Write the equation of the line perpendicular to 5y - 5x = -20 that passes through the point (-3, 8)

y = 5x - 8 is in slope intercept format:

y = mx + b where m is slope, b is y-intercept.

Slope is 5

Perpendicular lnes have slopes that are

negative reciprocals... mp = -1/m

You need equation of line through (2, 2)

with slope -1/5

y = (-1/5)x + b where (2,2) is on the line.

2 = (-1/5)(2) + b

2 = -2/5 + b

b = 2 2/5 or 2.4 or 12/5

Your equation is: y = (-1/5)x + 12/5

This can be reworked to standard form or left

as is depending on your needs.

5y = -x + 12

x + 5y = 12

x+y-5=0

Step-by-step explanation:

let eqn Perpendicular to eqn 5y-5x=-20 be -5x-5y+k=0(1)

eqn (1) passes through (-3,8)

-5×-3-5×8+k=0

15-40+k=0

K=25

since required eqn perpendicular to 5y-5x=-20 is -5x-5y+25=0

x+y-5=0ans.

step-by-step explanation:

y = | -2(-2) |     →     y = | 4 |     →     y = 4       ⇒   (-2, 4)

y = | -2(-1) |     →     y = | 2 |     →       y =   2       ⇒   (-1, 2)    

y = | -2(0) |     →     y = | 0 |     →       y = 0         ⇒   (0, 0)

y = | -2(1) |   c →     y = | -2 |     →     y =   2         ⇒   (1, 2)    

y = | -2(2) |     →     y = | - 4 |     →     y =   4         ⇒   (2, 4)