What Are The $x$- And $y$-coordinates Of Point $P$ On The Directed Line Segment From $A$ To $B$ Such That $P$ Is $\frac{1}{3}$ The Length Of The Line Segment From

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What are the $x$- and $y$-coordinates of point $P$ on the directed line segment from $A$ to $B$ such that $P$ is $\frac{1}{3}$ the length of the line segment from $A$ to $B$?

Understanding the Problem

When dealing with directed line segments, it's essential to understand the concept of a point being a fraction of the total length of the line segment. In this case, we're looking for the coordinates of point $P$, which is $\frac{1}{3}$ the length of the line segment from $A$ to $B$. This means that point $P$ divides the line segment into three equal parts, with $A$ being one end and $B$ being the other end.

Mathematical Representation

To solve this problem, we can use the concept of vectors and the parametric equation of a line. 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 represent the directed line segment from $A$ to $B$ as a vector $\vec{AB} = \langle x_2 - x_1, y_2 - y_1 \rangle$.

Finding the Coordinates of Point $P$

Since point $P$ is $\frac{1}{3}$ the length of the line segment from $A$ to $B$, we can use the concept of scalar multiplication to find the coordinates of point $P$. Let's assume that the parameter $t$ represents the fraction of the line segment from $A$ to $B$ that point $P$ divides. In this case, $t = \frac{1}{3}$.

Using the parametric equation of a line, we can write the coordinates of point $P$ as:

xP=x1+t(x2βˆ’x1)x_P = x_1 + t(x_2 - x_1)

yP=y1+t(y2βˆ’y1)y_P = y_1 + t(y_2 - y_1)

Substituting $t = \frac{1}{3}$, we get:

xP=x1+13(x2βˆ’x1)x_P = x_1 + \frac{1}{3}(x_2 - x_1)

yP=y1+13(y2βˆ’y1)y_P = y_1 + \frac{1}{3}(y_2 - y_1)

Simplifying the Equations

We can simplify the equations by distributing the fraction $\frac{1}{3}$:

xP=x1+x2βˆ’x13x_P = x_1 + \frac{x_2 - x_1}{3}

yP=y1+y2βˆ’y13y_P = y_1 + \frac{y_2 - y_1}{3}

Further Simplification

We can further simplify the equations by multiplying the numerator and denominator by 3:

xP=x1+x2βˆ’x13x_P = x_1 + \frac{x_2 - x_1}{3}

yP=y1+y2βˆ’y13y_P = y_1 + \frac{y_2 - y_1}{3}

xP=3x1+x2βˆ’x13x_P = \frac{3x_1 + x_2 - x_1}{3}

yP=3y1+y2βˆ’y13y_P = \frac{3y_1 + y_2 - y_1}{3}

Final Simplification

We can finally simplify the equations by combining like terms:

xP=2x1+x23x_P = \frac{2x_1 + x_2}{3}

yP=2y1+y23y_P = \frac{2y_1 + y_2}{3}

Conclusion

In conclusion, the $x$- and $y$-coordinates of point $P$ on the directed line segment from $A$ to $B$ such that $P$ is $\frac{1}{3}$ the length of the line segment from $A$ to $B$ are given by the equations:

xP=2x1+x23x_P = \frac{2x_1 + x_2}{3}

yP=2y1+y23y_P = \frac{2y_1 + y_2}{3}

These equations provide a general solution for finding the coordinates of point $P$ on a directed line segment, given the coordinates of points $A$ and $B$.

Example

Let's consider an example to illustrate the concept. Suppose we have two points $A(2, 3)$ and $B(6, 9)$. We want to find the coordinates of point $P$, which is $\frac{1}{3}$ the length of the line segment from $A$ to $B$.

Using the equations we derived earlier, we get:

xP=2(2)+63=4+63=103x_P = \frac{2(2) + 6}{3} = \frac{4 + 6}{3} = \frac{10}{3}

yP=2(3)+93=6+93=153=5y_P = \frac{2(3) + 9}{3} = \frac{6 + 9}{3} = \frac{15}{3} = 5

Therefore, the coordinates of point $P$ are $(\frac{10}{3}, 5)$.

Applications

The concept of finding the coordinates of a point on a directed line segment has numerous applications in mathematics, physics, and engineering. Some of the applications include:

  • Geometry: Finding the coordinates of a point on a line segment is essential in geometry, where it is used to determine the length and midpoint of a line segment.
  • Physics: In physics, the concept of finding the coordinates of a point on a line segment is used to describe the motion of objects in two-dimensional space.
  • Engineering: In engineering, the concept of finding the coordinates of a point on a line segment is used to design and analyze mechanical systems, such as bridges and buildings.

Conclusion

In conclusion, the $x$- and $y$-coordinates of point $P$ on the directed line segment from $A$ to $B$ such that $P$ is $\frac{1}{3}$ the length of the line segment from $A$ to $B$ are given by the equations:

xP=2x1+x23x_P = \frac{2x_1 + x_2}{3}

yP=2y1+y23y_P = \frac{2y_1 + y_2}{3}

These equations provide a general solution for finding the coordinates of point $P$ on a directed line segment, given the coordinates of points $A$ and $B$. The concept of finding the coordinates of a point on a directed line segment has numerous applications in mathematics, physics, and engineering.
Q&A: Finding the Coordinates of Point $P$ on a Directed Line Segment

Q: What is the concept of finding the coordinates of a point on a directed line segment?

A: The concept of finding the coordinates of a point on a directed line segment is a fundamental idea in mathematics, where we want to determine the coordinates of a point $P$ on a line segment that is a fraction of the total length of the line segment.

Q: How do we represent the directed line segment from $A$ to $B$?

A: We can represent the directed line segment from $A$ to $B$ as a vector $\vec{AB} = \langle x_2 - x_1, y_2 - y_1 \rangle$, where $(x_1, y_1)$ are the coordinates of point $A$ and $(x_2, y_2)$ are the coordinates of point $B$.

Q: What is the parametric equation of a line?

A: The parametric equation of a line is a way of representing a line in terms of a parameter $t$. The parametric equation of a line can be written as:

x=x1+t(x2βˆ’x1)x = x_1 + t(x_2 - x_1)

y=y1+t(y2βˆ’y1)y = y_1 + t(y_2 - y_1)

Q: How do we find the coordinates of point $P$ on a directed line segment?

A: To find the coordinates of point $P$ on a directed line segment, we can use the parametric equation of a line and substitute the value of $t$, which represents the fraction of the line segment from $A$ to $B$ that point $P$ divides.

Q: What is the formula for finding the coordinates of point $P$ on a directed line segment?

A: The formula for finding the coordinates of point $P$ on a directed line segment is:

xP=2x1+x23x_P = \frac{2x_1 + x_2}{3}

yP=2y1+y23y_P = \frac{2y_1 + y_2}{3}

Q: What are some applications of finding the coordinates of a point on a directed line segment?

A: Some applications of finding the coordinates of a point on a directed line segment include:

  • Geometry: Finding the coordinates of a point on a line segment is essential in geometry, where it is used to determine the length and midpoint of a line segment.
  • Physics: In physics, the concept of finding the coordinates of a point on a line segment is used to describe the motion of objects in two-dimensional space.
  • Engineering: In engineering, the concept of finding the coordinates of a point on a line segment is used to design and analyze mechanical systems, such as bridges and buildings.

Q: Can you provide an example of finding the coordinates of a point on a directed line segment?

A: Let's consider an example. Suppose we have two points $A(2, 3)$ and $B(6, 9)$. We want to find the coordinates of point $P$, which is $\frac{1}{3}$ the length of the line segment from $A$ to $B$.

Using the formula we derived earlier, we get:

xP=2(2)+63=4+63=103x_P = \frac{2(2) + 6}{3} = \frac{4 + 6}{3} = \frac{10}{3}

yP=2(3)+93=6+93=153=5y_P = \frac{2(3) + 9}{3} = \frac{6 + 9}{3} = \frac{15}{3} = 5

Therefore, the coordinates of point $P$ are $(\frac{10}{3}, 5)$.

Q: What are some common mistakes to avoid when finding the coordinates of a point on a directed line segment?

A: Some common mistakes to avoid when finding the coordinates of a point on a directed line segment include:

  • Not using the correct formula: Make sure to use the correct formula for finding the coordinates of a point on a directed line segment.
  • Not substituting the correct values: Make sure to substitute the correct values for the coordinates of points $A$ and $B$.
  • Not simplifying the equations: Make sure to simplify the equations to get the final coordinates of point $P$.

Q: Can you provide some tips for finding the coordinates of a point on a directed line segment?

A: Here are some tips for finding the coordinates of a point on a directed line segment:

  • Read the problem carefully: Make sure to read the problem carefully and understand what is being asked.
  • Use the correct formula: Use the correct formula for finding the coordinates of a point on a directed line segment.
  • Substitute the correct values: Substitute the correct values for the coordinates of points $A$ and $B$.
  • Simplify the equations: Simplify the equations to get the final coordinates of point $P$.

Q: What are some real-world applications of finding the coordinates of a point on a directed line segment?

A: Some real-world applications of finding the coordinates of a point on a directed line segment include:

  • GPS navigation: GPS navigation systems use the concept of finding the coordinates of a point on a directed line segment to determine the location of a vehicle.
  • Surveying: Surveyors use the concept of finding the coordinates of a point on a directed line segment to determine the location of landmarks and boundaries.
  • Computer-aided design: Computer-aided design (CAD) software uses the concept of finding the coordinates of a point on a directed line segment to create and manipulate geometric shapes.

Conclusion

In conclusion, finding the coordinates of a point on a directed line segment is a fundamental concept in mathematics that has numerous applications in various fields. By understanding the concept and using the correct formula, we can find the coordinates of a point on a directed line segment and apply it to real-world problems.