Math Problem Statement

State if the lines are parallel, perpendicular, or neither for the given equations, and calculate distances between specific points or lines as described in the image.

Solution

Let’s work through the problems on the page step-by-step. Here's how we address each section:

28) Determine if the lines are parallel, perpendicular, or neither:

  1. Line 1: y=23x4y = \frac{2}{3}x - 4
    • Slope: m=23m = \frac{2}{3}
  2. Line 2: 3y=2x363y = 2x - 36
    • Rewrite as y=23x12y = \frac{2}{3}x - 12
    • Slope: m=23m = \frac{2}{3}

Conclusion: The slopes are equal, so the lines are parallel.


29) Determine if the lines are parallel, perpendicular, or neither:

  1. Line 1: 5x4y=25x - 4y = -2
    • Rewrite as y=54x+12y = \frac{5}{4}x + \frac{1}{2}
    • Slope: m=54m = \frac{5}{4}
  2. Line 2: 5y=4x+355y = -4x + 35
    • Rewrite as y=45x+7y = -\frac{4}{5}x + 7
    • Slope: m=45m = -\frac{4}{5}

Conclusion: The product of the slopes is 5445=1\frac{5}{4} \cdot -\frac{4}{5} = -1, so the lines are perpendicular.


30) Determine if the lines are parallel, perpendicular, or neither:

  1. Line 1: y=27x+4y = \frac{2}{7}x + 4
    • Slope: m=27m = \frac{2}{7}
  2. Line 2: 7y=2x77y = -2x - 7
    • Rewrite as y=27x1y = -\frac{2}{7}x - 1
    • Slope: m=27m = -\frac{2}{7}

Conclusion: The slopes are neither equal nor negative reciprocals, so the lines are neither.


31–33) Find the distance between points or lines:

  • 31) Points x=4x = -4 and x=7x = 7

    • Distance: 7(4)=11|7 - (-4)| = 11.
  • 32) Points y=3y = 3 and y=5y = -5

    • Distance: 3(5)=8|3 - (-5)| = 8.
  • 33) Points (4,1)(-4, 1) and (7,2)(7, -2)

    • Use the distance formula: d=(x2x1)2+(y2y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} d=(7(4))2+(21)2=112+(3)2=121+9=130d = \sqrt{(7 - (-4))^2 + (-2 - 1)^2} = \sqrt{11^2 + (-3)^2} = \sqrt{121 + 9} = \sqrt{130}

Distance: 130\sqrt{130} or approximately 11.411.4.


Bonus: Distance between two lines

A) y=2x4y = 2x - 4 and y=2x+9y = 2x + 9

  • Distance between parallel lines: d=c2c11+m2d = \frac{|c_2 - c_1|}{\sqrt{1 + m^2}} Here, m=2m = 2, c1=4c_1 = -4, and c2=9c_2 = 9: d=9(4)1+22=1355.81d = \frac{|9 - (-4)|}{\sqrt{1 + 2^2}} = \frac{13}{\sqrt{5}} \approx 5.81

B) y=34x5y = \frac{3}{4}x - 5 and 3x4y=483x - 4y = -48

  • Rewrite 3x4y=483x - 4y = -48 as y=34x+12y = \frac{3}{4}x + 12.
  • Distance between parallel lines: d=12(5)1+(34)2=172516=1745=13.6d = \frac{|12 - (-5)|}{\sqrt{1 + (\frac{3}{4})^2}} = \frac{17}{\sqrt{\frac{25}{16}}} = \frac{17 \cdot 4}{5} = 13.6

Would you like any additional details or further breakdowns for any part of this problem?


Questions to Explore:

  1. How do you derive the formula for the distance between two parallel lines?
  2. What happens when the slopes are the same for both lines?
  3. How does the perpendicular condition m1m2=1m_1 \cdot m_2 = -1 arise geometrically?
  4. What is the significance of the distance formula in coordinate geometry?
  5. Can the distance between two skew lines in 3D space be calculated similarly?

Tip: Always verify line equations are in slope-intercept form before comparing their slopes!

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Math Problem Analysis

Mathematical Concepts

Coordinate Geometry
Slopes of Lines
Parallel Lines
Perpendicular Lines
Distance Formula

Formulas

Slope-intercept form of a line: y = mx + b
Condition for parallel lines: m1 = m2
Condition for perpendicular lines: m1 * m2 = -1
Distance formula: d = sqrt((x2 - x1)^2 + (y2 - y1)^2)
Distance between parallel lines: d = |c2 - c1| / sqrt(1 + m^2)

Theorems

Parallel and Perpendicular Line Theorems
Pythagorean Theorem

Suitable Grade Level

Grades 9-12