Problem 64 - Calculus Early Transcendentals: Pearson New International Edition

Question

Use this result to find the distance from the given point to the given line. (4,-1) ; 2 x-2 y+4=0

Step-by-Step Solution

Answer

\(\frac{7\sqrt{2}}{2}\)

Step 1: Identify the line equation

The equation of the line is given as \(2x - 2y + 4 = 0\). This is in the standard form \(Ax + By + C = 0\) where \(A = 2\), \(B = -2\), and \(C = 4\).

Step 2: Identify the point coordinates

The coordinates of the point from which we want to find the distance to the line are \((4, -1)\), meaning \(x_1 = 4\) and \(y_1 = -1\).

Step 3: Apply the distance formula

The distance \(d\) from a point \((x_1, y_1)\) to a line \(Ax + By + C = 0\) is given by: \[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]Substitute \(A = 2\), \(B = -2\), \(C = 4\), \(x_1 = 4\), and \(y_1 = -1\) into the formula:

Step 4: Substitute values into the formula

Calculate the numerator: \[|2(4) - 2(-1) + 4| = |8 + 2 + 4| = |14| = 14\]Calculate the denominator: \[\sqrt{2^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}\]

Step 5: Perform the final calculation

Plug the computed values back into the distance formula:\[ d = \frac{14}{2\sqrt{2}} = \frac{7}{\sqrt{2}} \]To rationalize the denominator, multiply the numerator and denominator by \(\sqrt{2}\):\[ d = \frac{7\sqrt{2}}{2} \]

Step 6: Present the final result

The distance from the point \((4, -1)\) to the line \(2x - 2y + 4 = 0\) is \(\frac{7\sqrt{2}}{2}\).

Key Concepts

Distance Formula

Finding the distance from a point to a line can seem challenging, but it's made much easier by using the distance formula. This formula helps us determine how far a point is from a straight line in a geometric context. For a point \(x_1, y_1\) and a line represented as \Ax + By + C = 0\, the formula is:

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

This formula might look complex at first glance, but breaking it down helps us see its straightforward nature. The numerator \|Ax_1 + By_1 + C|\ is the absolute value of plugging the point coordinates into the line equation. The denominator \sqrt{A^2 + B^2}\ accounts for the line's slope components, essentially normalizing the vector component that reflects the line direction.
By substituting the values of \(A\), \(B\), and \(C\) from the line equation, and the known point’s coordinates, you can efficiently find the shortest distance from the point to the line.

Line Equation

Understanding the line equation helps in applying mathematical concepts that involve points and straight lines. A line equation in the form \Ax + By + C = 0\ is known as the standard form of a linear equation.

• \(A\), \(B\) represent the coefficients of \(x\) and \(y\) respectively.
• \(C\) is the constant term.

In this form, any point \(x, y\) that satisfies the equation lies exactly on the line. This linear equation describes a straight path where each solution \(x, y\) pair is a point along that path. To find the distance of any exterior point to this line, we first identify these coefficients which will be key inputs for our distance formula.
This understanding enables us to manipulate the line equation further, like converting it to slope-intercept form if needed; however, for distance purposes, the standard form is all we need.

Rationalizing the Denominator

Rationalizing the denominator is a crucial step in simplifying the final form of our calculated distance. When you end up with a square root in the denominator, mathematicians prefer to have expressions without radicals for simplicity and clarity.
The process involves:

• Multiplying the numerator and denominator by the radical found in the denominator.
• In our example, \(\frac{14}{2\sqrt{2}}\), we multiply by \(\frac{\sqrt{2}}{\sqrt{2}}\).

This step results in the simplified form, \(\frac{7\sqrt{2}}{2}\), which is more legible and often easier for further operations. The rationale is to ensure that expressions meet conventional formats and are consistent for use in advancing mathematical models. By performing this step, you're effectively moving the radical to the numerator and clearing up the denominator for a neat appearance.