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Partitioning A Line Segment Formula

Written By Sunday, October 30, 2022 Add Comment Edit

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

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

Permit us kickoff review some terminology.

Definition: Line Segment

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

We can represent the line segment between two distinct points, 𝐴 and 𝐡 , using the note 𝐴 𝐡 . 𝐴 𝐡 contains all the points on the direct line between 𝐴 and 𝐡 .

To help united states of america empathize this definition, we can consider a line segment 𝐴 𝐡 drawn on the coordinate aeroplane with endpoints 𝐴 ( − 4 , 5 ) and 𝐡 ( 2 , − iii ) .

The midpoint of a line segment is the center bespeak of the segment, the bespeak that is equidistant between the 2 endpoints. We can find the coordinates of the midpoint of 𝐴 𝐡 by halving each of the horizontal and vertical distances between 𝐴 and 𝐡 .

Epitomize: The Midpoint of a Line Segment

We can find the midpoint, 𝑀 , of a line segment between ( π‘₯ , 𝑦 ) and ( π‘₯ , 𝑦 ) using 𝑀 = π‘₯ + π‘₯ 2 , 𝑦 + 𝑦 2 .

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 , nine ) and ( ix , 9 ) respectively. Determine the coordinates of the points that split 𝐴 𝐡 into four equal parts.

Answer

We can begin by sketching the line segment 𝐴 𝐡 and showing the points that divide it into 4 equal parts. We can ascertain these points every bit 𝑃 , 𝑀 , and 𝑄 .

As 𝐴 𝐡 is divided into four equal parts, we can arroyo this question by firstly finding the midpoint, 𝑀 , of 𝐴 𝐡 and so finding the midpoints of 𝐴 𝑀 and 𝑀 𝐡 .

We recollect that the midpoint, 𝑀 , of a line segment betwixt coordinates ( π‘₯ , 𝑦 ) and ( π‘₯ , 𝑦 ) is given by 𝑀 = π‘₯ + π‘₯ 2 , 𝑦 + 𝑦 2 .

To discover the midpoint, 𝑀 , of 𝐴 𝐡 , nosotros substitute the coordinates of 𝐴 ( 1 , 9 ) for the ( π‘₯ , 𝑦 ) values and the coordinates of 𝐡 ( nine , 9 ) for the ( π‘₯ , 𝑦 ) values, giving 𝑀 = 1 + 9 2 , 9 + 9 2 = 1 0 two , 1 8 two = ( 5 , 9 ) . Thus, the coordinates of 𝑀 are ( five , 9 ) .

Side by side, we detect the midpoint, 𝑃 , of 𝐴 𝑀 . Substituting the coordinates 𝐴 ( 1 , nine ) and 𝑀 ( v , 9 ) for the ( π‘₯ , 𝑦 ) , and ( π‘₯ , 𝑦 ) values, respectively, gives 𝑃 = 1 + 5 2 , 9 + ix 2 = 6 2 , 1 8 2 = ( three , 9 ) .

Finally, we find the midpoint, 𝑄 , of 𝑀 𝐡 . Using the coordinates 𝑀 ( five , 9 ) and 𝐡 ( nine , 9 ) gives 𝑃 = 5 + 9 ii , nine + 9 2 = one 4 2 , 1 viii 2 = ( 7 , 9 ) .

Thus, we have found the coordinates of 𝑃 , 𝑀 , and 𝑄 , which separate 𝐴 𝐡 into 4 equal parts, every bit ( 3 , 9 ) , ( five , 9 ) , ( 7 , 9 ) .

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

Example ii: Finding the Ratio past Which a Betoken Divides a Line Segment

If 𝐢 ∈ 𝐴 𝐡 and 𝐴 𝐡 = 3 𝐢 𝐡 , then 𝐢 divides 𝐡 𝐴 past the ratio .

  1. ii ∶ i
  2. 1 ∶ 2
  3. i ∶ 3
  4. three ∶ 1

Answer

We consider that we have a line segment 𝐴 𝐡 . Somewhere along this line segment volition exist betoken 𝐢 .

Since we need to accept into consideration the management of movement from 𝐴 to 𝐡 , we utilise the vector 𝐴 𝐡 .

The movement from point 𝐢 to bespeak 𝐡 is the vector 𝐢 𝐡 .

The magnitudes of vectors 𝐴 𝐡 and 𝐢 𝐡 are their lengths. Nosotros are given that 𝐴 𝐡 = 3 𝐢 𝐡 ; therefore, we can write that 50 e north g t h o f l e n g t h o f 𝐴 𝐡 = 3 × πΆ 𝐡 .

We can carve up 𝐴 𝐡 into 3 equal pieces.

However, nosotros demand to establish at which of these points 𝐢 will lie. If 𝐢 is closer to 𝐴 than 𝐡 , and then the length of 𝐢 𝐡 would be two-thirds the length of 𝐴 𝐡 .

Thus, it would non be true that 𝐴 𝐡 = 3 𝐢 𝐡 . Therefore, 𝐢 must be at the point that is closer to 𝐡 than 𝐴 .

