Question

Find the distance from (0,0) to the line 4x-4y=9

Answer 1

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

Answer 2

nope, distance is always positive

Answer 3

i know thats what bothers mee

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

the point given (0,0)..

Answer 8

OOH

Answer 9

then this gotta be simple right?

Answer 10

but i keep getting a negative distance

Answer 11

\[\frac{9}{\sqrt{4^2+4^2}}\]

Answer 12

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

Answer 13

so 1.5909 would be the distance??

Answer 14

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

Answer 15

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

Answer 16

i didnt say

Answer 17

http://www.wolframalpha.com/input/?i=9%2F%28sqrt%2832%29%29

Answer 18

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

Answer 19

@sleepyhead314 .. wat do u think?

Answer 20

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

Answer 21

where did u get that equation frm..?

Answer 22

did u derive that ?

Answer 23

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

Answer 24

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

Answer 25

I just plugged in
A = -k
B = 1
C = 9
x = 0
y = 0

because y = km + 9 --> -km + y = 9

Answer 26

ok so we now solve for K?

Answer 27

yep :)

Answer 28

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

Answer 29

that's what I got :)

Answer 30

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

Answer 31

http://www.wolframalpha.com/input/?i=distance+between+%280%2C0%29+and+y+%3D+%28sqrt%285%29%2F2%29x+%2B+9

Answer 32

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...^^

Answer 33

I suppose you could plug in
A = (sqrt(5)/2)
B = 1
C = 9
x = 0
y = 0

and make sure you get 6 ?

Answer 34

ok cool.. thanks to all