Lesson Explainer: Partitioning a Line Segment on the Coordinate Plane Mathematics

In this explainer, we will learn how to find the coordinates of a point that divides a line segment on the coordinate plane with a ratio using the section formula.

Let us first review some terminology.

Definition: Line Segment

A line segment is a part of a line bounded by two distinct endpoints.

We can represent the line segment between two distinct points, 𝐴 and 𝐡, using the notation 𝐴𝐡. 𝐴𝐡 contains all the points on the straight line between 𝐴 and 𝐡.

To help us understand this definition, we can consider a line segment 𝐴𝐡 drawn on the coordinate plane with endpoints 𝐴(βˆ’4,5) and 𝐡(2,βˆ’3).

The midpoint of a line segment is the middle point of the segment, the point that is equidistant between the two endpoints. We can find the coordinates of the midpoint of 𝐴𝐡 by halving each of the horizontal and vertical distances between 𝐴 and 𝐡.

Recap: The Midpoint of a Line Segment

We can find the midpoint, 𝑀, of a line segment between (π‘₯,𝑦) and (π‘₯,𝑦) using 𝑀=ο€Όπ‘₯+π‘₯2,𝑦+𝑦2.

We will now look at a variety of questions on dividing, or partitioning, line segments in a number of different ways.

Example 1: Dividing a Line Segment into Four Equal Parts

The coordinates of 𝐴 and 𝐡 are (1,9) and (9,9) respectively. Determine the coordinates of the points that divide 𝐴𝐡 into four equal parts.

Answer

We can begin by sketching the line segment 𝐴𝐡 and showing the points that divide it into 4 equal parts. We can define these points as 𝑃, 𝑀, and 𝑄.

As 𝐴𝐡 is divided into four equal parts, we can approach this question by firstly finding the midpoint, 𝑀, of 𝐴𝐡 and then finding the midpoints of 𝐴𝑀 and 𝑀𝐡.

We recall that the midpoint, 𝑀, of a line segment between coordinates (π‘₯,𝑦) and (π‘₯,𝑦) is given by 𝑀=ο€Όπ‘₯+π‘₯2,𝑦+𝑦2.

To find the midpoint, 𝑀, of 𝐴𝐡, we substitute the coordinates of 𝐴(1,9) for the (π‘₯,𝑦) values and the coordinates of 𝐡(9,9) for the (π‘₯,𝑦) values, giving 𝑀=ο€Ό1+92,9+92=ο€Ό102,182=(5,9). Thus, the coordinates of 𝑀 are (5,9).

Next, we find the midpoint, 𝑃, of 𝐴𝑀. Substituting the coordinates 𝐴(1,9) and 𝑀(5,9) for the (π‘₯,𝑦), and (π‘₯,𝑦) values, respectively, gives 𝑃=ο€Ό1+52,9+92=ο€Ό62,182=(3,9).

Finally, we find the midpoint, 𝑄, of 𝑀𝐡. Using the coordinates 𝑀(5,9) and 𝐡(9,9) gives 𝑃=ο€Ό5+92,9+92=ο€Ό142,182=(7,9).

Thus, we have found the coordinates of 𝑃, 𝑀, and 𝑄, which divide 𝐴𝐡 into 4 equal parts, as (3,9),(5,9),(7,9).

We will now look an example of how a line segment that has been partitioned by a point can be written in terms of a ratio.

Example 2: Finding the Ratio by Which a Point Divides a Line Segment

If 𝐢∈𝐴𝐡 and 𝐴𝐡=3οƒŸπΆπ΅, then 𝐢 divides 𝐡𝐴 by the ratio .

  1. 2∢1
  2. 1∢2
  3. 1∢3
  4. 3∢1

Answer

We consider that we have a line segment 𝐴𝐡. Somewhere along this line segment will be point 𝐢.

Since we need to take into consideration the direction of movement from 𝐴 to 𝐡, we use the vector 𝐴𝐡.

The movement from point 𝐢 to point 𝐡 is the vector οƒŸπΆπ΅.

