Practice exercise 5. Draw the following perpendicular lines on a Cartesian plane.
1. [tex]2 y+4 x=6[/tex] and [tex]y=\frac{x}{2}-3[/tex]
2. [tex]y-3 x=4[/tex] and [tex]3 y[/tex]
3. [tex]5 y=2 x+10[/tex] and [tex]2 y+5 x=29[/tex]
4. [tex]y+\frac{x}{2}=\frac{3}{2}[/tex] and [tex]y=[/tex]
5. [tex]y=-\frac{4}{3} x-6[/tex] and [tex]4 y=3 x+20[/tex]
Gradients of perpendicular lines
Group Activity 7
1. On a graph paper, draw a set of perpendicular lines.
2. Calculate the gradient of each line.
Answer (1)
Rewrite each equation in slope-intercept form ( y = m x + c ) and identify the slope ( m ).
Check if the product of the slopes of each pair of lines is -1.
If the product is -1, the lines are perpendicular.
The perpendicular lines are: 2 y + 4 x = 6 and y = 2 x − 3 ; 5 y = 2 x + 10 and 2 y + 5 x = 29 ; y = − 3 4 x − 6 and 4 y = 3 x + 20 .
Explanation
Problem Analysis Let's analyze the given pairs of lines to determine which are perpendicular. Two lines are perpendicular if the product of their slopes is -1. We'll rewrite each equation in slope-intercept form ( y = m x + c ) to easily identify the slope ( m ).
First Pair of Lines
The first pair of lines is 2 y + 4 x = 6 and y = 2 x − 3 .
Rewrite the first equation: 2 y = − 4 x + 6 ⟹ y = − 2 x + 3 . The slope is m 1 = − 2 .
The second equation is already in slope-intercept form: y = 2 1 x − 3 . The slope is m 2 = 2 1 .
Check if the lines are perpendicular: m 1 ∗ m 2 = − 2 ∗ 2 1 = − 1 . Therefore, these lines are perpendicular.
Second Pair of Lines
The second pair of lines is y − 3 x = 4 and 3 y . We will assume that the second equation is 3 y = x .
Rewrite the first equation: y = 3 x + 4 . The slope is m 1 = 3 .
Rewrite the second equation: 3 y = x ⟹ y = 3 1 x . The slope is m 2 = 3 1 .
Check if the lines are perpendicular: m 1 ∗ m 2 = 3 ∗ 3 1 = 1 = − 1 . Therefore, these lines are not perpendicular.
Third Pair of Lines
The third pair of lines is 5 y = 2 x + 10 and 2 y + 5 x = 29 .
Rewrite the first equation: 5 y = 2 x + 10 ⟹ y = 5 2 x + 2 . The slope is m 1 = 5 2 .
Rewrite the second equation: 2 y = − 5 x + 29 ⟹ y = − 2 5 x + 2 29 . The slope is m 2 = − 2 5 .
Check if the lines are perpendicular: m 1 ∗ m 2 = 5 2 ∗ − 2 5 = − 1 . Therefore, these lines are perpendicular.
Fourth Pair of Lines
The fourth pair of lines is y + 2 x = 2 3 and y = . We will assume that the second equation is y = x .
Rewrite the first equation: y = − 2 1 x + 2 3 . The slope is m 1 = − 2 1 .
The second equation is y = x . The slope is m 2 = 1 .
Check if the lines are perpendicular: m 1 ∗ m 2 = − 2 1 ∗ 1 = − 2 1 = − 1 . Therefore, these lines are not perpendicular.
Fifth Pair of Lines
The fifth pair of lines is y = − 3 4 x − 6 and 4 y = 3 x + 20 .
The first equation is already in slope-intercept form: y = − 3 4 x − 6 . The slope is m 1 = − 3 4 .
Rewrite the second equation: 4 y = 3 x + 20 ⟹ y = 4 3 x + 5 . The slope is m 2 = 4 3 .
Check if the lines are perpendicular: m 1 ∗ m 2 = − 3 4 ∗ 4 3 = − 1 . Therefore, these lines are perpendicular.
Conclusion In summary, the perpendicular lines are:
2 y + 4 x = 6 and y = 2 x − 3
5 y = 2 x + 10 and 2 y + 5 x = 29
y = − 3 4 x − 6 and 4 y = 3 x + 20
Examples
Understanding perpendicular lines is crucial in various real-world applications. For instance, architects use perpendicular lines to design buildings, ensuring walls meet floors at right angles for structural stability. Similarly, in navigation, understanding perpendicular relationships helps determine accurate routes and bearings, especially when using coordinate systems. In computer graphics, perpendicular lines are fundamental in creating 3D models and rendering images, ensuring objects appear correctly in space.
