Find the equation of the line perpendicular to x−5y=15 that passes through the point (−2,5).

Question

please answer

Answer 1

Find the equation of the line perpendicular to x−5y=15 that passes through the point (−2,5).

 

first we write equation in slope intercept form

 

x -5y =15

 

y = x/5 -3

 

 slope of given line is 1/5

 

so slope of perpendicular line is -5

 

y = -5x + b

 

as line goes through (−2,5).

so x = -2 and  y = 5

 

5 = 5(-2) +b

 

5 = -10 +b

 

b = 15

 

equation of perpendicular line is

y=-5x +15

 

 

Answer 2

Its slope must be the negative reciprocal of the slope of the given line.

 

Given line:

 

m = -1/-5 = 1/5

 

Perpendicular line:

 

m' = -1/m = -1/(1/5) = -5

 

y = m'x + b

 

y = -5x + b

 

Use the given point to find b.

 

5 = -5(-2) + b

 

Solve for b, plug it into

 

y = -5x + b,

 

and you are done.  Finish from here.

Answer 3

Deleted - I used the wrong sign on the slope!