In summary, a line is perpendicular to x + 3y = -3 if its slope is 3.
The subject requested is about finding the slope of a line that is perpendicular to the line specified by the equation x + 3y = -3. To find the slope, we'll first have to rewrite the given equation in slope-intercept form, y = mx + b, where m is the slope and b is the y-intercept.
Rewriting x + 3y = -3, we get y = -x/3 - 1. So the slope of the given line is -1/3. Now, two lines are perpendicular if and only if the product of their slopes is -1. Therefore, the slope of the line perpendicular to the given line will be -1 / original slope = -1 / (-1/3) = 3.
Thus, the slope of a line perpendicular to the line x + 3y = -3 is 3.
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Can you please help me with this question.
A. $4.50
B. $5.00
C. $5.50
D. $6.00
Answer:
5.00 (or B)
Step-by-step explanation:
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167 cm³
209 cm³
670 cm³
502 cm³
Answer:
The equation line is y + 3x = -10.
Step-by-step explanation:
In this question, first we have to find the slope of the equation:
The formula which we use to find the slope is:
slope = m = (y2-y1)/(x2-x1)
m = (8 + 1) / (-6 + 3)
m = 9 / (-3)
m = -3.
Now, put the value of slope and the first points (-3, -1). You can use any of the given points:
y - y1 = m(x - x1)
y + 1 = -3(x + 3)
y + 1 = -3x - 9
y + 3x = -10
This is the eqution line.
You can check if this line is true or not by putting the other points in this equation which will give the same points like,
y + 3(-6) = -10
y - 18 = -10
y = 8
So, we get the same old points (-6, 8) which we have given.