Math Problem Statement

In Exercises 5–8, use the given conditions to write an equation for each line in point-slope form and slope-intercept form. 5. Passing through (-8, -10) and parallel to the line whose equation is y = -4x + 3. 6. Passing through (-8, -7) and parallel to the line whose equation is y = -4x + 4. 7. Passing through (2, -3) and perpendicular to the line whose equation is y = 1/5x + 6. 8. Passing through (-4, 2) and perpendicular to the line whose equation is y = 3/5x + 7.

Solution

Let's solve each of the problems from Exercises 5 to 8 based on the provided conditions.

5. Passing through (-8, -10) and parallel to the line whose equation is y=4x+3y = -4x + 3

  • Step 1: Identify the slope of the given line: The given line is y=4x+3y = -4x + 3, so its slope mm is -4.

  • Step 2: Use the point-slope form equation: yy1=m(xx1)y - y_1 = m(x - x_1) Substituting m=4m = -4, x1=8x_1 = -8, and y1=10y_1 = -10: y(10)=4(x(8))y - (-10) = -4(x - (-8)) Simplifying: y+10=4(x+8)y + 10 = -4(x + 8) Expanding: y+10=4x32y + 10 = -4x - 32 Solving for yy: y=4x42y = -4x - 42 This is the slope-intercept form.

  • Point-slope form: y+10=4(x+8)y + 10 = -4(x + 8)

  • Slope-intercept form: y=4x42y = -4x - 42


6. Passing through (-8, -7) and parallel to the line whose equation is y=4x+4y = -4x + 4

  • Step 1: Identify the slope of the given line: The slope is m=4m = -4.

  • Step 2: Use the point-slope form equation: y(7)=4(x(8))y - (-7) = -4(x - (-8)) Simplifying: y+7=4(x+8)y + 7 = -4(x + 8) Expanding: y+7=4x32y + 7 = -4x - 32 Solving for yy: y=4x39y = -4x - 39 This is the slope-intercept form.

  • Point-slope form: y+7=4(x+8)y + 7 = -4(x + 8)

  • Slope-intercept form: y=4x39y = -4x - 39


7. Passing through (2, -3) and perpendicular to the line whose equation is y=15x+6y = \frac{1}{5}x + 6

  • Step 1: Find the slope of the perpendicular line: The slope of the given line is m=15m = \frac{1}{5}. For perpendicular lines, the slope is the negative reciprocal. So, the new slope m=5m = -5.

  • Step 2: Use the point-slope form: y(3)=5(x2)y - (-3) = -5(x - 2) Simplifying: y+3=5(x2)y + 3 = -5(x - 2) Expanding: y+3=5x+10y + 3 = -5x + 10 Solving for yy: y=5x+7y = -5x + 7 This is the slope-intercept form.

  • Point-slope form: y+3=5(x2)y + 3 = -5(x - 2)

  • Slope-intercept form: y=5x+7y = -5x + 7


8. Passing through (-4, 2) and perpendicular to the line whose equation is y=35x+7y = \frac{3}{5}x + 7

  • Step 1: Find the slope of the perpendicular line: The slope of the given line is m=35m = \frac{3}{5}. The slope of the perpendicular line is the negative reciprocal, so m=53m = -\frac{5}{3}.

  • Step 2: Use the point-slope form: y2=53(x(4))y - 2 = -\frac{5}{3}(x - (-4)) Simplifying: y2=53(x+4)y - 2 = -\frac{5}{3}(x + 4) Expanding: y2=53x203y - 2 = -\frac{5}{3}x - \frac{20}{3} Adding 2 (which is 63\frac{6}{3}) to both sides: y=53x203+63y = -\frac{5}{3}x - \frac{20}{3} + \frac{6}{3} Simplifying: y=53x143y = -\frac{5}{3}x - \frac{14}{3} This is the slope-intercept form.

  • Point-slope form: y2=53(x+4)y - 2 = -\frac{5}{3}(x + 4)

  • Slope-intercept form: y=53x143y = -\frac{5}{3}x - \frac{14}{3}


Let me know if you need more details or clarifications on any step!

Related Questions:

  1. How do you find the slope of a line perpendicular to another line?
  2. What is the general process to convert from point-slope form to slope-intercept form?
  3. How do you find the equation of a line passing through a point and parallel to another line?
  4. Why do you use the negative reciprocal of the slope for perpendicular lines?
  5. What is the relationship between slope and the steepness of a line?

Tip:

Always ensure that you simplify the final form of the equation, especially when working with fractions in slopes.

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Math Problem Analysis

Mathematical Concepts

Linear Equations
Parallel Lines
Perpendicular Lines
Point-Slope Form
Slope-Intercept Form

Formulas

Point-Slope Form: y - y1 = m(x - x1)
Slope-Intercept Form: y = mx + b
Slope of Parallel Lines: m1 = m2
Slope of Perpendicular Lines: m1 * m2 = -1

Theorems

Parallel and Perpendicular Line Theorems
Linear Equation Theory

Suitable Grade Level

Grades 9-12