Problem 22 - Differential Equations and Linear Algebra

Question

For Problems \(22-27,\) find the distance from the given point \(P\) to the given line \(L$$P(-8,0) ;\) Line \(L\) with equation \(y=3 x-4\)

Step-by-Step Solution

Answer

The distance from point P(-8, 0) to the line L with the equation y = 3x - 4 is 2\(\sqrt{10}\).

Step 1: Write the equation of the line in the form Ax + By + C = 0

The equation of the line L is given as y = 3x - 4. To write this in the form Ax + By + C = 0, we need to subtract 3x from both sides: -3x + y - 4 = 0 So, A = -3, B = 1, and C = -4.

Step 2: Substitute the coordinates of point P and the coefficients A, B, and C into the distance formula

Now we will substitute the values of A, B, C and the coordinates of point P (-8, 0) into the distance formula: Distance = \(\frac{|-3(-8) + 1(0) -4|}{\sqrt{(-3)^2 + 1^2}}\)

Step 3: Compute the distance

Next, we will compute the distance from point P to the line L: Distance = \(\frac{|24 - 4|}{\sqrt{9 + 1}}\) Distance = \(\frac{20}{\sqrt{10}}\) Distance = \(\frac{20}{\sqrt{10}} \cdot \frac {\sqrt{10}}{\sqrt{10}}\) (Rationalize the denominator) Distance = \(\frac{20\sqrt{10}}{10}\) Distance = 2\(\sqrt{10}\) So, the distance from point P(-8, 0) to the line L with the equation y = 3x - 4 is 2\(\sqrt{10}\).

Key Concepts

Distance Formula

The distance formula is a crucial tool in coordinate geometry. It helps find the shortest distance between a point and a line. In our exercise, this formula is used to see how far point P = (-8, 0) is from the line with equation y = 3x - 4.
To use the distance formula, the line must be expressed in the form Ax + By + C = 0, which is also known as the general form of a line. This allows us to plug the values into the formula as follows:

• Identify the coefficients A, B, and C from the line equation.
• Substitute A, B, C, and the coordinates of point P (x_0, y_0) into the formula:
• Distance = \( \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \)

This formula gives a straightforward way to determine how far the point is from the line by effectively calculating perpendicular distance, which is always the shortest path.

Coordinate Geometry

Coordinate geometry, or analytic geometry, involves placing geometric figures in a coordinate plane via algebraic equations. This exercise provides a solid example of how points and lines interact within this plane. Here,

• Point P = (-8, 0) is identified by its x and y coordinates, representing its position on the plane.
• The line y = 3x - 4 represents an infinite set of points forming a linear path.

To understand the distance from the point to the line, think about how a line extends in both directions from the coordinate origin, whereas a point is a fixed spot. Coordinate geometry aids in visualizing these relationships and computing precise values like distances using algebraic expressions.

Rationalizing Denominators

In mathematics, rationalizing denominators is a process used to eliminate radicals from the bottom of a fraction. A radical denominator makes it difficult to understand or work with expressions in a simplified form. Our task employs this process.
After calculating the initial distance \( \frac{20}{\sqrt{10}} \), the denominator is irrational. To rationalize it, the expression is multiplied by \( \frac{\sqrt{10}}{\sqrt{10}} \).

• This changes the denominator to a whole number, 10, since \( \sqrt{10} \cdot \sqrt{10} = 10 \).
• Consequently, the numerator becomes 20\sqrt{10}.

By rewriting fractions in this clear and concise manner, calculations become more manageable, aiding more accurate and understandable results in mathematics.