Problem 24 - The Calculus with Analytic Geometry

Question

Prove by the method of this section that the shortest distance from the point \(P_{1}\left(x_{1}, y_{1}\right)\) to the line \(l\), having the equation \(A x+B y+C=0\), is \(\left|A x_{1}+B y_{1}+C\right| / \sqrt{A^{2}+\bar{B}^{2}} .\) (HINT: If \(s\) is the number of units from \(P_{1}\) to a point \(P(x, y)\) on \(l\), then \(s\) will be an absolute minimum when \(s^{2}\) is an absolute minimum.)

Step-by-Step Solution

Answer

The shortest distance is \(|A x_{1} + B y_{1} + C|/ \sqrt{A^{2} + B^{2}} \).

Step 1 - Understand the Problem

We need to find the shortest distance from a point \(P_{1}(x_{1}, y_{1})\) to the line \(l: A x + B y + C = 0\). We're given a hint that the problem can be solved by minimizing \(s^{2}\).

Step 2 - Set Up the Distance Formula

The distance \(s\) from point \(P_{1}(x_{1}, y_{1})\) to any point \(P(x, y)\) on the line is given by: \[s = \sqrt{(x - x_{1})^{2} + (y - y_{1})^{2}}\]. To minimize this distance, we minimize its square \(s^{2} = (x - x_{1})^{2} + (y - y_{1})^{2}\).

Step 3 - Express \(y\) in terms of \(x\)

Since \(P(x, y)\) is on the line \(l,\) we have \(A x + B y + C = 0\). Thus, we can express \(y\) in terms of \(x\) by solving for \(y\): \[y = -\frac{A}{B} x - \frac{C}{B}\].

Step 4 - Substitute \(y\) in the Distance Formula

Substitute \(y = -\frac{A}{B} x - \frac{C}{B}\) into the distance formula: \[s^{2} = (x - x_{1})^{2} + \left(-\frac{A}{B} x - \frac{C}{B} - y_{1}\right)^{2}\].

Step 5 - Differentiate and Minimize

Differentiate \(s^{2}\) with respect to \(x\) and set the derivative to zero to find the minimum point. Solve \(\frac{d(s^{2})}{dx} = 2(x - x_{1}) + 2\left(-\frac{A}{B} x - \frac{C}{B} - y_{1}\right)(-\frac{A}{B}) = 0\) for \(x\).

Step 6 - Solving the Derivative

Simplify the derivative equation to derive the value of \(x\). Let \(N = -\frac{A}{B} x - \frac{C}{B}\) and solve for \(x: 2(x - x_{1}) + 2N(-\frac{A}{B}) = 0\).This is simplified to find the point where distance is minimum.

Step 7 - Substitute Back to Find \(x\) and \(y\)

Find \(x\) and then use the earlier equation to find \(y\). Plug back these values into the original distance formula to find the minimum distance.

Step 8 - Derive the Shortest Distance Formula

Use algebra to simplify and show that the shortest distance from \(P_{1}(x_{1}, y_{1})\) to the line \(l\) is: \[\frac{|A x_{1} + B y_{1} + C|}{\sqrt{A^{2} + B^{2}}}\].

Key Concepts

Minimization in Calculus

Minimizing a function is a core concept in calculus. When you need to find the minimum or maximum value of a function, you typically follow a process that involves differentiation and solving for critical points. The first step is identifying the function you want to minimize. In this case, we wish to minimize the squared distance from a point to a line. By minimizing the squared distance, we avoid dealing with the square root, which simplifies our calculations.

We initialize with the distance formula and then substitute any constraints into it. After this, we differentiate the squared distance function and find where its derivative (slope) is zero, which indicates a potential minimum or maximum. Finally, these critical points are checked against the function to confirm they provide the minimal distance.

Distance Formula

The distance formula originates from the Pythagorean theorem. For two points \(P_1(x_1, y_1)\) and \(P(x, y)\), the distance formula is: \[ s = \sqrt{(x - x_1)^2 + (y - y_1)^2} \]. This gives the straight-line distance between the points.

When dealing with the shortest distance from a point to a line, we use this formula a bit differently. Instead of the distance directly, we often work with its square in order to simplify differentiation. This approach helps to eliminate the square root, making our derivative calculations easier to handle.
We then substitute the line equation's expression for \(y\) back into the distance equation. This replaces \(y\) with a term in \(x\), allowing us to work with a single-variable function in terms of \(x\).

Differentiation

Differentiation is the process of finding the derivative of a function. The derivative represents the function's rate of change or slope at any given point. For finding the shortest distance, differentiation helps us find the critical points of the squared distance function. These points are where the slope is zero, indicating a potential minimum or maximum.

Starting with \( s^2 \) from our earlier substitute, we differentiate it with respect to \( x \). This derivative can be set to zero to find critical points: \[ \frac{d(s^2)}{dx} = 2(x - x_1) + 2(-\frac{A}{B}x - \frac{C}{B} - y_1)( -\frac{A}{B}) = 0 \]. Solving this equation gives us the x-coordinate where the distance is minimized. To find \( y \), substitute back into the line equation. The distance at these points represents the minimum distance from the point to the line.