Question

write the equation of a line that is perpendicular to the given line and that passes through the given point.
-x+5y=14; (-5,-2)

Answer 1

do you know the point-slope formula?

Answer 2

yes I do. I'm up to y-(2)=5x-25 I don't know what to do next.

Answer 3

get into the form y=mx+b
then perpendicular would be the negative reciprocal of the slope m
and then you use
y-y1=m(x-x1)
using x1 and y1 to be the coordinates given

Answer 4

@amorfide I know this. I'm stuck at one part.

Answer 5

the first step is to first find the slope of the given line. To do this, put it in slope intercept form:
-x+5y=14
5y=x+14
y=1/5x+14/5
So the slope of this line is 1/5.
If it is perpendicular to this, you know the slope will be the negative inverse, which would be -5

then plug what you know into the equation
y-y1=m(x-x1)
y-(-2)=-5(x+5)
y+2= -5x-25
y=-5x-27

Answer 6

where the heck does the 27 come from?

Answer 7

you have to put it back into slope intercept form
from y+2=-5x-25, you have to subtract 2 from each side to isolate y

Answer 8

y-y1=m(x-x1)
x1=-5
y1=-2
m=-5

y--2=-5(x--5)
y+2=-5(x+5)

y+2=-5x-25

since we expanded the brackets

so just subtract 2 from both sides to get
y=-5x-27


this is the step by step with an added step graze missed out

you should understand it now hopefully