## LynFran one year ago Find the distance from (0,0) to the line 4x-4y=9

1. LynFran

ok so i work this problem ..but keep getting a negative distance...can that be??

2. Vocaloid

nope, distance is always positive

3. LynFran

i know thats what bothers mee

4. anonymous

probably be easier if you write the line as $y=x-\frac{9}{4}$

5. LynFran

but dont we have to use the normal/ perpendicular formula $d=\frac{ Ax+By+C }{ \pm \sqrt{A^2+B^2} }$

6. anonymous

oh okay not familiar with that what are x and y in that formula ?

7. LynFran

the point given (0,0)..

8. anonymous

OOH

9. anonymous

then this gotta be simple right?

10. LynFran

but i keep getting a negative distance

11. anonymous

$\frac{9}{\sqrt{4^2+4^2}}$

12. anonymous

the numerator should be the absolute value your formula has a silly $$\pm$$ in it which just means make sure it is positive

13. LynFran

so 1.5909 would be the distance??

14. anonymous

$d=\frac{ Ax+By+C }{ \pm \sqrt{A^2+B^2} }$ should be $d=\frac{ |Ax_0+By_0+C| }{ \sqrt{A^2+B^2} }$

15. anonymous

idk are you supposed to give a decimal approximation, or the answer as a radical?

16. LynFran

i didnt say

17. anonymous
18. LynFran

ok i have another question Determine K so that the distance from the origin to the line y=km+9; p=6 will be stated

19. LynFran

@sleepyhead314 .. wat do u think?

well for your new question, if you plug in what they gave into the equation... $\frac{ 9 }{ \sqrt{k^2 + 1} } = 6$

21. LynFran

where did u get that equation frm..?

22. LynFran

did u derive that ?

ehh? the equation you gave before...? $\frac{ Ax + By + C}{ \sqrt{A^2 + B^2} }$

24. LynFran

o its the same so 9=C but what about A and B?

I just plugged in A = -k B = 1 C = 9 x = 0 y = 0 because y = km + 9 --> -km + y = 9

26. LynFran

ok so we now solve for K?

yep :)

28. LynFran

$9=6(\sqrt{K^2+1})$$\frac{ 3 }{ 2}=\sqrt{k^2+1}$$\frac{ 9 }{ 4 }=k^2+1$$\frac{ 5 }{ 4 }=K^2$$\frac{ \sqrt{5} }{ 2 }=k$

that's what I got :)

30. LynFran

ok is there a way to check if its correct though?

32. LynFran

like when we solve equation we can sub the value we find back into the equation to see that its correct..eg 4x+2y=0 so y=0 and x=0 4(0)+2(0)=0 like this...^^

I suppose you could plug in A = (sqrt(5)/2) B = 1 C = 9 x = 0 y = 0 and make sure you get 6 ?

34. LynFran

ok cool.. thanks to all