00:02
We're asked to find the minimum distance between the line defined by this equation and the point 1 minus 2.
00:11
So here's our point of interest.
00:13
Here's our line.
00:15
So the distance, you know, is given by the square root, you know, this thing here.
00:20
And then this is our constraint.
00:23
So we can solve this for x in terms of y and plug that into here.
00:28
And we get this for d.
00:31
Then we can take the derivative of this respect to y and set y equal y b.
00:35
One, our optimal value, and we get this.
00:39
Now, let's see here, we have, what do i want to say? we have, let's see, we wind up with, now we need to set this equal to zero, and we get, let's see here, we just get one solution.
00:58
Obviously, this denominator is irrelevant, so we get, you know, with this, which is that make this equal to zero, and that means y1 equals minus 58 over 25.
01:09
Now, it could have been a little easier, and they suggest this in the next problem, to actually minimize d squared.
01:14
If we minimize d squared, we're also minimizing d.
01:18
So, you know, we've got rid of the square root here...