Question

MEDAL!

Find parametric equations to describe the line 3x+4y = 12. Use your equations to find coordinates for the point that is three-fifths of the way from (4, 0) to (0, 3). By calculating some distances, verify that you have the correct point.

Answer 1

@zepdrix @agent0smith @mathmate @mathmale

Answer 2

easiest is to make \(x=t\) since it is just a line

Answer 3

then solve for \(3t+4y=12\) for \(y\) and you will have \(y\) as a function of \(t\)

Answer 4

I got the parametric equations:
\(x = 4-3t\)
\(y = 4t\)

Answer 5

i believe you but don't know how you got it

Answer 6

it is not correct

Answer 7

if x = 4-3t, then 3x = 12-9t
if y = 4t , then 4y = 12t
-----------------
3x+4y = 12 + 3t
you don't get the original back!!

Answer 8

Isn't 4(y=4t) as 4y=16t

Answer 9

oh, my bad. but even though 4y = 16t, you still not get the original back!!

Answer 10

I asked a question with this equation of 3x+4y=12 to @phi this morning and those are the parametric equations that we got:
http://openstudy.com/users/calculusxy#/updates/57ed1bb6e4b037122978e53b

Answer 11

You can let x = 4t +4, then 3x = 12t +12
y = -3t, then 4y = -12 t
so

3x + 4y =12, it is ok

Answer 12

Yes that actually makes sense.
x = 4 -4t
y = -3t

Answer 13

there are many parametric equations. You can use any of them. Like what @satellite73 suggested, you can let x =t, then solve for y. That works!!

Answer 14

let x = t,
y = 3-(3/4) t
it works!!

Answer 15

how would i do the remaining parts of the problem? especially with the distance.

Answer 16

for the point (4,0) that is x =4, y =0, right?

Answer 17

yes

Answer 18

if you use my parametric x = 4-4t, then when x =4, 4-4t =4, -4t = 0 --> t =0

Answer 19

well what point would be 3/5 of the way from (4,0) to (0,3)

Answer 20

@agent0smith Can you plz help me ?

Answer 21

"three-fifths of the way from (4, 0) to (0, 3)"

Find what t value gives (4, 0) and what t value gives (0, 3).
Subtract the t-values.
Multiply the result by 3/5.
Add it to the t-value that gave (4, 0).
Plug the resulting t-value into your parametric equations.

Answer 22

Just graph it to make sure it works/makes sense. You might have to subtract in the second last step, not add.

Answer 23

@calculuxy
My parametric gives you t <0 when replace back and it is hard to find out the point.
Hence, I use x= 4 -4t
y = 3t
at (4,0) , t = 0
at (0,3) , t =1
Hence the point 3/5 from (4,0) to (0,3) is at t = 3/5
Now, you apply to the parametric above to get
x= 4-4(3/5) = 8/5
y =3(3/5) =9/5

Answer 24

Now, test back by distance formula
with ratio k = 3/5
\(x=x_1+k(x_2-x_1)\), and \(y= y_1+k(y_2-y_1)\), we have the point is

\(x = 4+\dfrac{3}{5}(0-4) =8/5\) and \(y= 0+\dfrac{3}{5}(3-0) =\dfrac{9}{5}\)
They are matched.