Answer:
Step-by-step explanation:
To find the equation of a line parallel to x+y=3x+y=3 that passes through the point (−1,8)(−1,8), follow these steps:
Determine the slope of the given line x+y=3x+y=3.
Use the obtained slope to write the equation of the parallel line.
Let's find the slope of the given line:
x+y=3x+y=3 can be rearranged into the slope-intercept form y=mx+by=mx+b, where mm is the slope.
y=−x+3y=−x+3
Now, the slope of the given line is −1−1.
Since the parallel line will have the same slope, the slope of the parallel line is also −1−1.
Now, use the point-slope form of a linear equation to find the equation of the parallel line:
The point-slope form is given by y−y1=m(x−x1)y−y1=m(x−x1), where (x1,y1)(x1,y1) is a point on the line, and mm is the slope.
Using (−1,8)(−1,8) as the point and −1−1 as the slope, substitute these values into the point-slope form:
y−8=−1(x−(−1))y−8=−1(x−(−1))
Simplify:
y−8=−1(x+1)y−8=−1(x+1)
Distribute the −1−1:
y−8=−x−1y−8=−x−1
Add 88 to both sides:
y=−x+7y=−x+7
So, the equation of the line parallel to x+y=3x+y=3 that passes through the point (−1,8)(−1,8) is y=−x+7y=−x+7.