 In other courses, you found the distance between two points. The distance between other objects is a little more complicated. We could introduce formulas, but we won't. Remember, learn concepts, not formulas. For example, let's try to find the distance between the point 3, 7, and the line y equals 2x minus 8. So a key idea, start with what you know. We know how to find the distance between any two points, so let's consider the distance between the point 3, 7, and some point x, y on the line. As we move along the line, this distance changes. And remember, calculus is the mathematics of changing quantities. Now, what we call the distance between the point and the line is going to be the least value of these distances. Now this is an optimization problem, so let's set it up. The distance between 3, 7, and x, y on the line can be given by the distance formula. But remember, equals means replaceable. On the line, y equals 2x minus 7, so we can replace. And this gives us a function for the distance between a point and the line. To find the extreme values, we'll differentiate. This differentiating a square root gets complicated will square and use implicit differentiation. So s squared equals differentiating and solving. And since the critical values occur when the derivative is zero or undefined, they'll occur where the denominator is zero, which would make the derivative undefined, or where the numerator is zero, which would make the derivative zero. So remember, s is the distance between the point and the line, so s equals zero would require the line to pass through the point. But 3, 7 is not on the line, so s can never be zero. So the critical value must be where the numerator is zero, solving gives us, and you should probably reduce this, but this isn't an arithmetic class, focus on the calculus. Now remember, we should always verify the critical value corresponds to an extreme value. So we could use the first or second derivative test to verify that our solution 62 tenths corresponds to a minimum distance. But from geometry, we know that there will be no maximum distance between a point and the line, so this critical value must correspond to a minimum. Sometimes we want the actual distance we'll substitute into the equation and find, and this gives us the distance between the point and the line, and again you can reduce this if you want to, but this is a calculus class, not arithmetic. One of the advantages to the calculus approach is that while we use this to find the distance between a point and a line, nothing changes if we use a different curve. So let's try to find the distance between a point and a parabola. So again, the distance between our point and some point on the parabola will be, but equals means replaceable, so y squared equals 8x, and so we can rewrite this as, and again we want to find the minimum distance, so we'll find s squared and differentiate. While s equals zero would be a critical value, the parabola does not pass through the point, so s can never be zero, and so the only critical value will occur when the numerator is zero. So the other critical value will be the solution too, and since we want the actual distance we'll substitute into our distance formula to find as the distance between the point and the parabola.