Go back, I just saw this below and replied.
http://www.jiskha.com/display.cgi?id=1406309814
where is the origin located in this equation?
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
6 answers
Also what is the x and y intercept for this equation and origin? 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
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
If y = x + 8
the x axis intercept for that is when y = 0
0 = x + 8
x = -8 so (-8,0)
the y axis intercept for that is when x = 0
which is b in y = m x + b
so 8 so at (0, 8)
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I thought I did that other one too.
This time you did the problem my way and you got the right answer.
If we did it your way:
y = (-1/2) x + 1
slope = m = -1/2
SO
slope of perpendicular = -1/m = -1/(-1/2)
= 2
then your way
2 = (y -y1) / (x - x1)
2 = (y -2)/ (x-4)
2 x - 8 = y - 2
y = 2 x - 6
the x axis intercept for that is when y = 0
0 = x + 8
x = -8 so (-8,0)
the y axis intercept for that is when x = 0
which is b in y = m x + b
so 8 so at (0, 8)
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I thought I did that other one too.
This time you did the problem my way and you got the right answer.
If we did it your way:
y = (-1/2) x + 1
slope = m = -1/2
SO
slope of perpendicular = -1/m = -1/(-1/2)
= 2
then your way
2 = (y -y1) / (x - x1)
2 = (y -2)/ (x-4)
2 x - 8 = y - 2
y = 2 x - 6
Your talk of origins does not make a lot of sense to me since the origin is at (0,0) for all your problems.
However perhaps your class is trying to work toward a vector version of this material where you specify a point and move out from there at a certain slope. In that case your "origin" might be the given point you start out from, (x1,y1).
This is quite a bit further than where you are (see https://www.khanacademy.org/math/linear-algebra/vectors_and_spaces/vectors/v/linear-algebra--parametric-representations-of-lines but might be what your class is working toward.
However perhaps your class is trying to work toward a vector version of this material where you specify a point and move out from there at a certain slope. In that case your "origin" might be the given point you start out from, (x1,y1).
This is quite a bit further than where you are (see https://www.khanacademy.org/math/linear-algebra/vectors_and_spaces/vectors/v/linear-algebra--parametric-representations-of-lines but might be what your class is working toward.
That is:
https://www.khanacademy.org/math/linear-algebra/vectors_and_spaces/vectors/v/linear-algebra--parametric-representations-of-lines
https://www.khanacademy.org/math/linear-algebra/vectors_and_spaces/vectors/v/linear-algebra--parametric-representations-of-lines
Thank you so much! Algebra is bit confusing for me
1 answer