Math Problem Statement

(a)    −3x + 4y + 4 = 0

(b)    −4x − 3y = 0

(c)    −3x − 4y = 3

(d)    4x + 3y + 5 = 0

(e)    3x − 4y + 9 = 0

(f )    −4x + 3y − 4 = 0

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

Solution

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 1-1.

First, we rearrange each equation into the slope-intercept form y=mx+by = mx + b, where mm is the slope.

Line (a):

3x+4y+4=0-3x + 4y + 4 = 0 Rearrange to solve for yy: 4y=3x44y = 3x - 4 y=34x1y = \frac{3}{4}x - 1 Slope ma=34m_a = \frac{3}{4}

Line (b):

4x3y=0-4x - 3y = 0 Rearrange to solve for yy: 3y=4x-3y = 4x y=43xy = -\frac{4}{3}x Slope mb=43m_b = -\frac{4}{3}

Line (c):

3x4y=3-3x - 4y = 3 Rearrange to solve for yy: 4y=3x3-4y = 3x - 3 y=34x+34y = -\frac{3}{4}x + \frac{3}{4} Slope mc=34m_c = -\frac{3}{4}

Line (d):

4x+3y+5=04x + 3y + 5 = 0 Rearrange to solve for yy: 3y=4x53y = -4x - 5 y=43x53y = -\frac{4}{3}x - \frac{5}{3} Slope md=43m_d = -\frac{4}{3}

Line (e):

3x4y+9=03x - 4y + 9 = 0 Rearrange to solve for yy: 4y=3x9-4y = -3x - 9 y=34x+94y = \frac{3}{4}x + \frac{9}{4} Slope me=34m_e = \frac{3}{4}

Line (f):

4x+3y4=0-4x + 3y - 4 = 0 Rearrange to solve for yy: 3y=4x+43y = 4x + 4 y=43x+43y = \frac{4}{3}x + \frac{4}{3} Slope mf=43m_f = \frac{4}{3}

Checking for Perpendicularity

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

  • ma×mb=(34)×(43)=1m_a \times m_b = \left(\frac{3}{4}\right) \times \left(-\frac{4}{3}\right) = -1
  • ma×md=(34)×(43)=1m_a \times m_d = \left(\frac{3}{4}\right) \times \left(-\frac{4}{3}\right) = -1
  • ma×mf=(34)×(43)=1m_a \times m_f = \left(\frac{3}{4}\right) \times \left(\frac{4}{3}\right) = 1 (not perpendicular)
  • mc×mf=(34)×(43)=1m_c \times m_f = \left(-\frac{3}{4}\right) \times \left(\frac{4}{3}\right) = -1
  • me×mb=(34)×(43)=1m_e \times m_b = \left(\frac{3}{4}\right) \times \left(-\frac{4}{3}\right) = -1
  • me×md=(34)×(43)=1m_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 23\frac{2}{3} and 23-\frac{2}{3} be perpendicular?
  3. What is the slope of a line perpendicular to y=5x7y = 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+by = mx + b to easily identify the slope.

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Math Problem Analysis

Mathematical Concepts

Algebra
Slope of a Line
Perpendicular Lines

Formulas

Slope-intercept form: y = mx + b
Perpendicularity condition: m1 * m2 = -1

Theorems

Slope-Intercept Theorem
Condition for Perpendicular Lines

Suitable Grade Level

Grades 9-12