Problem 66 - Calculus Early Transcendentals: Pearson New International Edition

Question

Use this result to find the distance from the given point to the given line. (3,-1) ; y=2 x-5

Step-by-Step Solution

Answer

The distance from the point to the line is \(\frac{2\sqrt{5}}{5}\) units.

Step 1: Identify the Equation of the Line

The given line is in slope-intercept form: \( y = 2x - 5 \). Here, the slope \( m \) is 2 and the y-intercept \( b \) is -5.

Step 2: Use the Distance Formula for a Point to a Line

The formula to find the distance from a point \((x_1, y_1)\) to a line \(Ax + By + C = 0\) is given by: \[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \] Convert the line equation \(y = 2x - 5\) to standard form \( -2x + y + 5 = 0 \). The coefficients are \( A = -2 \), \( B = 1 \), and \( C = 5 \).

Step 3: Substitute the Point into the Distance Formula

Substitute the point \((3, -1)\) and the coefficients \(A = -2\), \(B = 1\), and \(C = 5\) into the distance formula: \[ d = \frac{|-2(3) + 1(-1) + 5|}{\sqrt{(-2)^2 + (1)^2}} \] Calculate the numerator: \[ |-6 - 1 + 5| = |-2| = 2 \]

Step 4: Calculate the Denominator

Calculate the denominator: \[ \sqrt{(-2)^2 + (1)^2} = \sqrt{4 + 1} = \sqrt{5} \]

Step 5: Compute the Distance

Substitute the values back into the distance formula:\[ d = \frac{2}{\sqrt{5}} \] For a rationalized form, multiply by \(\frac{\sqrt{5}}{\sqrt{5}}\): \[ d = \frac{2\sqrt{5}}{5} \]

Step 6: Concluding Step: Interpret the Result

The distance from the point \((3, -1)\) to the line \(y = 2x - 5\) is \(\frac{2\sqrt{5}}{5}\) units.

Key Concepts

Distance Formula

The distance formula is a mathematical expression that helps us measure how far a point is from a line. For lines in the coordinate plane, the formula to calculate the distance from a point \((x_1, y_1)\) to a line defined by the equation \(Ax + By + C = 0\) is:

• \(d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}\)

Here, \(d\) represents the perpendicular distance from the point to the line. The numerator involves the absolute value of a linear combination of the point's coordinates and the line's constants. The denominator normalizes this expression, considering the line's slope. This formula helps in transforming the line equation into a usable form for distance computation. It's important to ensure the line equation is in its correct standard form before using the formula.

Slope-Intercept Form

The slope-intercept form is a way of expressing the equation of a line so you can easily identify its slope and y-intercept. It is written as:

• \(y = mx + b\)

In this formula, \(m\) is the slope of the line, and it tells you how steep the line is. The slope shows how much the \(y\)-value increases or decreases as the \(x\)-value increases by 1. The \(b\) is the y-intercept, the point where the line crosses the y-axis. This form is particularly useful for quickly graphing lines or when you need to convert into other forms, such as the standard form, to utilize formulas like the distance formula.

Standard Form of Line

The standard form of a line's equation is another way to express linear equations, often used for analysis and calculation. It is structured as:

• \(Ax + By + C = 0\)

With \(A\), \(B\), and \(C\) as constants, the equation is arranged to have both variables on the left and the constant term on the right. Converting a line from the slope-intercept form \(y = mx + b\) to this form means rearranging the terms to achieve:

• \(-mx + y - b = 0\)

It makes solving certain problems easier, such as finding the distance from a point to the line using the distance formula.

Rationalizing Fractions

Rationalizing fractions involves manipulating a fraction to eliminate any square roots from its denominator. This makes the expression simpler and easier to interpret. Consider the fraction:

• \(\frac{2}{\sqrt{5}}\)

To rationalize the denominator, multiply both the numerator and the denominator by the same square root:

• \(\frac{2}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} = \frac{2\sqrt{5}}{5}\)

This process is fundamental in mathematics as it allows expressions to convert to a more standardized form, enabling clear communication and simplification in further calculations or derivations.