Problem 24 - Differential Equations and Linear Algebra

Question

Find the distance from the given point \(P\) to the given line \(L\). \(P(-6,4) ;\) Line \(L\) with equation \(x-y=3\)

Step-by-Step Solution

Answer

The distance from the point \(P(-6,4)\) to the line \(L\) with equation \(x-y=3\) is \( \frac{13\sqrt{2}}{2} \).

Step 1: Rewrite the line equation

Since we are given the line equation as \(x-y=3\), we can rewrite it as the general form of a line equation: \(Ax + By + C = 0\). In our case, we have \(A=1\), \(B=-1\), and \(C=-3\). So the general form of our line equation becomes \(1\cdot x - 1\cdot y - 3 = 0\).

Step 2: Apply the distance formula

To find the distance from the point P to the line L, we make use of the following distance formula: \( d = \frac{|Ax_{1} + By_{1} + C|}{\sqrt{A^2 + B^2}} \) Plugging in our values from the given point \(P(-6,4)\) and our line with coefficients \(A=1\), \(B=-1\), \(C=-3\), we get: \( d = \frac{|1\cdot(-6) - 1\cdot4 - 3|}{\sqrt{1^2 + (-1)^2}} \)

Step 3: Solve for the distance

Now, we can solve for the distance: \( d = \frac{|-6 - 4 - 3|}{\sqrt{1^2 + (-1)^2}} = \frac{|-13|}{\sqrt{2}} = \frac{13}{\sqrt{2}} \) Now, we can rationalize the denominator: \( d = \frac{13\sqrt{2}}{2} \) So the distance from the point \(P(-6,4)\) to the line \(L\) with equation \(x-y=3\) is \( \frac{13\sqrt{2}}{2} \).

Key Concepts

Distance Formula

The distance formula is a powerful mathematical tool that helps us determine the distance between a point and a line. It is given by: \[d = \frac{|Ax_{1} + By_{1} + C|}{\sqrt{A^2 + B^2}} \] This formula uses the coefficients from the line's equation in its general form: \(Ax + By + C = 0\), and the coordinates from the point \(P(x_1, y_1)\). Understanding how this formula works is essential:

• The numerator \(|Ax_{1} + By_{1} + C|\) calculates the absolute value of the linear expression for the given point, determining how far the point stands from the line in a direction perpendicular to it.
• The denominator \(\sqrt{A^2 + B^2}\) normalizes this distance in terms of the coefficients of the line, essentially giving the length of the perpendicular distance from the point to the line in a unitless form.

This formula is extremely handy since it allows us to efficiently calculate distances without needing the detailed trigonometric processes.

Line Equation

To use the distance formula effectively, we must first express our line in its general form, which takes the structure \(Ax + By + C = 0\). For instance, the line \(x-y=3\) should be rewritten to align with this form: \(1 \cdot x - 1 \cdot y - 3 = 0\), where \(A=1\), \(B=-1\), and \(C=-3\).

• This standardization is critical so the coefficients can be comfortably used with the distance formula.
• Once transformed into this form, it offers us a systematic way to plug into related equations and solve real-world geometry problems easier.

By converting line equations into this general form, mathematics becomes more accessible and calculations become a lot smoother, maintaining consistency across different problems.

Point-Line Distance

Determining the point-line distance involves calculating how far a given point is from a specified line. This type of problem typically engages the distance formula to achieve accurate results. If we use a point \(P(-6,4)\) and a line with equation \(x-y=3\):

• Substitute \(x = -6\) and \(y = 4\) into the absolute value expression \(|Ax_{1} + By_{1} + C|\). For our example, this becomes: \[ |-6 - 4 - 3| = |-13| = 13 \]
• Compute the denominator \(\sqrt{A^2 + B^2}\), which in this case is: \[ \sqrt{1^2 + (-1)^2} = \sqrt{2} \]

Upon plugging these into our formula, we derive the initial distance: \( \frac{13}{\sqrt{2}} \). This value represents the length of the shortest path a perpendicular line would take from the point to the original line.

Rationalizing the Denominator

Rationalizing the denominator is a helpful algebraic technique that involves adjusting the expression so that the denominator becomes a simple rational number. When we have a distance \( \frac{13}{\sqrt{2}} \), the radical \(\sqrt{2}\) in the denominator makes it less clean for precise applications.

• We multiply the numerator and the denominator by \(\sqrt{2}\) to eliminate the root from the denominator and maintain the value's equality: \[ \frac{13}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{13\sqrt{2}}{2} \]
• This results in \(\frac{13\sqrt{2}}{2}\), a neater form convenient for additional calculations or discussions.

Rationalizing aids in producing a straightforward representation, which builds a foundation for real-life problem solving and provides clearer insights during analysis.