(a)    −3*x* + 4*y* + 4 = 0

(b)    −4*x* − 3*y* = 0

(c)    −3*x* − 4*y* = 3

(d)    4*x* + 3*y* + 5 = 0

(e)    3*x* − 4*y* + 9 = 0

(f )    −4*x* + 3*y* − 4 = 0

Which of the given lines are perpendicular to each other? (Select all that apply.)

To determine which of the given lines are perpendicular to each other, we need to examine their slopes. Lines are perpendicular if the product of their slopes is \(-1\).

First, we rearrange each equation into the slope-intercept form \(y = mx + b\), where \(m\) is the slope.

### Line (a):
\[
-3x + 4y + 4 = 0
\]
Rearrange to solve for \(y\):
\[
4y = 3x - 4
\]
\[
y = \frac{3}{4}x - 1
\]
Slope \(m_a = \frac{3}{4}\)

### Line (b):
\[
-4x - 3y = 0
\]
Rearrange to solve for \(y\):
\[
-3y = 4x
\]
\[
y = -\frac{4}{3}x
\]
Slope \(m_b = -\frac{4}{3}\)

### Line (c):
\[
-3x - 4y = 3
\]
Rearrange to solve for \(y\):
\[
-4y = 3x - 3
\]
\[
y = -\frac{3}{4}x + \frac{3}{4}
\]
Slope \(m_c = -\frac{3}{4}\)

### Line (d):
\[
4x + 3y + 5 = 0
\]
Rearrange to solve for \(y\):
\[
3y = -4x - 5
\]
\[
y = -\frac{4}{3}x - \frac{5}{3}
\]
Slope \(m_d = -\frac{4}{3}\)

### Line (e):
\[
3x - 4y + 9 = 0
\]
Rearrange to solve for \(y\):
\[
-4y = -3x - 9
\]
\[
y = \frac{3}{4}x + \frac{9}{4}
\]
Slope \(m_e = \frac{3}{4}\)

### Line (f):
\[
-4x + 3y - 4 = 0
\]
Rearrange to solve for \(y\):
\[
3y = 4x + 4
\]
\[
y = \frac{4}{3}x + \frac{4}{3}
\]
Slope \(m_f = \frac{4}{3}\)

### Checking for Perpendicularity

Lines are perpendicular if the product of their slopes is \(-1\). Let’s check the products of the slopes:

- \(m_a \times m_b = \left(\frac{3}{4}\right) \times \left(-\frac{4}{3}\right) = -1\)
- \(m_a \times m_d = \left(\frac{3}{4}\right) \times \left(-\frac{4}{3}\right) = -1\)
- \(m_a \times m_f = \left(\frac{3}{4}\right) \times \left(\frac{4}{3}\right) = 1\) (not perpendicular)
- \(m_c \times m_f = \left(-\frac{3}{4}\right) \times \left(\frac{4}{3}\right) = -1\)
- \(m_e \times m_b = \left(\frac{3}{4}\right) \times \left(-\frac{4}{3}\right) = -1\)
- \(m_e \times m_d = \left(\frac{3}{4}\right) \times \left(-\frac{4}{3}\right) = -1\)

Therefore, the lines that are perpendicular to each other are:

- (a) and (b)
- (a) and (d)
- (c) and (f)
- (e) and (b)
- (e) and (d)

Would you like any more details or have any questions? Here are some related questions:

1. How do you find the slope of a line given its equation in standard form?
2. Can two lines with slopes of \(\frac{2}{3}\) and \(-\frac{2}{3}\) be perpendicular?
3. What is the slope of a line perpendicular to \(y = 5x - 7\)?
4. If two lines are parallel, what can be said about their slopes?
5. How do you find the equation of a line perpendicular to a given line and passing through a specific point?

**Tip:** Always simplify the equation of a line to the slope-intercept form \(y = mx + b\) to easily identify the slope.

Which of the given lines are perpendicular to each other? (Select all that apply.)
(a) −3*x* + 4*y* + 4 = 0
(b) −4*x* − 3*y* = 0
(c) −3*x* − 4*y* = 3
(d) 4*x* + 3*y* + 5 = 0
(e) 3*x* − 4*y* + 9 = 0
(f) −4*x* + 3*y* − 4 = 0

Discover how to identify perpendicular lines using the slope-intercept form. This problem explores key algebraic concepts like determining slopes and the condition for perpendicularity. Suitable for students in Grades 9-12.

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