Parametric Equation of a Line

Oct 13, 2025 by recoveryshake.com 29 views

Introduction

When dealing with directed line segments, it's often necessary to find the coordinates of a point that lies on the segment. This can be a challenging task, especially when the point is a fraction of the total length of the line. In this article, we'll explore how to find the ${$ x $}$- and ${$ y $}$-coordinates of a point ${$ P $}$ on a directed line segment from ${$ A $}$ to ${$ B $}$ such that ${$ P $}$ is ${$ \frac{2}{3}$}$ the length of the line.

Understanding the Problem

To solve this problem, we need to understand the concept of a directed line segment and how to find the coordinates of a point on the segment. A directed line segment is a line segment with a specific direction, which is often represented by an arrow. The coordinates of a point on the line segment can be found using the parametric equation of the line.

Parametric Equation of a Line

The parametric equation of a line is given by:

${$ x = x_1 + t(x_2 - x_1) $}{$ y = y_1 + t(y_2 - y_1) $}$

where ${$ (x_1, y_1) $}$ is the starting point of the line, ${$ (x_2, y_2) $}$ is the ending point of the line, and ${$ t $}$ is a parameter that varies from 0 to 1.

Finding the Coordinates of Point ${$ P $}$

To find the coordinates of point ${$ P $}$, we need to find the value of ${$ t $}$ such that ${$ P $}$ is ${$ \frac{2}{3}$}$ the length of the line. We can do this by using the distance formula:

${$ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $}$

where ${$ d $}$ is the distance between the two points.

Calculating the Distance

Let's assume that the coordinates of point ${$ A $}$ are ${$ (x_1, y_1) $}$ and the coordinates of point ${$ B $}$ are ${$ (x_2, y_2) $}$. We can calculate the distance between the two points using the distance formula:

${$ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $}$

Finding the Value of ${$ t $}$

To find the value of ${$ t $}$, we need to set up an equation using the distance formula and the fact that ${$ P $}$ is ${$ \frac{2}{3}$}$ the length of the line. We can do this by substituting the value of ${$ d $}$ into the equation:

${$ \frac{2}{3}d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $}$

Solving for ${$ t $}$

To solve for ${$ t $}$, we can square both sides of the equation and simplify:

${$ \left(\frac{2}{3}\right)^2d2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 $}$

${$ \frac{4}{9}d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 $}$

Substituting the Value of ${$ d $}$

We can substitute the value of ${$ d $}$ into the equation:

${$ \frac{4}{9}\left(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\right)2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 $}$

Simplifying the Equation

We can simplify the equation by canceling out the squared terms:

${$ \frac{4}{9}(x_2 - x_1)^2 + (y_2 - y_1)^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 $}$

Solving for ${$ t $}$

To solve for ${$ t $}$, we can divide both sides of the equation by ${$ (x_2 - x_1)^2 + (y_2 - y_1)^2 $}$:

${$ \frac{4}{9} = 1 $}$

Finding the Value of ${$ t $}$

We can find the value of ${$ t $}$ by substituting the value of ${$ d $}$ into the equation:

${$ t = \frac{2}{3} $}$

Finding the Coordinates of Point ${$ P $}$

Now that we have the value of ${$ t $}$, we can find the coordinates of point ${$ P $}$ using the parametric equation of the line:

${$ x = x_1 + t(x_2 - x_1) $}{$ y = y_1 + t(y_2 - y_1) $}$

Substituting the Value of ${$ t $}$

We can substitute the value of ${$ t $}$ into the equation:

${$ x = x_1 + \frac{2}{3}(x_2 - x_1) $}{$ y = y_1 + \frac{2}{3}(y_2 - y_1) $}$

Simplifying the Equation

We can simplify the equation by multiplying out the terms:

${$ x = \frac{1}{3}x_1 + \frac{2}{3}x_2 $}{$ y = \frac{1}{3}y_1 + \frac{2}{3}y_2 $}$

Conclusion

In this article, we've explored how to find the ${$ x $}$- and ${$ y $}$-coordinates of a point ${$ P $}$ on a directed line segment from ${$ A $}$ to ${$ B $}$ such that ${$ P $}$ is ${$ \frac{2}{3}$}$ the length of the line. We've used the parametric equation of the line and the distance formula to find the value of ${$ t $}$ and the coordinates of point ${$ P $}$. This problem is a great example of how to use mathematical concepts to solve real-world problems.

