## anonymous 5 years ago equation of a line perpendicular to y=3x-1?

1. anonymous

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 $y = -\frac{1}{3}x$

2. anonymous

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

3. myininaya

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

4. anonymous

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

5. anonymous

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

6. myininaya

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

7. anonymous

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

8. anonymous

yes

9. myininaya

set 24/x=3x-1

10. myininaya

solve for x

11. anonymous

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???

12. myininaya

24/x is not a line its a hyperbola

13. anonymous

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

14. myininaya

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

15. myininaya

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

16. anonymous

thank you !

17. myininaya

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

18. myininaya

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

19. anonymous

ib math studies :)

20. myininaya

ib?

21. anonymous

International Baccalaureate its a higher level of math offered worldwide

22. myininaya

i am jealous

23. myininaya

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

24. anonymous

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

25. myininaya

so did you understand everything above?

26. anonymous

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

27. myininaya

i really liked your problem. its nice