Question

What is the equation of the line that passes through the point (-2, 2) and has a slope of 5?

Thank you to anyone who can explain how to do this!

Answer 1

1)
slope = 2, point (1, 6)
(y - 6) = 2(x - 1)
y - 6 = 2x - 2
y = 2x + 4

2)
slope = -1/5, point (5, 4)
(y - 4) = (-1/5)(x - 5)
y - 4 = (-1/5)x + 1
y = (-1/5)x + 5


3)
slope = 4/3, point (12, 1)
(y - 1) = (4/3)(x - 12)
y - 1 = (4/3)x - 16
y = (4/3)x - 15


4)
slope = 0, y-intercept -3
y = -3


5)
slope = 2, point (-3, -10)
(y + 10) = 2(x + 3)
y + 10 = 2x + 6
y = 2x - 4


6)
slope = 1/2, point = (-2, 4)
(y - 4) = (1/2)(x + 2)
y - 4 = (1/2)x + 1
y = (1/2)x + 5

7)
slope = 5/4, point (4, -8)
(y + 8) = (5/4)(x - 4)
y + 8 = (5/4)x - 5
y = (5/4)x - 13

There, you have your lines

Answer 2

use the point slope form.\[(y-y_{1})=m(x-x_{1})\]
plug in (-2,2)
for \[y_{1}\] and \[ x_{1}\]
m = slope
solve and isolate y to one side for y - intercept
then rewrite equation in y=mx+b form

Answer 3

you need to use the point slope formula because you have a set of points and a slope.
y - y1 = m(x - x1)
slope(m) = 5
(-2,2) x1 = -2 and y1 = 2

now just sub your info in the formula
y - 2 = 5(x - (-2)
y - 2 = 5(x + 2)
y - 2 = 5x + 10
y = 5x + 10 + 2
y = 5x + 12

Answer 4

Thank you for explaining. I originally had 2= -10 + 12. I was forgetting to add the two on the other side.
:P

Answer 5

easy mistake.....you would have eventually found your mistake and corrected it :)

Answer 6

Thank you again for the help (: