Problem 37 - Algebra and Trigonometry Real Mathematics, Real People

Question

Eliminate the parameter and obtain the standard form of the rectangular equation. Line through \(\left(x_{1}, y_{1}\right)\) and \(\left(x_{2}, y_{2}\right)\): \(x=x_{1}+t\left(x_{2}-x_{1}\right)\) \(y=y_{1}+t\left(y_{2}-y_{1}\right)\)

Step-by-Step Solution

Answer

The standard form of the line passing through \(\left(x_{1}, y_{1}\right)\) and \(\left(x_{2}, y_{2}\right)\) is \(y - y_{1} = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}(x - x_{1})\). In other words the equation of line in standard form is obtained by eliminating parameter \( t \) from the given parametric equations.

Step 1: Solve for the Parameter t

First, isolate 't' in each equasion. For \(x = x_{1}+t(x_{2}-x_{1})\), we rearrange it to get \(t = \frac{x - x_{1}}{x_{2} - x_{1}}\). For \(y = y_{1}+t(y_{2}-y_{1})\), we rearrange it to get \(t = \frac{y - y_{1}}{y_{2}- y_{1}}\).

Step 2: Equate the Expressions for t

The two literal equations give us two expressions for 't'. Since 't' is the same in both equations, the expressions should be equal. Therefore, equate \(\frac{x - x_{1}}{x_{2} - x_{1}}\) and \(\frac{y - y_{1}}{y_{2}- y_{1}}\).

Step 3: Simplify to Obtain the Standard Form

After equating the expressions, simplify the equation by cross-multiplying and isolating y to one side of the equation. The standard form results in an equation y = mx + b, where m is the slope and b is the y-intercept of the line.

Key Concepts

Parameter Elimination

In mathematics, especially in the context of lines and curves, the process of parameter elimination helps us convert parametric equations into a single, more familiar form. This process is all about removing the parameter ‘t’ to obtain a rectangular equation. The original problem provides parametric equations, where each point on a line corresponds to a parameter 't':

• \(x = x_{1}+t(x_{2}-x_{1})\)
• \(y = y_{1}+t(y_{2}-y_{1})\)

To eliminate 't', you first solve each equation individually for 't'. For the x-equation, isolate 't': \(t = \frac{x - x_{1}}{x_{2} - x_{1}}\). Similarly, solve the y-equation: \(t = \frac{y - y_{1}}{y_{2} - y_{1}}\).Since 't' is the same in both expressions, equate them: \(\frac{x - x_{1}}{x_{2} - x_{1}} = \frac{y - y_{1}}{y_{2} - y_{1}}\). This vital step sets the stage for eliminating the parameter, leading us toward drafting the line's equation in a standard rectangular form.

Standard Form of a Line

When working with the familiar equation of a line, we often strive to present it in various forms for simplicity and convenience. One such format is the standard form, which appears as \(Ax + By = C\). In this state, A, B, and C are integers, and the equation allows easy interpretation and use in geometry. After equating and cross-multiplying the expressions derived from eliminating 't', you get a single equation. For instance, solving \(\frac{x - x_{1}}{x_{2} - x_{1}} = \frac{y - y_{1}}{y_{2} - y_{1}}\) involves multiplying both sides by the denominators to simplify the equation.The outcome is: \[(y_{2} - y_{1})(x - x_{1}) = (x_{2} - x_{1})(y - y_{1})\]Expand and rearrange this equation to derive the line's standard form. Ensure that A, B, and C are integers by multiplying through if necessary. The standard form emphasizes the relation between x and y, lending clarity to the line's geometric representation.

Slope-Intercept Form

The slope-intercept form is another friendly form of linear equations, appearing as \(y = mx + b\). This form is particularly useful because it clearly displays the slope \(m\) and the y-intercept \(b\), making it perfect for graphing. From the standard form, \(Ax + By = C\), you can easily transform the equation into slope-intercept form by solving for y.Steps to follow:

• Isolate the term containing y: subtract \(Ax\) from both sides to get \(By = -Ax + C\).
• Divide every term by B to solve for y: \(y = -\frac{A}{B}x + \frac{C}{B}\).

Here, the slope \(-\frac{A}{B}\) indicates the rate at which y changes relative to x, and \(\frac{C}{B}\) is where the line crosses the y-axis. This transformation makes it simpler to understand how a line behaves in a graph, while still relating back to the more comprehensive standard form.