Problem 130 - Elementary and Intermediate Algebra

Question

Analytical Geometry. The length of the perpendicular segment drawn from \((-2,2)\) to the line with equation \(2 x-4 y=4\) is given by $$ L=\frac{|2(-2)+(-4)(2)+(-4)|}{\sqrt{(2)^{2}+(-4)^{2}}} $$ Find \(L\). Express the result in simplified radical form. Then give an approximation to the nearest tenth.

Step-by-Step Solution

Answer

The length of the perpendicular is \(\frac{8\sqrt{5}}{5}\), approximately 3.6.

Step 1: Substitute into the Formula

Begin by substituting the given point and line equation coefficients into the formula for the distance from a point to a line. The given formula is already set up, so the point (-2, 2) and line coefficients 2, -4, and -4 are substituted correctly.

Step 2: Calculate the Numerator

Calculate the numerator: \[|2(-2) + (-4)(2) + (-4)| = |-4 - 8 - 4| = |-16| = 16.\]

Step 3: Calculate the Denominator

Calculate the denominator by finding the square root of the sum of the squares of the line coefficients:\[\sqrt{(2)^2 + (-4)^2} = \sqrt{4 + 16} = \sqrt{20}.\]

Step 4: Simplify the Denominator

Simplify \(\sqrt{20}\) to its simplest radical form:\[\sqrt{20} = \sqrt{4 \times 5} = \sqrt{4} \times \sqrt{5} = 2\sqrt{5}.\]

Step 5: Compute the Final Answer

Substitute the simplified denominator back into the fraction and compute:\[L = \frac{16}{2\sqrt{5}} = \frac{16}{2} \times \frac{1}{\sqrt{5}} = 8 \times \frac{\sqrt{5}}{5} = \frac{8\sqrt{5}}{5}.\]

Step 6: Approximating the Value

To get an approximate value correct to the nearest tenth, calculate \(\frac{8\sqrt{5}}{5}\) using an approximate value for \(\sqrt{5} \approx 2.236\):\[L \approx \frac{8 \times 2.236}{5} = \frac{17.888}{5} \approx 3.6.\]

Key Concepts

Distance from a Point to a Line

Understanding how to find the distance from a point to a line in analytical geometry involves a special formula. This formula calculates the shortest path from a point, directly perpendicular, to the line.

Here's how it works:

• A point in space is given, denoted by coordinates, like ext{(-2, 2)} in our exercise.
• A line is represented by an equation, such as ext{2x - 4y = 4}.
• The distance formula is ext{\(\frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}\) }, where ext{(x_1, y_1)} are the coordinates of the point, and ext{A, B, C} are the coefficients from the line equation ext{Ax + By + C = 0}.

By substituting the point coordinates and the line coefficients into this formula, the distance is calculated. This distance is the length of the perpendicular segment from the point to the line, helping us find out how far the point is from the line directly and accurately.

Simplified Radical Form

Once we calculate the distance using the formula, the result is often an expression involving a square root. To make this easier to work with, it's important to simplify the radical.

In our exercise:

• The denominator is ext{\(\sqrt{20}\)} from the distance computation.
• Simplifying ext{\(\sqrt{20}\)} involves breaking it into factors: ext{\(\sqrt{4 \times 5}\)}.
• We know ext{\(\sqrt{4}\)} is 2, so the expression becomes ext{\(2\sqrt{5}\)}.

This step turns a somewhat complex square root into a simpler form, ext{\(2\sqrt{5}\)}, making it easier to perform further calculations. Simplifying radicals can greatly aid in solving and interpreting mathematical problems in analytical geometry.

Approximation to Nearest Tenth

To express the distance in a more practical way, we convert the simplified radical form into an approximation, especially useful when a decimal expression is needed.

Here's the process:

• Use the simplified form ext{\(\frac{8\sqrt{5}}{5}\)} from the calculation.
• First, approximate ext{\(\sqrt{5}\)} as 2.236, a common rounded value.
• Multiply: ext{8 \times 2.236} to get approximately 17.888.
• Finally, divide by 5: ext{\(\frac{17.888}{5} \approx 3.6\)}.

Thus, the distance of ext{\(\frac{8\sqrt{5}}{5}\)} is approximately 3.6 when expressed to the nearest tenth. This step is crucial if you need to communicate the distance in day-to-day language or practical situations.