The magnitudes of vectors 𝐴𝐡 and οƒŸπΆπ΅ are their lengths. We are given that 𝐴𝐡=3οƒŸπΆπ΅; therefore, we can write that lengthoflengthof𝐴𝐡=3×𝐢𝐡.

We can divide 𝐴𝐡 into 3 equal pieces.

However, we need to establish at which of these points 𝐢 will lie. If 𝐢 is closer to 𝐴 than 𝐡, then the length of 𝐢𝐡 would be two-thirds the length of 𝐴𝐡.

Thus, it would not be true that 𝐴𝐡=3οƒŸπΆπ΅. Therefore, 𝐢 must be at the point that is closer to 𝐡 than 𝐴.

In this way, οƒŸπΆπ΅=13𝐴𝐡 and 𝐴𝐡=3οƒŸπΆπ΅.

To find the ratio, as 𝐴𝐡 is divided into 3 parts, there will be 2 parts of the total in 𝐴𝐢 and 1 part in 𝐢𝐡.

We could write that 𝐢 divides 𝐴𝐡 in the ratio 2∢1. However, we were asked how 𝐢 divides 𝐡𝐴; therefore, the solution is the ratio given in answer option B: 1∢2.

We will now investigate how we can find the coordinates of a point on a line segment that splits the line into a given ratio.

Vectors can be useful when partitioning line segments in a ratio. Recall that vectors represent direction and magnitude, rather than position on a coordinate plane. Given two distinct points 𝐴 and 𝐡, vector 𝐴𝐡 tells us the relative direction of point 𝐡 with respect to point 𝐴, as well as the distance between the two points. In particular, 𝐴𝐡 does not have to begin at 𝐴 or end at 𝐡, as long as it has the same direction and magnitude. This flexibility of vectors is an advantage when we work with geometric problems such as partitioning a line.

Let us consider how to identify coordinate points. If point 𝑃 partitions 𝐴𝐡 in the ratio π‘šβˆΆπ‘›, this means that point 𝑃 lies on the line segment 𝐴𝐡 and the ratio of the magnitudes of the vectors satisfies β€–β€–οƒŸπ΄π‘ƒβ€–β€–βˆΆβ€–β€–οƒŸπ‘ƒπ΅β€–β€–=π‘šβˆΆπ‘›.

In other words, if β€–β€–οƒŸπ΄π‘ƒβ€–β€– is π‘š length units, ‖‖𝐴𝐡‖‖ would be equal to π‘š+𝑛 length units, which leads to β€–β€–οƒŸπ΄π‘ƒβ€–β€–=π‘šπ‘š+𝑛‖‖𝐴𝐡‖‖.

In general, it would be difficult to use only the equation above to find the coordinates of the partitioning point 𝑃. However, this equation does not contain the information that 𝑃 lies on the line segment 𝐴𝐡. In particular, this means that οƒŸπ΄π‘ƒ has the same direction as 𝐴𝐡. Recall that two vectors have the same direction if one vector is obtained by multiplying the other vector by some positive constant. Considering the equation above, we can see that this positive constant is given by π‘šπ‘š+𝑛. This leads to οƒŸπ΄π‘ƒ=π‘šπ‘š+𝑛𝐴𝐡.

We can use this property to find the coordinates of a point that partitions a directed line segment in a given ratio. To achieve this, we first write οƒŸπ΄π‘ƒ and 𝐴𝐡 each as a difference of two position vectors: οƒŸπ΄π‘ƒ=οƒŸπ‘‚π‘ƒβˆ’οƒ π‘‚π΄,𝐴𝐡=οƒŸπ‘‚π΅βˆ’οƒ π‘‚π΄.

Substituting these expressions into the original formula leads to οƒŸπ‘‚π‘ƒβˆ’οƒ π‘‚π΄=π‘šπ‘š+π‘›ο€ΊοƒŸπ‘‚π΅βˆ’οƒ π‘‚π΄ο†.

