-3y = -5x - 3 \implies y = \frac{5}{3}x + 1
\]
-
From the equation \(y - 2 = \frac{3}{4}(x + 1)\):
- This is already in point-slope form. Rearranging it:
\[
y - 2 = \frac{3}{4}x + \frac{3}{4} \implies y = \frac{3}{4}x + \frac{11}{4}
\]
- Slope \(m_3 = \frac{3}{4}\).
- This is already in point-slope form. Rearranging it:
\[
y - 2 = \frac{3}{4}x + \frac{3}{4} \implies y = \frac{3}{4}x + \frac{11}{4}
\]
-
From the equation \(y = -\frac{3}{4}x + 12\):
- This is already in slope-intercept form.
- Slope \(m_4 = -\frac{3}{4}\).
- This is already in slope-intercept form.
Now, let's check for perpendicular pairs:
-
\(m_1 = \frac{4}{3}\) and \(m_4 = -\frac{3}{4}\):
- The product is \(\frac{4}{3} \times -\frac{3}{4} = -1\). These lines are perpendicular.
-
\(m_2 = \frac{5}{3}\) and \(m_4 = -\frac{3}{4}\):
- The product is \(\frac{5}{3} \times -\frac{3}{4} \neq -1\). These lines are not perpendicular.
-
\(m_3 = \frac{3}{4}\) and \(m_4 = -\frac{3}{4}\):
- The product is \(\frac{3}{4} \times -\frac{3}{4} \neq -1\). These lines are not perpendicular.
From this analysis, the TWO lines that are perpendicular are:
- \(4x - 3y = -6\)
- \(y = -\frac{3}{4}x + 12\)