What Is The \[$ X \$\]-coordinate Of The Point That Divides The Directed Line Segment From \[$ K \$\] To \[$ J \$\] Into A Ratio Of \[$ 1:3 \$\]?$\[ X = \left(\frac{m}{m+n}\right)\left(x_2-x_1\right)+x_1

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What is the { x $}$-coordinate of the point that divides the directed line segment from { K $}$ to { J $}$ into a ratio of { 1:3 $}$?

In geometry, a line segment is a part of a line that is bounded by two distinct end points. A directed line segment is a line segment with a specific direction, often represented by an arrow. When a line segment is divided into a ratio, it means that the line segment is divided into two parts, with the ratio of the lengths of the two parts being given. In this article, we will discuss how to find the { x $}$-coordinate of the point that divides a directed line segment from { K $}$ to { J $}$ into a ratio of { 1:3 $}$.

The section formula is a mathematical formula used to find the coordinates of a point that divides a line segment into a given ratio. The formula is given by:

x=(mm+n)(x2βˆ’x1)+x1x = \left(\frac{m}{m+n}\right)\left(x_2-x_1\right)+x_1

where { x $}$ is the { x $}$-coordinate of the point that divides the line segment, { m $}$ and { n $}$ are the ratios of the two parts of the line segment, and { x_1 $}$ and { x_2 $}$ are the { x $}$-coordinates of the two end points of the line segment.

To find the { x $}$-coordinate of the point that divides the directed line segment from { K $}$ to { J $}$ into a ratio of { 1:3 $}$, we can use the section formula. Let { x_1 $}$ be the { x $}$-coordinate of point { K $}$ and { x_2 $}$ be the { x $}$-coordinate of point { J $}$. The ratio of the two parts of the line segment is { 1:3 $}$, so we can let { m = 1 $}$ and { n = 3 $}$.

Calculating the { x $}$-coordinate

Substituting the values of { m $}$, { n $}$, { x_1 $}$, and { x_2 $}$ into the section formula, we get:

x=(11+3)(x2βˆ’x1)+x1x = \left(\frac{1}{1+3}\right)\left(x_2-x_1\right)+x_1

Simplifying the equation, we get:

x=(14)(x2βˆ’x1)+x1x = \left(\frac{1}{4}\right)\left(x_2-x_1\right)+x_1

Now, we can substitute the values of { x_1 $}$ and { x_2 $}$ into the equation. Let { x_1 = a $}$ and { x_2 = b $}$. Substituting these values into the equation, we get:

x=(14)(bβˆ’a)+ax = \left(\frac{1}{4}\right)\left(b-a\right)+a

Simplifying the equation, we get:

x=bβˆ’a4+ax = \frac{b-a}{4}+a

x=b4+3a4x = \frac{b}{4}+\frac{3a}{4}

In this article, we discussed how to find the { x $}$-coordinate of the point that divides a directed line segment from { K $}$ to { J $}$ into a ratio of { 1:3 $}$. We used the section formula to find the { x $}$-coordinate of the point that divides the line segment. The section formula is a mathematical formula used to find the coordinates of a point that divides a line segment into a given ratio. We also provided a step-by-step solution to the problem, using the section formula to find the { x $}$-coordinate of the point that divides the line segment.

Here are some example problems that you can try to practice the concept of finding the { x $}$-coordinate of the point that divides a directed line segment into a given ratio.

  • Find the { x $}$-coordinate of the point that divides the directed line segment from { A $}$ to { B $}$ into a ratio of { 2:3 $}$.
  • Find the { x $}$-coordinate of the point that divides the directed line segment from { C $}$ to { D $}$ into a ratio of { 3:2 $}$.
  • Find the { x $}$-coordinate of the point that divides the directed line segment from { E $}$ to { F $}$ into a ratio of { 1:4 $}$.

Here are the solutions to the example problems.

  • Find the { x $}$-coordinate of the point that divides the directed line segment from { A $}$ to { B $}$ into a ratio of { 2:3 $}$.

