Question

Show the following pairs of lines intersect.
Determine the coordinates of the point of intersection and the angles formed by this lines.
a. L1: x = 5 + 2t, y =-3 + t and L2: x = 23 - 2s, y = 6 - s

What I did so far is:
L1:
coordinates (5, -3)
direction vector (2, 1)

L2:
coordinates (23 , 6)
direction vector ( -2, -1)

I'm thinking that, this two lines are collinear because direction vectors are scalar multiple and therefore the angle form will be 0.
Am I right?

Answer 1

@Psymon

Answer 2

@thomaster

Answer 3

Yeah, the direction vector is a scalar multiple, so parallel at the very least.

Answer 4

Tagged me just as I logged on, haha

Answer 5

ok, thanks again:)
lol..of course, i know you can help me..

Answer 6

For something we're learning at the same time, maybe, lol.

Answer 7

yes..I'm almost done with this chapter, i just need to finish all this questions..
my next topic will be "The intersection of a line with a Plane and the intersection of two lines"

Answer 8

Yeah, I need to keep up with ya. Well, assuming we didn't notice that they were scalar multiples, do you know how wed check?

Answer 9

...no, i'm not sure, there's a question like that and i'm doing it now.. trying to figure it out..

Answer 10

Im just taking out my text, just got home, lol.

Answer 11

oh you are attending school?

Answer 12

But I believe we'd have to solve a system of equations. So if the lines intersect, then there will be some t that will cause things to be equal.
And yeah, Im a college student. School doesnt start till the 26th, though. This is just me doing my own calc 3 preview, lol.

Answer 13

x= 5 + 2t y = -3 + t
x = 23-2s y = 6 - s

Answer 14

So we have to solve for s and t in this system by setting the x parametrics and the y parametrics equal:
5+2t = 23-2s
-3+t = 6-s

We good on how to solve systems of equations?

Answer 15

like for example:
\[L1: \frac{ x+3 }{ 3 }=\frac{ y+1 }{ 4 }\]
\[L2:\frac{ x-6 }{3}=\frac{ y-2 }{ -2 }\]
do you think I still need to change it to Cartesian equation?

Answer 16

oopsss!, i did not know you post something

Answer 17

xDD

Answer 18

Oh.. ok, I get it now

Answer 19

do you still have any reminders that I need to know??:)

Answer 20

Not really. Im still learning it, too xD

Answer 21

wait a sec. i'll try to solve another problem like this..

Answer 22

Alrighty. I need more examples anyway

Answer 23

Lot of typing there xD

Answer 24

I'm confused again..:]
for the example I gave above,
I change it to parametric equation:
\[L1: x=-3+3s, y=-1+4s\]
\[L2: x=6 + 3t, y=2-2t\]
then I followed what you said:

x-values:
\[-3+3s=6+3t\]
\[3s - 3t =9\]
\[s= t+3\]---> equation 1

y-values:
\[-1+4s=2-2t\]
\[4s+2t=3\]
--->equation 2

I tried to solved it, using substitution:
substitute equation 1 in equation 2
then if I will try to solve for s, it is impossible