Find the equation of the line that passes through (2,2) and (4,-2). Write the answer in both point-slope form-

-and slope-intercept form.

the point slope form formula is
y - y1 = m (x - x1)

find first the slope:
m = (y2-y1) / (x2-x1)
= (2- - 2) / (2-4)
= (2+2)/(2-4)
=4/-2
= -2

hence, y - y1=m(x-x1), you can use any of the given coordinates as your x1 and y1 . . .
so,
y - 2 = -2(x - 2) . . or . . y + 2 = -2(x - 4) . .answer

in slope - intercept form . .
y - 2 = -2(x -2)
y - 2 = -2x + 4
y = -2x + 4 + 2
y = -2x + 6 . ..answer
where m = -2 . .and y-intercept is 6

Find the equation of the line that passes through (2,2) and (4,-2). Write the answer in both point-slope form-?

(2,2) and (4,-2)

standard equation of a line is:

y = mx + c
y is a y value
m is the gradient or slope
c is the y intercept of where the line cuts the y axis

equation for gradient or slope(m):

(y2 - y1) / (x2 - x1)

(2,2)
let the first 2 be x1
let the second 2 be y1

(4,-2)
let the 4 be x2
let the -2 be y2

so lets use the equation

(-2 - 2) / (4 - 2)
= (-4) / (2)
= -4 / 2
= -2

therefore the gradient or slope(m) = -2

lets find for c, the y intercept and lets use (4, -2)
4 is your x value
-2 is your y value

recall; y = mx + c and we found m your slope to be -2

so using the formula above, we have:
-2 = -2(4) + c
-2 = -8 + c
- 2 + 8 = c
6 = c
c = 6

so now using the standard for of the equation, we have:

y = -2x + 6

m = [2-(-2)]/(2-4) = 4/-2 = -2

y-2=-2(x-2)
y=-2x+4+2
y=-2x+6