Problem 72 - Precalculus

Question

Find the distance between the point and the line. $$\begin{array}{cc}\text{Point} && \text{Line} \\ (2,-3) && 4 x-5 y=-2\end{array}$$

Step-by-Step Solution

Answer

The distance between the given point (2,-3) and the line 4x - 5y + 2 = 0 is \(\frac{25\sqrt{41}}{41}\).

Step 1: Identify the Values from the Problem

For the given point (2,-3), take \(x_1 = 2\) and \(y_1 = -3\). For the standard form of the line equation 4x - 5y + 2 = 0, take \(A = 4\), \(B = -5\) and \(C = 2\).

Step 2: Substitution into the Distance Formula

Substitute the identified values from Step 1 into the distance formula. This gives the expression \[d = \frac{|4(2) - 5(-3) + 2|}{\sqrt{4^2 + (-5)^2}}\]. Simplify this expression to find the distance.

Step 3: Simplify the Expression

Upon simplifying the expression from Step 2, the calculation becomes \[d = \frac{|8 + 15 + 2|}{\sqrt{16 + 25}}\] which further simplifies to \[d = \frac{25}{\sqrt{41}}\].

Step 4: Simplify the Denominator

To simplify the denominator, take the square root of 41. So, the distance \(d\) between the point and the line is \[d = \frac{25}{\sqrt{41}} = \frac{25\sqrt{41}}{41}\].

Key Concepts

Distance Formula

To find the distance from a point to a line in a 2D coordinate system, we use the distance formula derived from algebra and geometry principles.
For a point \( (x_1, y_1) \) and a line expressed in the standard form \( Ax + By + C = 0 \), the distance \( d \) is computed using the formula:

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

Here's how it works:- **Numerator:** The absolute value \( |Ax_1 + By_1 + C| \) is used to ensure the distance is always a positive value, as distance can't be negative.- **Denominator:** \( \sqrt{A^2 + B^2} \) ensures the formula accounts for the line's orientation in the plane.
Using this formula, you can find how far any point is from a line, which is a fundamental aspect of geometry applications like analyzing graphs or optimizing paths.

Equations of Lines

Understanding line equations is crucial in calculating distances and their properties. Lines in a plane can be expressed in various forms:

• **Standard Form:** \( Ax + By + C = 0 \), where \( A, B, \) and \( C \) are constants.
• **Slope-Intercept Form:** \( y = mx + b \), where \( m \) is the slope and \( b \) is the y-intercept.
• **Point-Slope Form:** \( y - y_1 = m(x - x_1) \), suitable when you know a point on the line and the slope.

For finding the distance from a point to a line, the standard form works best because it splits the line’s geometric and algebraic properties effectively. In our example, the line is given by \( 4x - 5y = -2 \), transforming to \( 4x - 5y + 2 = 0 \) to fit the standard form.Knowing these forms helps you switch between them especially when fitting them into calculations for geometrical applications.

Simplification of Expressions

Simplifying expressions is a necessary step in solving problems efficiently. For instance, after substituting the values into the distance formula, you'll often end up with expressions that need simplification.Here's a simplified approach:- **Numerator:** Break down the arithmetic step by step. For example, evaluate the addition and subtraction within the absolute value first before dividing.- **Denominator:** Compute the square root and any powers present.- **Fractional Form:** Reduce the fraction if possible to simplify the result further.
In the example provided, simplifying \( |8 + 15 + 2| \) led to \( 25 \) in the numerator, while simplifying \( \sqrt{16 + 25} \) in the denominator gives \( \sqrt{41} \). This fraction \( \frac{25}{\sqrt{41}} \) can be further simplified by multiplying by \( \frac{\sqrt{41}}{\sqrt{41}} \) to rationalize the denominator, providing a neater final form.Correctly simplifying expressions not only gives you the right answers but also deepens your understanding of algebraic operations.