Example

Let's assume that the coordinates of point ${$ A $}$ are ${$ (1, 2) $}$ and the coordinates of point ${$ B $}$ are ${$ (4, 6) $}$. We can use the parametric equation of the line to find the coordinates of point ${$ P $}$:

${$ x = 1 + \frac{2}{3}(4 - 1) $}{$ y = 2 + \frac{2}{3}(6 - 2) $}$

${$ x = 1 + \frac{2}{3}(3) $}{$ y = 2 + \frac{2}{3}(4) $}$

${$ x = 1 + 2 $}{$ y = 2 + \frac{8}{3} $}$

${$ x = 3 $}{$ y = \frac{14}{3} $}$

Therefore, the coordinates of point ${$ P $}$ are ${$ (3, \frac{14}{3}) $}$.

Final Answer

The final answer is ${$ (3, \frac{14}{3}) $}$.

Introduction

In our previous article, we explored how to find the ${$ x $}$- and ${$ y $}$-coordinates of a point ${$ P $}$ on a directed line segment from ${$ A $}$ to ${$ B $}$ such that ${$ P $}$ is ${$ \frac{2}{3}$}$ the length of the line. In this article, we'll answer some common questions related to this topic.

Q: What is the parametric equation of a line?

A: The parametric equation of a line is given by:

${$ x = x_1 + t(x_2 - x_1) $}{$ y = y_1 + t(y_2 - y_1) $}$

where ${$ (x_1, y_1) $}$ is the starting point of the line, ${$ (x_2, y_2) $}$ is the ending point of the line, and ${$ t $}$ is a parameter that varies from 0 to 1.

Q: How do I find the value of ${$ t $}$?

A: To find the value of ${$ t $}$, you need to set up an equation using the distance formula and the fact that ${$ P $}$ is ${$ \frac{2}{3}$}$ the length of the line. You can do this by substituting the value of ${$ d $}$ into the equation:

${$ \frac{2}{3}d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $}$

Q: What is the distance formula?

A: The distance formula is given by:

${$ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $}$

where ${$ d $}$ is the distance between the two points.

Q: How do I find the coordinates of point ${$ P $}$?

A: To find the coordinates of point ${$ P $}$, you need to substitute the value of ${$ t $}$ into the parametric equation of the line:

${$ x = x_1 + t(x_2 - x_1) $}{$ y = y_1 + t(y_2 - y_1) $}$

Q: What if the value of ${$ t $}$ is not a simple fraction?

A: If the value of ${$ t $}$ is not a simple fraction, you can use a calculator or a computer program to find the coordinates of point ${$ P $}$. Alternatively, you can use the parametric equation of the line to find the coordinates of point ${$ P $}$.

Q: Can I use this method to find the coordinates of a point on a line segment that is not ${$ \frac{2}{3}$}$ the length of the line?

A: Yes, you can use this method to find the coordinates of a point on a line segment that is not ${$ \frac{2}{3}$}$ the length of the line. Simply substitute the value of ${$ d $}$ into the equation and solve for ${$ t $}$.

Q: What if the line segment is vertical or horizontal?

A: If the line segment is vertical or horizontal, you can use the parametric equation of the line to find the coordinates of point ${$ P $}$. For a vertical line segment, the equation is:

${$ x = x_1 + t(x_2 - x_1) $}{$ y = y_1 $}$

For a horizontal line segment, the equation is:

${$ x = x_1 $}{$ y = y_1 + t(y_2 - y_1) $}$

Q: Can I use this method to find the coordinates of a point on a line segment that is not a straight line?

A: No, this method is only applicable to straight line segments. If the line segment is not a straight line, you will need to use a different method to find the coordinates of point ${$ P $}$.

Conclusion

In this article, we've answered some common questions related to finding the ${$ x $}$- and ${$ y $}$-coordinates of a point ${$ P $}$ on a directed line segment. We've also provided some examples and explanations to help you understand the concept. If you have any further questions, feel free to ask.

Example

${$ x = 1 + \frac{2}{3}(4 - 1) $}{$ y = 2 + \frac{2}{3}(6 - 2) $}$

${$ x = 1 + \frac{2}{3}(3) $}{$ y = 2 + \frac{2}{3}(4) $}$

${$ x = 1 + 2 $}{$ y = 2 + \frac{8}{3} $}$

${$ x = 3 $}{$ y = \frac{14}{3} $}$

Therefore, the coordinates of point ${$ P $}$ are ${$ (3, \frac{14}{3}) $}$.

Final Answer

The final answer is ${$ (3, \frac{14}{3}) $}$.