given this kind of standard question, I must assume that you have learned how to find the equation.
What part of the process do you not understand?
Which method do you use?
- the point-slope form ?
- the y-intercept slope form?
- ... ?
Can someone help me figure this out for my review?
Find the equation of the line that passes through (2,-3) and has a slope of (5/2)
4 answers
I normally use point slope. I'm used to having to sets of points not just one.
I've done.
y-(-3)=(5/2)(x-(-3))
y-(-3)=(5/2(-15/2)
y=(5/2x)-(-15/2)+3
y=(5/2x)+(9/2)
Is that correct?
y-(-3)=(5/2)(x-(-3))
y-(-3)=(5/2(-15/2)
y=(5/2x)-(-15/2)+3
y=(5/2x)+(9/2)
Is that correct?
no, you made an error in the first line
should have been
y-(-3)=(5/2)(x-(2))
y + 3 = (5/2)(x-2) ---->
times 2
2y + 6 = 5(x-2)
2y + 6 = 5x - 10
16 = 5x - 2y
or
5x - 2y = 16 OR 5x - 2y - 16 = 0
or
from ---->
y + 3 = (5/2)(x-2)
y = (5/2)x - 5 - 3
y = (5/2)x - 8
should have been
y-(-3)=(5/2)(x-(2))
y + 3 = (5/2)(x-2) ---->
times 2
2y + 6 = 5(x-2)
2y + 6 = 5x - 10
16 = 5x - 2y
or
5x - 2y = 16 OR 5x - 2y - 16 = 0
or
from ---->
y + 3 = (5/2)(x-2)
y = (5/2)x - 5 - 3
y = (5/2)x - 8