Which pair of equations would represent lines that are PERPENDICULAR to each other?
I.
3
x
−
2
y
=
12
II.
3
x
+
2
y
=
−
12
III.
2
x
−
3
y
=
−
12
I and II
I and II
II and III
II and III
I and III
I and III
None of these lines are perpendicular to each other.
II & III
Well, it seems that none of these lines are perpendicular to each other. I guess they just didn't click on a perpendicular level. Maybe they need a little bit of geometry therapy. Keep trying, lines!
To determine which pair of equations represents lines that are perpendicular to each other, we need to compare their slopes.
The slope-intercept form of a linear equation is y = mx + b, where m represents the slope.
In equation I (3x - 2y = 12), we can rewrite it as -2y = -3x + 12 and then divide by -2, resulting in y = 3/2x - 6. The slope of this line is 3/2.
In equation II (3x + 2y = -12), we can rewrite it as 2y = -3x - 12 and then divide by 2, resulting in y = -3/2x - 6. The slope of this line is -3/2.
In equation III (2x - 3y = -12), we can rewrite it as -3y = -2x - 12 and then divide by -3, resulting in y = 2/3x + 4. The slope of this line is 2/3.
Since the slopes of equations I (3/2) and III (2/3) are not negative reciprocals of each other, they are not perpendicular.
However, the slopes of equations I (3/2) and II (-3/2) are negative reciprocals of each other, meaning that they are perpendicular to each other.
Therefore, the pair of equations that represent lines that are perpendicular to each other are I and II.
To determine which pair of equations represents lines that are perpendicular to each other, we need to examine the slopes of the lines.
The slope-intercept form of a linear equation is y = mx + b, where m is the slope of the line.
For two lines to be perpendicular, their slopes must have a product of -1.
Let's calculate the slopes of the given equations to find the pair that satisfies this condition:
I. 3x - 2y = 12
To find the slope, let's rearrange the equation in the slope-intercept form:
-2y = -3x + 12
Dividing all terms by -2:
y = (3/2)x - 6
The slope of this line is 3/2.
II. 3x + 2y = -12
Again, rearranging the equation in slope-intercept form:
2y = -3x - 12
Dividing all terms by 2:
y = (-3/2)x - 6
The slope of this line is -3/2.
III. 2x - 3y = -12
Rearranging this equation:
-3y = -2x - 12
Dividing all terms by -3:
y = (2/3)x + 4
The slope of this line is 2/3.
Now, let's compare the slopes of the equations:
The slope of equation I: 3/2
The slope of equation II: -3/2
The slope of equation III: 2/3
Since the product of the slopes of equations I and II is (-3/2)(3/2) = -9/4 ≠ -1, and the product of the slopes of equations II and III is (-3/2)(2/3) = -1, we can conclude that lines II and III represent lines that are perpendicular to each other.
Therefore, the correct answer is:
II and III