Answer:
1/2
Step-by-step explanation:
First convert the equation to slope-intercept form.
3x+3y=63
3y=-3y+63
y=-x+21
The slope for a parallel line would be -1.
To find the slope for a perpendicular line you need to find the opposite reciprocal. Therefore the slope would be 1/2.
The slope of the line that is perpendicular to another line whose equation is 3x + 3y = 63 is 1
From the question,
We are to determine the slope of a line that is perpendicular to another line whose equation is 3x + 3y = 63.
NOTE: The slopes of two perpendicular lines are negative reciprocals of each other.
That is,
If a line is perpendicular to another line that has slope m, then the slope of the line is [tex]-\frac{1}{m}[/tex]
First, we will determine the slope of the line whose equation is 3x + 3y = 63
To do this,
We will rearrange the equation in the slope-intercept form of an equation of a straight line and then compare
The slope-intercept form of an equation of a straight line is
y = mx +c
Where m is the slope
and c is the y-intercept
Now, we will arrange the equation
3x + 3y = 63
First, divide through by 3 to get
x + y = 21
∴ y = -x + 21
Now, compare this to the slope-intercept form of an equation of a straight line ( y = mx +c)
∴ m = -1 and c = 21
Thus, the slope of the line whose equation is 3x + 3y = 63 is -1
Now, for the slope of the line that is perpendicular to line whose equation is 3x + 3y = 63
The slope of the line = [tex]-\frac{1}{-1}[/tex]
∴ The slope of the line = 1
Hence, the slope of the line that is perpendicular to another line whose equation is 3x + 3y = 63 is 1
Learn more here: https://brainly.com/question/20719015
