91Ó°ÊÓ

When calculating the distance between two points in a plane, we use the distance formula. It is derived from the Pythagorean theorem and is written as follows:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
This formula computes the straight-line (or Euclidean) distance between any two points \[ (x_1, y_1) \] and \[ (x_2, y_2) \] in the coordinate plane.

To understand this better, imagine a right triangle where the difference in x-coordinates represents the base and the difference in y-coordinates represents the height. The distance formula gives the hypotenuse of this triangle.

Derivatives are a powerful tool in calculus, used to find rates of change and to optimize functions. When we want to find the minimum or maximum value of a function, derivatives help us identify points where these extrema occur.

In this exercise, we used the derivative to minimize the distance formula, giving us the point on the line closest to a specific point.

The derivative of a function is found using the rules of differentiation. For example, to differentiate \[ f(x) = x^2 \], we use the power rule which results in \[ f'(x) = 2x \].

By setting the derivative equal to zero and solving for x, we find the critical points where the function could have a minimum or maximum value.

Critical points are places on a graph where the derivative is zero or undefined. These points are important because they can indicate where a function has a maximum, a minimum, or a point of inflection.

To find critical points:

• Take the derivative of the function.
• Set the derivative equal to zero.
• Solve for the variable.

In our exercise, after finding the derivative of the distance squared function, we set it to zero and solved for x. This identified the critical point, which helped in determining the closest point on the line to a given point.

Line equations are used to describe straight lines in algebra and geometry. A common form is the slope-intercept form, written as \[ y = mx + b \], where \( m \) is the slope and \( b \) is the y-intercept.

In our exercise, the given line equation was \[ y = x + 1 \]. The slope here is 1, and the y-intercept is 1. This tells us the line has a steady rise and passes through the point (0,1).

Understanding line equations is crucial for solving many geometric problems, as they provide a way to connect algebraic expressions with visual graphs. Knowing how to manipulate and understand these equations can greatly simplify finding relationships and solving problems involving linear functions.