Problem 72
Question
For each of the following pairs of equations, (1) predict whether they represent parallel lines, perpendicular lines, or lines that intersect but are not perpendicular, and (2) graph each pair of lines to check your prediction. (a) \(5.2 x+3.3 y=9.4\) and \(5.2 x+3.3 y=12.6\) (b) \(1.3 x-4.7 y=3.4\) and \(1.3 x-4.7 y=11.6\) (c) \(2.7 x+3.9 y=1.4\) and \(2.7 x-3.9 y=8.2\) (d) \(5 x-7 y=17\) and \(7 x+5 y=19\) (e) \(9 x+2 y=14\) and \(2 x+9 y=17\) (f) \(2.1 x+3.4 y=11.7\) and \(3.4 x-2.1 y=17.3\)
Step-by-Step Solution
Verified Answer
(a) Parallel, (b) Parallel, (c) Intersect but not perpendicular, (d) Perpendicular, (e) Intersect but not perpendicular, (f) Perpendicular.
1Step 1: Analyze Equations for Parallelism
For any two lines, if their equations have the same coefficients for both the x and y terms after being rewritten to the form \( ax + by = c \), they are parallel. Check this for each pair of equations given.(a) Both equations have coefficients \( 5.2 \) and \( 3.3 \), making them parallel.(b) Similarly, \( 1.3 \) and \( -4.7 \) are the same for both equations, indicating parallelism here too.
2Step 2: Analyze Perpendicularity and Intersection
To check for perpendicularity, multiply the slopes of the lines. If the product is \(-1\), the lines are perpendicular. Otherwise, if they have different slopes, then they intersect but are not perpendicular. Convert the equations into the slope-intercept form \( y = mx + b \) if needed.(c) Line 1's slope is \(-\frac{2.7}{3.9}\) and Line 2's slope is \(\frac{2.7}{-3.9}\), making them reflections across the y-axis but not perpendicular.(d) Convert to slope-intercept form: - For first equation: slope = \(\frac{5}{7}\) - For second equation: slope = \(-\frac{7}{5}\) - The product is \(-1\), indicating perpendicularity.(e) Convert to slope-intercept form: - First equation slope \( = -\frac{9}{2} \) - Second equation slope \( = -\frac{2}{9} \), different slopes indicate they intersect but are not perpendicular.(f) Slope of first: \(-\frac{2.1}{3.4}\), Slope of second: \(\frac{3.4}{2.1}\). Their product is \(-1\), so lines are perpendicular.
3Step 3: Graph Equations for Confirmation
Graph the lines for each pair.
- (a) The lines are parallel because they have the same slope and different y-intercepts.
- (b) Also parallel with distinct y-intercepts, no intersection.
- (c) The intersection of non-parallel, non-perpendicular lines.
- (d) The graph shows intersections at right angles denoting perpendicularity.
- (e) Graph shows they intersect but do not cross perpendicularly.
- (f) Graph confirms perpendicular intersection.
Key Concepts
Slope of a LineLinear EquationsGraphing Lines
Slope of a Line
The slope of a line is like the guide that tells us how steep the line is on a graph. In mathematical terms, the slope is often represented by the letter \( m \). It is calculated as the change in the vertical direction (up/down), divided by the change in the horizontal direction (left/right) between any two points on the line. In a formula, it looks like this:\[m = \frac{{y_2 - y_1}}{{x_2 - x_1}} \]
- When the slope is positive, the line goes uphill from left to right.
- When the slope is negative, it goes downhill.
- If the slope is zero, the line is perfectly horizontal.
- An undefined slope indicates a vertical line.
- Two lines are parallel if their slopes are equal. For instance, in the original exercise, equations (a) and (b) demonstrate the parallelism by having identical slopes.
- Lines are perpendicular if the product of their slopes is \(-1\). For example, in part (d) of the exercise, slopes are reciprocals and opposite, indicating that lines are perpendicular.
Linear Equations
Linear equations are fundamental in algebra and appear as straight lines on a graph. They generally have the form \( ax + by = c \), where \( a \), \( b \), and \( c \) are constants. In this form, \( x \) and \( y \) are variables representing coordinate points on the graph.
The form can also be rearranged to the more familiar slope-intercept form \( y = mx + b \):
For example, in question (d) of the original exercise, transforming both equations helped identify one as a positive slope and the other with a negative reciprocal, signifying perpendicularity. By putting equations in slope-intercept form, we can easily see if lines are parallel, perpendicular, or simply intersecting.
The form can also be rearranged to the more familiar slope-intercept form \( y = mx + b \):
- \( m \) represents the slope, telling us how steep the line is.
- \( b \) is the y-intercept, which is where the line crosses the y-axis.
For example, in question (d) of the original exercise, transforming both equations helped identify one as a positive slope and the other with a negative reciprocal, signifying perpendicularity. By putting equations in slope-intercept form, we can easily see if lines are parallel, perpendicular, or simply intersecting.
Graphing Lines
Graphing lines allows us to visually represent linear equations and understand the relationships between them. With the slope-intercept form \( y = mx + b \), plotting is straightforward because we know:
- The slope \( m \) dictates the tilt of the line.
- The y-intercept \( b \) establishes the starting point on the y-axis.
- Start by plotting the y-intercept \( b \) on the y-axis.
- Use the slope \( m \) to find another point. For instance, if \( m = 2 \), go up 2 units and to the right 1 unit from the y-intercept.
- Draw a straight line through these points, extending across the graph.
Other exercises in this chapter
Problem 71
Remember that we did some work with parallel lines back in the graphing calculator activities in Problem Set 7.1. Now let's do some work with perpendicular line
View solution Problem 72
How would you help a friend determine the equation of the line that is perpendicular to \(x-5 y=7\) and contains the point \((5,4)\) ?
View solution Problem 73
\text { Explain how you would find the slope of the line } y=4 \text {. }
View solution Problem 74
The equation of a line that contains the two points \(\left(x_{1}, y_{1}\right)\) and \(\left(x_{2}, y_{2}\right)\) is \(\frac{y-y_{1}}{x-x_{1}}=\frac{y_{2}-y_{
View solution