Rearranging the equation so that οƒŸπ‘‚π‘ƒ is the subject, we obtain οƒŸπ‘‚π‘ƒ=π‘šπ‘š+π‘›ο€ΊοƒŸπ‘‚π΅βˆ’οƒ π‘‚π΄ο†+οƒ π‘‚π΄οƒŸπ‘‚π‘ƒ=π‘šπ‘š+π‘›οƒŸπ‘‚π΅+ο€Όβˆ’π‘šπ‘š+𝑛+1οˆοƒ π‘‚π΄οƒŸπ‘‚π‘ƒ=π‘šπ‘š+π‘›οƒŸπ‘‚π΅+π‘›π‘š+𝑛𝑂𝐴.

Formula: Position Vector of a Point Partitioning a Line Segment by a Ratio

Let 𝑃 be a point on line segment 𝐴𝐡, partitioning it in the ratio π‘šβˆΆπ‘›. Then, the position vector οƒŸπ‘‚π‘ƒ is given by οƒŸπ‘‚π‘ƒ=π‘šπ‘š+π‘›οƒŸπ‘‚π΅+π‘›π‘š+𝑛𝑂𝐴.

Let us see how we can use this formula to obtain an expression for the Cartesian coordinates of the partitioning point. Let us denote the coordinates of the points 𝐴(π‘₯,𝑦) and 𝐡(π‘₯,𝑦). Then, we can write the corresponding position vectors 𝑂𝐴=(π‘₯,𝑦),οƒŸπ‘‚π΅=(π‘₯,𝑦).

Substituting these expressions into the formula above, we obtain οƒŸπ‘‚π‘ƒ=π‘šπ‘š+𝑛(π‘₯,𝑦)+π‘›π‘š+𝑛(π‘₯,𝑦)=ο€Όπ‘šπ‘š+𝑛π‘₯,π‘šπ‘š+π‘›π‘¦οˆ+ο€Όπ‘›π‘š+𝑛π‘₯,π‘›π‘š+π‘›π‘¦οˆ=ο€Όπ‘šπ‘š+𝑛π‘₯+π‘›π‘š+𝑛π‘₯,π‘šπ‘š+𝑛𝑦+π‘›π‘š+π‘›π‘¦οˆ=ο€Όπ‘šπ‘₯+𝑛π‘₯π‘š+𝑛,π‘šπ‘¦+π‘›π‘¦π‘š+π‘›οˆ.

We arrive at the following formula.

Theorem: The Section Formula

If we have distinct points 𝐴(π‘₯,𝑦) and 𝐡(π‘₯,𝑦) and the point π‘ƒβˆˆπ΄π΅ divides 𝐴𝐡 such that π΄π‘ƒβˆΆπ‘ƒπ΅=π‘šβˆΆπ‘›, then 𝑃 has the coordinates 𝑃=ο€Όπ‘šπ‘₯+𝑛π‘₯π‘š+𝑛,π‘šπ‘¦+π‘›π‘¦π‘š+π‘›οˆ.

We will now see how we can apply this formula in a few example questions.

Example 3: Finding the Coordinate That Divides a Line Segment Internally

If the coordinates of 𝐴 and 𝐡 are (5,5) and (βˆ’1,βˆ’4), respectively, find the coordinates of point 𝐢 that divides 𝐴𝐡 internally by the ratio 2∢1.

Answer

We can sketch this directed line segment 𝐴𝐡 as shown.

We can apply the following formula to find point 𝐢 that divides 𝐴𝐡 internally in the ratio 2∢1. This means that πΆβˆˆοƒ π΄π΅ and the ratio will be given as 𝐴𝐢∢𝐢𝐡=2∢1.

We can then apply the formula to partition a line segment in a given ratio.

If 𝐴(π‘₯,𝑦) and 𝐡(π‘₯,𝑦) and point 𝑃 divides 𝐴𝐡 such that π΄π‘ƒβˆΆπ‘ƒπ΅=π‘šβˆΆπ‘›, then 𝑃 has the coordinates 𝑃=ο€Όπ‘šπ‘₯+𝑛π‘₯π‘š+𝑛,π‘šπ‘¦+π‘›π‘¦π‘š+π‘›οˆ.

For our problem, 𝐴 has coordinates (5,5) and 𝐡 has coordinates (βˆ’1,βˆ’4). We can substitute these coordinates into the formula for (π‘₯,𝑦) and (π‘₯,𝑦) respectively.

The ratio values 2∢1 can be substituted for π‘š and 𝑛 respectively.

Therefore, we will have the coordinates of 𝐢 as 𝐢=ο€½2(βˆ’1)+1(5)2+1,2(βˆ’4)+1(5)2+1.

Simplifying, we have 𝐢=ο€Ό33,βˆ’33=(1,βˆ’1).

Thus, the coordinates of point 𝐢 that divides 𝐴𝐡 internally by the ratio 2∢1 are (1,βˆ’1).

We will now see how we can partition a line segment externally in a given ratio.

So far, we have observed how to identify the coordinates of a point that divides a line segment in a given ratio. We refer to such problems as internal partitioning problems since the point that we are looking for lies within the line segment.

Let us now consider a different type of problems, known as external division problems. In these problems, the point that divides the line segment does not lie within but rather on an extension of the line segment as shown on the diagram below.

Considering the diagram above, we say that point 𝑃 divides 𝐴𝐡 externally in the ratio π‘šβˆΆπ‘›, where π‘š>𝑛. We can solve such external division problems by slightly modifying our previous approach to internal partitioning problems. The main difference for this case is that οƒŸπ΄π‘ƒ, with magnitude π‘š length units, is the larger vector compared to οƒŸπ΅π‘ƒ, with magnitude 𝑛 length units. By subtraction, we can see that the length of 𝐴𝐡 is equal to π‘šβˆ’π‘› units. This leads to β€–β€–οƒŸπ΄π‘ƒβ€–β€–=π‘šπ‘šβˆ’π‘›β€–β€–οƒ π΄π΅β€–β€–.

As in the previous context, οƒŸπ΄π‘ƒ and 𝐴𝐡 have the same direction, so we can write οƒŸπ΄π‘ƒ=π‘šπ‘šβˆ’π‘›οƒ π΄π΅.

As we have done for the internal partitioning problem, we can calculate the formula for the position vector of 𝑃: οƒŸπ‘‚π‘ƒβˆ’οƒ π‘‚π΄=π‘šπ‘šβˆ’π‘›ο€ΊοƒŸπ‘‚π΅βˆ’οƒ π‘‚π΄ο†οƒŸπ‘‚π‘ƒ=π‘šπ‘šβˆ’π‘›ο€ΊοƒŸπ‘‚π΅βˆ’οƒ π‘‚π΄ο†+οƒ π‘‚π΄οƒŸπ‘‚π‘ƒ=π‘šπ‘šβˆ’π‘›οƒŸπ‘‚π΅+ο€»βˆ’π‘šπ‘šβˆ’π‘›+1ο‡οƒ π‘‚π΄οƒŸπ‘‚π‘ƒ=π‘šπ‘šβˆ’π‘›οƒŸπ‘‚π΅βˆ’π‘›π‘šβˆ’π‘›οƒ π‘‚π΄.

We note that this formula closely resembles the one obtained previously for internal division problems. The notable differences are as follows:

  • The two expressions are subtracted rather than added.
  • The π‘š+𝑛 expression is replaced above by π‘šβˆ’π‘›.

On the right-hand side above, the expression π‘š+𝑛 for internal partitioning is replaced by π‘šβˆ’π‘› for external partitioning. Let’s derive the formula for the Cartesian coordinates of the partitioning point, given the coordinates 𝐴(π‘₯,𝑦) and 𝐡(π‘₯,𝑦): οƒŸπ‘‚π‘ƒ=π‘šπ‘šβˆ’π‘›(π‘₯,𝑦)βˆ’π‘›π‘šβˆ’π‘›(π‘₯,𝑦)=ο€»π‘šπ‘šβˆ’π‘›π‘₯,π‘šπ‘šβˆ’π‘›π‘¦ο‡βˆ’ο€»π‘›π‘šβˆ’π‘›π‘₯,π‘›π‘šβˆ’π‘›π‘¦ο‡=ο€»π‘šπ‘šβˆ’π‘›π‘₯βˆ’π‘›π‘šβˆ’π‘›π‘₯,π‘šπ‘šβˆ’π‘›π‘¦βˆ’π‘›π‘šβˆ’π‘›π‘¦ο‡=ο€»π‘šπ‘₯βˆ’π‘›π‘₯π‘šβˆ’π‘›,π‘šπ‘¦βˆ’π‘›π‘¦π‘šβˆ’π‘›ο‡.

This leads to the following formula.

Theorem: The Section Formula with External Division

If we have distinct points 𝐴(π‘₯,𝑦) and 𝐡(π‘₯,𝑦) and point π‘ƒβˆ‰π΄π΅ divides 𝐴𝐡 such that π΄π‘ƒβˆΆπ‘ƒπ΅=π‘šβˆΆπ‘›, then 𝑃 has the coordinates 𝑃=ο€»π‘šπ‘₯βˆ’π‘›π‘₯π‘šβˆ’π‘›,π‘šπ‘¦βˆ’π‘›π‘¦π‘šβˆ’π‘›ο‡.

We will now see how we can apply this formula in the following example.

Example 4: Finding the Coordinates of a Point That Divides a Line Segment Externally into a Given Ratio

If 𝐴(3,βˆ’2) and 𝐡(βˆ’2,4), find in vector form the coordinates of point 𝐢 that divides 𝐴𝐡 externally in the ratio 4∢3.

Answer

We can begin by sketching the points 𝐴 and 𝐡 and extending the directed line segment 𝐴𝐡 to point 𝐢 that divides 𝐴𝐡 externally. We can write that 𝐴𝐢∢𝐢𝐡=4∢3.

We recall the section formula for external division.

If we have distinct points 𝐴(π‘₯,𝑦) and 𝐡(π‘₯,𝑦) and point π‘ƒβˆ‰π΄π΅ divides 𝐴𝐡 such that π΄π‘ƒβˆΆπ‘ƒπ΅=π‘šβˆΆπ‘›, then 𝑃 has the coordinates 𝑃=ο€»π‘šπ‘₯βˆ’π‘›π‘₯π‘šβˆ’π‘›,π‘šπ‘¦βˆ’π‘›π‘¦π‘šβˆ’π‘›ο‡.

We can substitute the values 𝐴(3,βˆ’2) and 𝐡(βˆ’2,4) for the (π‘₯,𝑦) and (π‘₯,𝑦) values, respectively, and the ratio values 4∢3 for π‘š and 𝑛 into the section formula to find the coordinates of 𝐢. This gives us 𝐢=ο€½4(βˆ’2)βˆ’3(3)4βˆ’3,4(4)βˆ’3(βˆ’2)4βˆ’3=ο€Όβˆ’8βˆ’91,16+61=(βˆ’17,22).

As we are asked to give our answer in vector form, we can give the position vector of 𝐢 as (βˆ’17,22).

We will now see an example of how we can use the section formula to find the ratio in which a line segment is divided.

Example 5: Finding the Ratio by Which the π‘₯-Axis Divides a Line Segment

Fill in the blank: Given that 𝐢(βˆ’3,3) and 𝐷(4,βˆ’2), the π‘₯-axis divides 𝐢𝐷 in the ratio .

  1. 3∢5
  2. 5∢3
  3. 2∢3
  4. 3∢2

Answer

We can begin by plotting the coordinates 𝐢 and 𝐷 and sketching the vector 𝐢𝐷.

In order to find how the π‘₯-axis divides 𝐢𝐷, we first need to find the point where 𝐢𝐷 crosses the π‘₯-axis. Given the coordinates of 𝐢 and 𝐷, we can find the equation of 𝐢𝐷, beginning by finding the slope of this line.