Let { x_1 $}$ be the { x $}$-coordinate of point { A $}$ and { x_2 $}$ be the { x $}$-coordinate of point { B $}$. The ratio of the two parts of the line segment is { 2:3 $}$, so we can let { m = 2 $}$ and { n = 3 $}$.

Substituting the values of { m $}$, { n $}$, { x_1 $}$, and { x_2 $}$ into the section formula, we get:

x=(22+3)(x2βˆ’x1)+x1x = \left(\frac{2}{2+3}\right)\left(x_2-x_1\right)+x_1

Simplifying the equation, we get:

x=(25)(x2βˆ’x1)+x1x = \left(\frac{2}{5}\right)\left(x_2-x_1\right)+x_1

Now, we can substitute the values of { x_1 $}$ and { x_2 $}$ into the equation. Let { x_1 = a $}$ and { x_2 = b $}$. Substituting these values into the equation, we get:

x=(25)(bβˆ’a)+ax = \left(\frac{2}{5}\right)\left(b-a\right)+a

Simplifying the equation, we get:

x=2b5βˆ’2a5+ax = \frac{2b}{5}-\frac{2a}{5}+a

x=2b5+3a5x = \frac{2b}{5}+\frac{3a}{5}

  • Find the { x $}$-coordinate of the point that divides the directed line segment from { C $}$ to { D $}$ into a ratio of { 3:2 $}$.

Let { x_1 $}$ be the { x $}$-coordinate of point { C $}$ and { x_2 $}$ be the { x $}$-coordinate of point { D $}$. The ratio of the two parts of the line segment is { 3:2 $}$, so we can let { m = 3 $}$ and { n = 2 $}$.

Substituting the values of { m $}$, { n $}$, { x_1 $}$, and { x_2 $}$ into the section formula, we get:

x=(33+2)(x2βˆ’x1)+x1x = \left(\frac{3}{3+2}\right)\left(x_2-x_1\right)+x_1

Simplifying the equation, we get:

x=(35)(x2βˆ’x1)+x1x = \left(\frac{3}{5}\right)\left(x_2-x_1\right)+x_1

Now, we can substitute the values of { x_1 $}$ and { x_2 $}$ into the equation. Let { x_1 = a $}$ and { x_2 = b $}$. Substituting these values into the equation, we get:

x=(35)(bβˆ’a)+ax = \left(\frac{3}{5}\right)\left(b-a\right)+a

Simplifying the equation, we get:

x=3b5βˆ’3a5+ax = \frac{3b}{5}-\frac{3a}{5}+a

x=3b5+2a5x = \frac{3b}{5}+\frac{2a}{5}

  • Find the { x $}$-coordinate of the point that divides the directed line segment from { E $}$ to { F $}$ into a ratio of { 1:4 $}$.

Let { x_1 $}$ be the { x $}$-coordinate of point { E $}$ and { x_2 $}$ be the { x $}$-coordinate of point { F $}$. The ratio of the two parts of the line segment is { 1:4 $}$, so we can let { m = 1 $}$ and { n = 4 $}$.

Substituting the values of { m $}$, { n $}$, { x_1 $}$,
Q&A: Finding the { x $}$-coordinate of the point that divides a directed line segment

A: The section formula is a mathematical formula used to find the coordinates of a point that divides a line segment into a given ratio. The formula is given by:

x=(mm+n)(x2βˆ’x1)+x1x = \left(\frac{m}{m+n}\right)\left(x_2-x_1\right)+x_1

where { x $}$ is the { x $}$-coordinate of the point that divides the line segment, { m $}$ and { n $}$ are the ratios of the two parts of the line segment, and { x_1 $}$ and { x_2 $}$ are the { x $}$-coordinates of the two end points of the line segment.

