7)
Given the line
![y=-x+2](https://tex.z-dn.net/?f=y%3D-x%2B2)
We know that the slope-intercept form of the line equation is
![y=mx+b](https://tex.z-dn.net/?f=y%3Dmx%2Bb)
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
![y-y_1=m\left(x-x_1\right)](https://tex.z-dn.net/?f=y-y_1%3Dm%5Cleft%28x-x_1%5Cright%29)
substituting the values m = -1 and the point (4, 0)
![y-0 = -1 (x-4)](https://tex.z-dn.net/?f=y-0%20%3D%20-1%20%28x-4%29)
Writing in the slope-intercept form
![y = -x+4](https://tex.z-dn.net/?f=y%20%3D%20-x%2B4)
Thus, the slope-intercept form of the equation of the line equation parallel to y=-x+ 2 will be:
8)
<em>Note: </em><em>Your line is a little bit unclear. But, I am assuming the</em>
<em>ine is: </em>y = -x-1
<em />
Given the assumed line
y = -x-1
We know that the slope-intercept form of the line equation is
![y=mx+b](https://tex.z-dn.net/?f=y%3Dmx%2Bb)
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
![y-y_1=m\left(x-x_1\right)](https://tex.z-dn.net/?f=y-y_1%3Dm%5Cleft%28x-x_1%5Cright%29)
substituting the values m = 1 and the point (4, 3)
![y - 3 = 1 (x-4)](https://tex.z-dn.net/?f=y%20-%203%20%3D%201%20%28x-4%29)
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: