Understanding the Shortest Distance from the Origin to a Line Using Different Approaches

In this article, we will explore the concept of finding the shortest distance from the origin to a line using a variety of mathematical approaches, including vector analysis, algebraic methods, and the application of the Pythagorean theorem.

Introduction

The shortest distance from the origin to the line given by the equation (3x - 4y 20) can be determined using different mathematical techniques. We will examine each approach in detail to provide a comprehensive understanding of the problem.

Method 1: Using the Distance Formula

The distance (d) from a point ((x_0, y_0)) to a line given by the equation (Ax By C 0) can be found using the formula:

[d frac{|Ax_0 By_0 C|}{sqrt{A^2 B^2}}] A 3 B -4 C -20 (x_0, y_0 (0, 0))

Substitute these values into the formula:

[d frac{|3(0) - 4(0) - 20|}{sqrt{3^2 (-4)^2}} frac{20}{sqrt{9 16}} frac{20}{sqrt{25}} frac{20}{5} 4]

Hence, the shortest distance from the origin to the line (3x - 4y 20) is 4 units.

Method 2: Using Vector Analysis

A result from vector analysis states that the distance from a point ((a, b)) to a line (Ax By C 0) can be calculated as:

[text{Distance} frac{|Aa Bb C|}{sqrt{A^2 B^2}}]

Using the given values:

[text{Distance} frac{|3(0) - 4(0) - 20|}{sqrt{3^2 (-4)^2}} frac{20}{sqrt{9 16}} frac{20}{sqrt{25}} frac{20}{5} 4text{ units}]

This confirms the distance is 4 units.

Method 3: Intersection Points and Right-Angle Triangle

The given line (3x - 4y 20) intersects the x-axis at (x frac{20}{3}) and the y-axis at (y -5). These points and the line form a right-angle triangle at the origin. Dropping a perpendicular from the origin to the line gives the height (h).

Using the inverse Pythagorean theorem:

[frac{1}{h^2} frac{1}{x^2} frac{1}{y^2} frac{3}{left(frac{20}{3}right)^2} frac{1}{5^2} frac{3 times 9}{400} frac{1}{25} frac{27}{400} frac{16}{400} frac{43}{400}]

Hence:

[h sqrt{frac{400}{43}} approx 4text{ units}]

This approach also confirms the distance is 4 units.

Method 4: Using the Point-Slope Form and Optimization

Convert the given line (3x - 4y 20) to the form (y mx c):

[3x - 4y 20 Rightarrow -4y -3x 20 Rightarrow y frac{3}{4}x - 5]

Consider a point ((x, y)) on the line such that (y frac{3}{4}x - 5). The distance (D) from this point to the origin is:

[D^2 x^2 left(-frac{3}{4}x 5right)^2]

To find the minimum distance, differentiate (D^2) with respect to (x) and set the derivative to zero:

[frac{d(D^2)}{dx} 2x 2left(-frac{3}{4}x 5right)left(-frac{3}{4}right) 2x - frac{3}{2}x 15 0]

Solving for (x):

[frac{1}{2}x 15 0 Rightarrow x 12]

Substitute (x 12) back into the line equation to find (y):

[y frac{3}{4}(12) - 5 9 - 5 4]

The distance from the origin to the point ((12, 4)) is:

[D sqrt{12^2 4^2} sqrt{144 16} sqrt{160} 4sqrt{10} approx 4text{ units}]