A: To apply the section formula, you need to follow these steps:

  1. Identify the ratios of the two parts of the line segment.
  2. Identify the { x $}$-coordinates of the two end points of the line segment.
  3. Substitute the values of the ratios and the { x $}$-coordinates into the section formula.
  4. Simplify the equation to find the { x $}$-coordinate of the point that divides the line segment.

A: The ratios of the two parts of the line segment are given by the values of { m $}$ and { n $}$ in the section formula. For example, if the ratio is { 2:3 $}$, then { m = 2 $}$ and { n = 3 $}$.

A: To find the { x $}$-coordinate of the point that divides the directed line segment from { A $}$ to { B $}$ into a ratio of { 2:3 $}$, you can use the section formula. Let { x_1 $}$ be the { x $}$-coordinate of point { A $}$ and { x_2 $}$ be the { x $}$-coordinate of point { B $}$. The ratio of the two parts of the line segment is { 2:3 $}$, so we can let { m = 2 $}$ and { n = 3 $}$.

Substituting the values of { m $}$, { n $}$, { x_1 $}$, and { x_2 $}$ into the section formula, we get:

x=(22+3)(x2βˆ’x1)+x1x = \left(\frac{2}{2+3}\right)\left(x_2-x_1\right)+x_1

Simplifying the equation, we get:

x=(25)(x2βˆ’x1)+x1x = \left(\frac{2}{5}\right)\left(x_2-x_1\right)+x_1

Now, we can substitute the values of { x_1 $}$ and { x_2 $}$ into the equation. Let { x_1 = a $}$ and { x_2 = b $}$. Substituting these values into the equation, we get:

x=(25)(bβˆ’a)+ax = \left(\frac{2}{5}\right)\left(b-a\right)+a

Simplifying the equation, we get:

x=2b5βˆ’2a5+ax = \frac{2b}{5}-\frac{2a}{5}+a

x=2b5+3a5x = \frac{2b}{5}+\frac{3a}{5}

A: Here are some common mistakes to avoid when using the section formula:

  • Make sure to identify the correct ratios of the two parts of the line segment.
  • Make sure to identify the correct { x $}$-coordinates of the two end points of the line segment.
  • Make sure to substitute the values of the ratios and the { x $}$-coordinates into the section formula correctly.
  • Make sure to simplify the equation correctly to find the { x $}$-coordinate of the point that divides the line segment.

A: Yes, you can use the section formula to find the { y $}$-coordinate of the point that divides a directed line segment. The section formula is given by:

y=(mm+n)(y2βˆ’y1)+y1y = \left(\frac{m}{m+n}\right)\left(y_2-y_1\right)+y_1

where { y $}$ is the { y $}$-coordinate of the point that divides the line segment, { m $}$ and { n $}$ are the ratios of the two parts of the line segment, and { y_1 $}$ and { y_2 $}$ are the { y $}$-coordinates of the two end points of the line segment.

A: Yes, you can use the section formula to find the coordinates of a point that divides a line segment into a ratio of { 1:1 $}$. In this case, the ratios of the two parts of the line segment are equal, so { m = n $}$. Substituting this value into the section formula, we get:

x=(11+1)(x2βˆ’x1)+x1x = \left(\frac{1}{1+1}\right)\left(x_2-x_1\right)+x_1

Simplifying the equation, we get:

x=(12)(x2βˆ’x1)+x1x = \left(\frac{1}{2}\right)\left(x_2-x_1\right)+x_1

Now, we can substitute the values of { x_1 $}$ and { x_2 $}$ into the equation. Let { x_1 = a $}$ and { x_2 = b $}$. Substituting these values into the equation, we get:

x=(12)(bβˆ’a)+ax = \left(\frac{1}{2}\right)\left(b-a\right)+a

Simplifying the equation, we get:

x=bβˆ’a2+ax = \frac{b-a}{2}+a

x=b2+a2x = \frac{b}{2}+\frac{a}{2}

This is the { x $}$-coordinate of the point that divides the line segment into a ratio of { 1:1 $}$.