Problem 63 - Precalculus

Question

Find the distance between the point and the line. $$\begin{array}{cc}\text{Point} && \text{Line} \\ (-2,6) && y=-x+5\end{array}$$

Step-by-Step Solution

Answer

The distance between the point (-2,6) and the line \(y=-x+5\) is \(\sqrt{2}/2\).

Step 1: Convert the Line Equation to Standard Form

We convert the line equation \(y=-x+5\) to the standard form, Ax+By+C=0. When we move the x component to the left side of the equation, we get the equation \(x+y-5=0\).

Step 2: Identify A, B, C, x_1 and y_1

We identify A, B, and C from the line equation, \(x+y-5=0\). Therefore, A=1, B=1, and C=-5. Then, we identify \(x_1\) and \(y_1\), which are -2 and 6 respectively, from the given point (-2,6).

Step 3: Apply the Distance Formula

We apply the distance formula, \(d = \left| Ax_1+By_1+C \right|/\sqrt{A^2 + B^2}\). Subbing the values A=1, B=1, C=-5, \(x_1=-2\), and \(y_1=6\), we get \(d = \left| 1*(-2) + 1*6 - 5 \right|/\sqrt{(1)^2 + (1)^2}\). Simplifying this, \(d = \left| -1 \right|/\sqrt{2} = 1/\sqrt{2}\). To rationalize the denominator, we multiply by \(\sqrt{2}/\sqrt{2}\), which gives \(d = \sqrt{2}/2\).

Key Concepts

Standard Form of Line Equation

The standard form of a line equation is an essential format that helps in solving various geometric problems, including finding distances. It is written as \( Ax + By + C = 0 \), where \( A \), \( B \), and \( C \) are constants. To convert any linear equation into this form:

• Move all the terms to one side of the equation, ensuring zero is on the other side.
• Rearrange it to make sure the terms are in the order of \( Ax + By + C \).
• Usually, \( A \) should be a positive integer for consistency and ease of calculations.

Understanding this form simplifies the process of applying other mathematical formulas, such as the distance formula. In our example, starting with \( y = -x + 5 \), we moved the \( x \) term to acquire \( x + y - 5 = 0 \), thus identifying \( A = 1 \), \( B = 1 \), and \( C = -5 \). This step sets the stage for further calculations, ensuring all necessary components are clear and accessible.

Distance Formula

The distance formula is crucial for finding how far a specific point is from a line. It is derived from the geometric concept of perpendicular distance from a point to a line in the coordinate plane. The formula is expressed as:\[d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}\]Here's how you can use it:

• Identify the coefficients \( A \), \( B \), and \( C \) from the line's standard equation \( Ax + By + C = 0 \).
• Take the coordinates of the given point (\( x_1 \), \( y_1 \)).
• Plug these values into the formula and simplify.

In our case, the formula becomes \( d = \frac{|1(-2) + 1(6) - 5|}{\sqrt{1^2 + 1^2}} \), simplifying to \( \frac{1}{\sqrt{2}} \). This form gives a clear and concise distance. The distance formula is a versatile tool in geometry, making it easier to handle line and point relationships without auxiliary constructs like projections or intersections.

Rationalizing Denominators

Rationalizing the denominator is a technique used to eliminate radicals from the denominators in fractions, thus achieving a clearer and sometimes more aesthetically pleasing form.When you have a fraction like \( \frac{1}{\sqrt{2}} \), the goal is to make the denominator a whole number. This process involves:

• Multiplying both the numerator and the denominator by the radical present in the denominator.
• This process effectively squares the radical, thereby simplifying it into a non-radical form.

Applying this to \( \frac{1}{\sqrt{2}} \), you multiply by \( \frac{\sqrt{2}}{\sqrt{2}} \), leading to the simplified result \( \frac{\sqrt{2}}{2} \).Rationalized denominators often make further mathematical operations easier and results easier to interpret. It is a commonly adopted practice in mathematical expressions, particularly useful in educational settings and formal writing.