Question

I need to check and see if this is correct?

1). The first equation that I am using for my discussion is: y=x+4 and the parallel must past through this given point (-7, 1)

y=x=-4 is the equation and (-7, 1) is the ordered pair
y=x+4 Since I want the new line parallel, I need to change the slope
y-y1=m(x-x1) point of slope
y-y1=1(x-(-7) since it must have the same slope, it must look like
y = x + b,
plug in your new point, (-7, 1)
1 = -7 + 1 ---->b = 8
new equation: y = x+8

Answer 1

yes
I did not entirely follow all your steps though.
The way I do it
y = 1 x + 4
m = 1, so my new m must be 1 also if parallel ( I did not follow you here)

point is (-7, 1)
1 = 1(-7) + b
yes b = 8
y = x + 8

Answer 2

where is the origin located in this equation?

Answer 3

The origin is always at (0,0)
However none of your lines go through the origin.

Answer 4

Oh, I see what you are doing.
given
y = x + 4
we want parallel
and through (-7,1)

since y = 1 x + 4
has slope = m = 1
ANY parallel line ALSO has slope = 1

slope = 1 = (y-yp)/(x - x1)
or
1 = ( y - 1 ) / (x - -7)
1 = (y-1) / (x+7)
y-1 = x + 7
y = x + 8

I like my way better, but the slope method works.

Answer 5

typo yp should be y1

Answer 6

How about this equation, does it have an origin and what is an x and y intercept?

2). My second equation that I am using is: y=-1/2x+1, and the perpendicular must pass through this given point (4, 2).

y=1/2+1 equation and (4, 2) is the ordered pair
Slope m=-1/2
So the slope of perpendicular =-1/m=2
y=2x+b put in point
2=2(4) +b Simplify
b=-6
y=2x-6 my answer in slope-intercept form