Problem 52 - Precalculus with Limits

Question

In Exercises 51-58, find the distance between the point and the line. \(\textit{Point}\) \((0, 0)\) \(\textit{Line}\) \(2x - y = 4\)

Step-by-Step Solution

Answer

The shortest distance from the point (0,0) to the line \(2x - y = 4\) is \( \frac{4\sqrt{5}}{5} \)

Step 1: Reformat the line equation

We need to establish the equation of the line in the form Ax + By + C = 0. The given line is 2x - y = 4; so, our A, B and C values are 2, -1 and -4 respectively.

Step 2: Identify the point (x1, y1)

The point given is (0,0). Thus x1 is 0 and y1 is also 0.

Step 3: Substitute into the distance formula

We can now substitute all the identified values into the formula \( d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \). \n Plugging the values in, we get: \[d = \frac{| 2*0 - 1*0 - 4 |}{\sqrt{2^2 + (-1)^2}}\] which simplifies to: \[d = \frac{4}{\sqrt{5}}\]

Step 4: Simplification of the distance

It's always good practice to simplify expressions which include irrational numbers, like square roots. The distance can be simplified as: \[d = \frac{4}{\sqrt{5}} * \frac{\sqrt{5}}{\sqrt{5}} = \frac{4\sqrt{5}}{5}\]

Key Concepts

Point-Line Distance

When talking about the distance from a point to a line in coordinate geometry, we refer to the shortest route that connects the point to the line. Imagine you're standing at point (0, 0) and want to reach your friend who is fishing somewhere along the line described by the equation 2x - y = 4. You'd naturally take the shortest path, right? In mathematics, this translates to a perpendicular drop from the point onto the line.

The formula used to calculate this distance is given by: \[d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}\]This may look complex at first, but it's quite approachable when broken down.

• A, B, C: These values are coefficients from the line equation reformulated as Ax + By + C = 0.
• (x_1, y_1): These point coordinates provide x and y inputs for calculation.

Plugging in these values will unveil how far you are from the line. In our earlier problem, the calculated distance from the point (0, 0) to the line is \( \frac{4\sqrt{5}}{5} \) units.

Distance Calculation

The task of distance calculation in coordinate geometry is essential. It helps us place any point in relation to lines. In terms of computation, understanding each step within the formula is crucial.Here's how the distance is carefully calculated:

• Start by rewriting the linear equation in standard form (Ax + By + C = 0). This sets the stage for the calculation.
• Identify individual components that need to be plugged into the distance formula. Extract values of A, B, and C, as well as the point coordinates.
• Insert these values into the formula:
  \(d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}\)
  which measures the perpendicular distance.
• Simplify the expression by rationalizing the denominator, if necessary, to render a more interpretable result.

In our example, adhering these steps results in a final distance of \( \frac{4\sqrt{5}}{5} \) units, which is a precise assessment of the point's separation from the line.

Coordinate Geometry

Coordinate geometry is a mathematical system using a coordinate plane to explore the relationships between points and lines. This is extremely powerful as it turns spatial queries into manageable algebraic problems. One of its core applications includes determining distances using coordinates. The following concepts are instrumental:

• Points: Defined as pairs (x, y) indicating position on a plane.
• Lines: Expressed linearly with standard equations Ax + By + C = 0 .

Together, these elements allow us to explore distances. By leveraging the point-line distance formula, a host of complex geometric relationships can be unraveled. In our studied exercise, the ability to discern and compute the distance from (0,0) to 2x - y = 4 demonstrates this idea vividly. Now, it's not just about finding how far one thing is from another; it's about understanding spatial interactions within the coordinate plane.