The slope, π‘š, of a line joining two points (π‘₯,𝑦) and (π‘₯,𝑦) can be found using π‘š=π‘¦βˆ’π‘¦π‘₯βˆ’π‘₯.

Therefore, the slope between 𝐢(βˆ’3,3) and 𝐷(4,βˆ’2) is given by π‘š=βˆ’2βˆ’34βˆ’(βˆ’3)=βˆ’57=βˆ’57.

We can then use the point–slope form of a line such that, given a point (π‘₯,𝑦) and the slope, π‘š, we can write the equation of the line as π‘¦βˆ’π‘¦=π‘š(π‘₯βˆ’π‘₯).

We can substitute the coordinates of either 𝐢 or 𝐷 into this form, so using 𝐢(βˆ’3,3) for the (π‘₯,𝑦) values and π‘š=βˆ’57, we have π‘¦βˆ’3=βˆ’57(π‘₯βˆ’(βˆ’3))π‘¦βˆ’3=βˆ’57(π‘₯+3).

We can then multiply both sides by 7 and expand the parentheses on the right-hand side, giving 7π‘¦βˆ’21=βˆ’5(π‘₯+3)7π‘¦βˆ’21=βˆ’5π‘₯βˆ’15.

Rearranging to write this in the general form of the equation of a line, π‘Žπ‘₯+𝑏𝑦+𝑐=0, we have 5π‘₯+7π‘¦βˆ’6=0.

We recall that a line crosses the π‘₯-axis when 𝑦=0, so substituting this into the equation 5π‘₯+7π‘¦βˆ’6=0 and simplifying gives 5π‘₯+7(0)βˆ’6=0,5π‘₯βˆ’6=0,5π‘₯=6,π‘₯=65.

We have now calculated that the line segment 𝐢𝐷 crosses the π‘₯-axis at the point ο€Ό65,0.

We now need to find the ratio by which this coordinate, ο€Ό65,0, divides 𝐢𝐷.

To do this, we can use the section formula for internal division of a line segment. If we have distinct points 𝐴(π‘₯,𝑦) and 𝐡(π‘₯,𝑦) and point π‘ƒβˆˆπ΄π΅ divides 𝐴𝐡 such that π΄π‘ƒβˆΆπ‘ƒπ΅=π‘šβˆΆπ‘›, then 𝑃 has the coordinates 𝑃=ο€Όπ‘šπ‘₯+𝑛π‘₯π‘š+𝑛,π‘šπ‘¦+π‘›π‘¦π‘š+π‘›οˆ.

In this question, we know points 𝐢(βˆ’3,3) and 𝐷(4,βˆ’2) and point 𝑃65,0 that divides 𝐢𝐷. We need to calculate the ratio values of π‘š and 𝑛.

Substituting 𝐢(βˆ’3,3) and 𝐷(4,βˆ’2) for the (π‘₯,𝑦) and (π‘₯,𝑦) values, respectively, into the section formula, we have 𝑃=ο€½π‘š(4)+𝑛(βˆ’3)π‘š+𝑛,π‘š(βˆ’2)+𝑛(3)π‘š+𝑛=ο€Ό4π‘šβˆ’3π‘›π‘š+𝑛,βˆ’2π‘š+3π‘›π‘š+π‘›οˆ.

We know that 𝑃 has coordinates ο€Ό65,0, so we can write ο€Ό65,0=ο€Ό4π‘šβˆ’3π‘›π‘š+𝑛,βˆ’2π‘š+3π‘›π‘š+π‘›οˆ.

Evaluating the π‘₯-coordinates, we have 65=4π‘šβˆ’3π‘›π‘š+𝑛.

We can cross multiply and simplify to write an expression for π‘š in terms of 𝑛 as 5(4π‘šβˆ’3𝑛)=6(π‘š+𝑛)20π‘šβˆ’15𝑛=6π‘š+6𝑛14π‘š=21π‘›π‘šπ‘›=2114π‘šπ‘›=32.

The ratio of πΆπ‘ƒβˆΆπ‘ƒπ·=π‘šβˆΆπ‘›, so πΆπ‘ƒβˆΆπ‘ƒπ·=3∢2.

Thus, we can give the answer that the π‘₯-axis divides 𝐢𝐷 in the ratio 3∢2.

As a check of our answer, we could find the distance of 𝐢𝑃 and the distance of 𝑃𝐷 and find the ratio of πΆπ‘ƒβˆΆπ‘ƒπ· directly.

We recall that the distance formula for finding the distance, 𝑑, between two points (π‘₯,𝑦) and (π‘₯,𝑦) is given by 𝑑=(π‘₯βˆ’π‘₯)+(π‘¦βˆ’π‘¦).

To find the length of 𝐢𝑃, β€–β€–οƒŸπΆπ‘ƒβ€–β€–, we substitute the values of 𝐢(βˆ’3,3) and 𝑃65,0 for the (π‘₯,𝑦) and (π‘₯,𝑦) values to give β€–β€–οƒŸπΆπ‘ƒβ€–β€–=ο„Ÿο€Ό65βˆ’(βˆ’3)+(0βˆ’3)=ο„Ÿο€Ό215+3=ο„ž66625=3√745.

To find the length of 𝑃𝐷, β€–β€–οƒŸπ‘ƒπ·β€–β€–, we substitute 𝑃65,0 and 𝐷(4,βˆ’2) for the (π‘₯,𝑦) and (π‘₯,𝑦) values, which gives β€–β€–οƒŸπ‘ƒπ·β€–β€–=ο„Ÿο€Ό4βˆ’65+(βˆ’2βˆ’0)=ο„Ÿο€Ό145+(βˆ’2)=ο„ž29625=2√745.

We can then write the ratio of πΆπ‘ƒβˆΆπ‘ƒπ· as πΆπ‘ƒβˆΆπ‘ƒπ·=3√745∢2√745.

Multiplying both sides of the ratio by 5, and then dividing by √74, gives πΆπ‘ƒβˆΆπ‘ƒπ·=3√74∢2√74=3∢2.

Thus, we have confirmed our answer, that the π‘₯-axis divides 𝐢𝐷 by the ratio 3∢2.

In the next example, we can see a more complex problem involving the partitioning of a line segment.

Example 6: Solving a Word Problem by Dividing a Line Segment

A bus is traveling from city 𝐴(10,βˆ’10) to city 𝐡(βˆ’8,8). Its first stop is at 𝐢, which is halfway between the cities. Its second stop is at 𝐷, which is two-thirds of the way from 𝐴 to 𝐡. What are the coordinates of 𝐢 and 𝐷?

Answer

We are given that city 𝐴 has coordinates (10,βˆ’10) and city 𝐡 has coordinates (βˆ’8,8). Firstly, we need to find the coordinates of city 𝐢, halfway between these.

We can make use of the formula for the midpoint of a line. To find the midpoint, 𝑀, of a line segment between two points (π‘₯,𝑦) and (π‘₯,𝑦) we can use 𝑀=ο€Όπ‘₯+π‘₯2,𝑦+𝑦2.

Substituting (π‘₯,𝑦)=(10,βˆ’10) and (π‘₯,𝑦)=(βˆ’8,8) into this formula gives the midpoint, 𝐢, as 𝐢=ο€½10+(βˆ’8)2,βˆ’10+82=ο€Ό22,βˆ’22=(1,βˆ’1).

Next, we need to find the coordinates of 𝐷, which is two-thirds of the way from 𝐴 to 𝐡. The direction, 𝐴 to 𝐡, is important as it indicates the position of 𝐷. 𝐷 will be closer to 𝐡 than 𝐴. We can think of the position by dividing 𝐴𝐡 into 3 equal pieces. We can write the ratio 𝐴𝐷∢𝐷𝐡 as 2∢1.

We can use the formula to partition a line segment in a given ratio. If 𝐴(π‘₯,𝑦) and 𝐡(π‘₯,𝑦) and point 𝑃 divides 𝐴𝐡 such that π΄π‘ƒβˆΆπ‘ƒπ΅=π‘šβˆΆπ‘›, then 𝑃 has the coordinates 𝑃=ο€Όπ‘šπ‘₯+𝑛π‘₯π‘š+𝑛,π‘šπ‘¦+π‘›π‘¦π‘š+π‘›οˆ.

In this question, we have 𝐴(10,βˆ’10), 𝐡(βˆ’8,8), and point 𝐷 that divides 𝐴𝐡 in the ratio 2∢1. Substituting these into the formula to find 𝐷 gives 𝐷=ο€½2(βˆ’8)+1(10)2+1,2(8)+1(βˆ’10)2+1=ο€Όβˆ’63,63=(βˆ’2,2).

As a useful check of our answer, we can consider the lengths of 𝐴𝐡 and 𝐴𝐷 by applying the distance formula. To find the distance, 𝑑, between two points (π‘₯,𝑦) and (π‘₯,𝑦), we calculate 𝑑=(π‘₯βˆ’π‘₯)+(π‘¦βˆ’π‘¦).

To calculate the length of 𝐴𝐷, we can substitute the coordinates 𝐴(10,βˆ’10) and 𝐷(βˆ’2,2) into the formula to give 𝑑=(βˆ’2βˆ’10)+(2βˆ’(βˆ’10))=(βˆ’12)+12=√288=12√2.

To calculate the length of 𝐴𝐡, we substitute 𝐴(10,βˆ’10) and 𝐡(βˆ’8,8) into the distance formula, giving 𝑑=(βˆ’8βˆ’10)+(8βˆ’(βˆ’10))=(βˆ’18)+18=√648=18√2.

Thus, we can write the ratio of lengths 𝐴𝐷∢𝐴𝐡 as 𝐴𝐷∢𝐴𝐡=12√2∢18√2=12∢18=2∢3.

We were given that 𝐷 is two-thirds of the way from 𝐴 to 𝐡. Therefore, we have confirmed that 𝐷 has the coordinates (βˆ’2,2).

We can give the answer that the coordinates of 𝐢 and 𝐷 are (1,βˆ’1)(βˆ’2,2).and

Key Points

  • The midpoint, 𝑀, of the line segment between (π‘₯,𝑦) and (π‘₯,𝑦) is given by 𝑀=ο€Όπ‘₯+π‘₯2,𝑦+𝑦2.
  • If we have distinct points 𝐴(π‘₯,𝑦) and 𝐡(π‘₯,𝑦) and point 𝑃 divides 𝐴𝐡 internally such that π΄π‘ƒβˆΆπ‘ƒπ΅=π‘šβˆΆπ‘›, then 𝑃 has the coordinates 𝑃=ο€Όπ‘šπ‘₯+𝑛π‘₯π‘š+𝑛,π‘šπ‘¦+π‘›π‘¦π‘š+π‘›οˆ.
  • When answering problems involving the partitioning of a line segment, we need to be careful to establish the correct order of the ratio. If 𝐴𝐡 is split by point 𝑃 in the ratio π‘šβˆΆπ‘›, then the ratio of π΄π‘ƒβˆΆπ‘ƒπ΅ will be π‘šβˆΆπ‘›. If 𝐡𝐴 is split by point 𝑃 in the ratio π‘šβˆΆπ‘›, then the ratio of π΄π‘ƒβˆΆπ‘ƒπ΅ will instead be π‘›βˆΆπ‘š.
  • If we have distinct points 𝐴(π‘₯,𝑦) and 𝐡(π‘₯,𝑦) and point π‘ƒβˆ‰π΄π΅ divides 𝐴𝐡 externally such that π΄π‘ƒβˆΆπ‘ƒπ΅=π‘šβˆΆπ‘›, then 𝑃 has the coordinates 𝑃=ο€»π‘šπ‘₯βˆ’π‘›π‘₯π‘šβˆ’π‘›,π‘šπ‘¦βˆ’π‘›π‘¦π‘šβˆ’π‘›ο‡.

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