7)
Given the line
[tex]y=-x+2[/tex]
We know that the slope-intercept form of the line equation is
[tex]y=mx+b[/tex]
here
m is the slope and b is the intercept
Thus, the slope = -1
We know that the parallel lines have the same. Thus, the slope of the parallel line is: -1
Using point-slope of the line equation
[tex]y-y_1=m\left(x-x_1\right)[/tex]
substituting the values m = -1 and the point (4, 0)
[tex]y-0 = -1 (x-4)[/tex]
Writing in the slope-intercept form
[tex]y = -x+4[/tex]
Thus, the slope-intercept form of the equation of the line equation parallel to y=-x+ 2 will be:
8)
Note: Your line is a little bit unclear. But, I am assuming the
ine is: y = -x-1
Given the assumed line
y = -x-1
We know that the slope-intercept form of the line equation is
[tex]y=mx+b[/tex]
here
m is the slope and b is the intercept
Thus, the slope = m = -1
We know that a line perpendicular to another line contains a slope that is the negative reciprocal of the slope of the other line, such as:
slope = m = -1
perpendicular slope = – 1/m = -1/-1 = 1
Using point-slope of the line equation
[tex]y-y_1=m\left(x-x_1\right)[/tex]
substituting the values m = 1 and the point (4, 3)
[tex]y - 3 = 1 (x-4)[/tex]
Writing in the slope-intercept form
y-3 = x-4
y = x-4+3
y = x - 1
Thus, the slope-intercept form of the equation of the line equation perpendicular to y = -x-1 will be:
