need some help

[abc123]

Hi

if I have the equations x = 1 + t and y = 10 - t, how would i get the equation path in terms of x and y. I'm dealing with a graph and this is the path of another object.

I think that I would need to do t = x - 1 and t = -y+10 but I'm not really sure. Confirmation would be great.

 

[Mrspi]

Hi

if I have the equations x = 1 + t and y = 10 - t, how would i get the equation path in terms of x and y. I'm dealing with a graph and this is the path of another object.

I think that I would need to do t = x - 1 and t = -y+10 but I'm not really sure. Confirmation would be great.

Ok...you now have two expressions that are EACH equal to t....set them equal to each other:

t = t
x - 1 = -y + 10

There's your equation in terms of x and y....you can put it into whichever form is required.

 

[abc123]

x - 1 = -y + 10

so then I could do x+y = 11 and graph that by doing -a over b for slope and using 11 as b?

 

[abc123]

sorry to double post but if x = 3 + 2t and y = 1 + t and I wanted the object to move twice as fast would I have to do x = 3 + 1/2 t and y = 1 + 1/2 t since its taking half the amount of time? (twice as fast)

 

[Mrspi]

x - 1 = -y + 10

so then I could do x+y = 11 and graph that by doing -a over b for slope and using 11 as b?

If your equation is

x + y = 11

and you're looking for the slope, solve the equation for y:

y = -x + 11

y = mx + b

"m" is the slope, and "b" is the y-intercept.

What is the slope if you have y = -x + 11?

 

[abc123]

I can find slope easily m = -1

I'm more worried about an earlier step now, thanks for help with this, I think I got it down.

I'm wondering if when i make an object move twice as fast if t has to become 1/2 t instead of 2 t? Help please.

I don't know if I should make x = 3 + 2t into x = 3 + t or x = 3 + 4t or x = 3 + 1/2 t.


thanks.