00:01
So are these two lines the same? let's rewrite them to separate their base point and their line component, so to speak.
00:09
I'm going to take out the constants 1 1 1 plus our t component, excuse me, that'll be 2 negative 3 1 times t, and then our l2 has our point 1 0 4 plus negative 4 6 negative 2 times t.
00:38
Now what i want you to notice is that if we define s to be negative 2 times t, we can put our s here, and that, well, i suppose t will be s over negative 2, so s, if we divide 3 by negative 2, this becomes 2 negative 3 1, just as established by a little change of parameters.
01:08
And so we conclude that because these two lines have the same directional component, they are parallel.
01:14
This means they're the same line only if they share a point.
01:18
Let's go ahead and see what that point might be.
01:21
Well, 1 1 1 is on l1, right, so let's see if 1 1 1 is on l2.
01:27
1 1 1 should ostensibly be equal to 1 0 4 plus 2 negative 3 1 times s...