Question

equation of a line perpendicular to y=3x-1?

Answer 1

perpendicular line is the negative reciprocal
so the slope of the line has to be
\[-\frac{1}{3}\]

since there's no point, you can put any constant or you can just leave it at zero

\[y = -\frac{1}{3}x\]

Answer 2

what do you mean since there's no point? what about the -1?

Answer 3

hes says your question did't say what point the perpendicular line was running through

Answer 4

the -1 is the intercept for y = 3x -1
but the question asks you for equation of A line, so ANY line
it can be y = -(1/3)x
or
y = -(1/3)x + 3
y = -(1/3)x + 5

it can be any number
because the question doesn't specify a specific point that the line has to go through

Answer 5

it says it has to pass through the point of intersection from these two equations:
y=24/x and y=3x-1

Answer 6

y=-(1/3)x+C where C is a constant is the general perpendicular line for the line y=3x+K where K can be any constant

Answer 7

it has to pass through the point where those two equations intersect?

Answer 8

yes

Answer 9

set 24/x=3x-1

Answer 10

solve for x

Answer 11

so set those equations equal to one another and solve for the point
24/x = 3x - 1
3x^2 -x = 24
3x^2 - x - 24 = 0
x = -(8/3), 3
myininaya, what am i doing wrong? how can i get two intersections for two lines???

Answer 12

24/x is not a line its a hyperbola

Answer 13

oh wait yeah, i didn't catch that..

Answer 14

so since we have two intersections and your question didn't specify which intersection we have 2 equations:
they both have the the form y=-(1/3)x+C
so we have 1 equation with x=-8/3 so y=3(-8/3)-1=-9 so we have -9=-(1/3)(-8/3)+b
solve for b to find the y intercept
and we also have equation 2 with x=3 so y=3(3)-1=8 so we have 8=-(1/3)(3)+b solve this for b
and you have your two equations

Answer 15

so the first equation we have -9-8/9=b so b=-89/9 so equation 1 is y=-(1/3)x-89/9 and the second equation we have 8+1=b so b=9 so equation 2 is y=-(1/3)x+9

Answer 16

thank you !

Answer 17

your welcome. what class is that? is that really algebra? That is hard question for algebra students. no offense. i would know since I teach algebra

Answer 18

i like it though. I might give it as a project hehe

Answer 19

ib math studies :)

Answer 20

ib?

Answer 21

International Baccalaureate its a higher level of math offered worldwide

Answer 22

i am jealous

Answer 23

i wish i was challenged more but I guess I know enough lol, but no one could really know enough so I take that back

Answer 24

its like college level math. well all of my courses are ib.

Answer 25

so did you understand everything above?

Answer 26

this is supposed to be an easy problem compared to the other problems we learn lol yes i did thank you !

Answer 27

i really liked your problem. its nice