In this way, 𝐢 𝐡 = 1 iii 𝐴 𝐡 and 𝐴 𝐡 = iii 𝐢 𝐡 .

To observe the ratio, as 𝐴 𝐡 is divided into iii parts, in that location volition be 2 parts of the full in 𝐴 𝐢 and 1 function in 𝐢 𝐡 .

We could write that 𝐢 divides 𝐴 𝐡 in the ratio two ∶ 1 . However, we were asked how 𝐢 divides 𝐡 𝐴 ; therefore, the solution is the ratio given in answer option B: i ∶ 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 ii singled-out points 𝐴 and 𝐡 , vector 𝐴 𝐡 tells united states the relative direction of bespeak 𝐡 with respect to betoken 𝐴 , likewise every bit the altitude between the ii points. In detail, 𝐴 𝐡 does not take to brainstorm at 𝐴 or end at 𝐡 , every bit long equally it has the same direction and magnitude. This flexibility of vectors is an reward when we work with geometric problems such every bit segmentation a line.

Let us consider how to place coordinate points. If point 𝑃 partitions 𝐴 𝐡 in the ratio π‘š ∶ 𝑛 , this means that point 𝑃 lies on the line segment 𝐴 𝐡 and the ratio of the magnitudes of the vectors satisfies ‖ ‖ 𝐴 𝑃 ‖ ‖ ∶ ‖ ‖ 𝑃 𝐡 ‖ ‖ = π‘š ∶ 𝑛 .

In other words, if ‖ ‖ 𝐴 𝑃 ‖ ‖ is π‘š length units, ‖ ‖ 𝐴 𝐡 ‖ ‖ would be equal to π‘š + 𝑛 length units, which leads to ‖ ‖ 𝐴 𝑃 ‖ ‖ = π‘š π‘š + 𝑛 ‖ ‖ 𝐴 𝐡 ‖ ‖ .

In general, information technology would be difficult to apply only the equation to a higher place to discover the coordinates of the partitioning signal 𝑃 . However, this equation does not incorporate the information that 𝑃 lies on the line segment 𝐴 𝐡 . In particular, this means that 𝐴 𝑃 has the same direction as 𝐴 𝐡 . Call up that two vectors accept the same management if 1 vector is obtained past multiplying the other vector by some positive abiding. Because the equation in a higher place, we can see that this positive constant is given by π‘š π‘š + 𝑛 . This leads to 𝐴 𝑃 = π‘š π‘š + 𝑛 𝐴 𝐡 .

We can use this property to find the coordinates of a indicate that partitions a directed line segment in a given ratio. To accomplish this, nosotros first write 𝐴 𝑃 and 𝐴 𝐡 each as a difference of two position vectors: 𝐴 𝑃 = 𝑂 𝑃 − 𝑂 𝐴 , 𝐴 𝐡 = 𝑂 𝐡 − 𝑂 𝐴 .

Substituting these expressions into the original formula leads to 𝑂 𝑃 − 𝑂 𝐴 = π‘š π‘š + 𝑛 𝑂 𝐡 − 𝑂 𝐴 .

Rearranging the equation so that 𝑂 𝑃 is the bailiwick, we obtain 𝑂 𝑃 = π‘š π‘š + 𝑛 𝑂 𝐡 − 𝑂 𝐴 + 𝑂 𝐴 𝑂 𝑃 = π‘š π‘š + 𝑛 𝑂 𝐡 + − π‘š π‘š + 𝑛 + one 𝑂 𝐴 𝑂 𝑃 = π‘š π‘š + 𝑛 𝑂 𝐡 + 𝑛 π‘š + 𝑛 𝑂 𝐴 .

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

Permit 𝑃 be a point on line segment 𝐴 𝐡 , partitioning information technology in the ratio π‘š ∶ 𝑛 . And so, the position vector 𝑂 𝑃 is given by 𝑂 𝑃 = π‘š π‘š + 𝑛 𝑂 𝐡 + 𝑛 π‘š + 𝑛 𝑂 𝐴 .

Let u.s. come across how we can apply this formula to obtain an expression for the Cartesian coordinates of the partitioning point. Let us announce the coordinates of the points 𝐴 ( π‘₯ , 𝑦 ) and 𝐡 ( π‘₯ , 𝑦 ) . Then, we can write the corresponding position vectors 𝑂 𝐴 = ( π‘₯ , 𝑦 ) , 𝑂 𝐡 = ( π‘₯ , 𝑦 ) .

Substituting these expressions into the formula higher up, we obtain 𝑂 𝑃 = π‘š π‘š + 𝑛 ( π‘₯ , 𝑦 ) + 𝑛 π‘š + 𝑛 ( π‘₯ , 𝑦 ) = π‘š π‘š + 𝑛 π‘₯ , π‘š π‘š + 𝑛 𝑦 + 𝑛 π‘š + 𝑛 π‘₯ , 𝑛 π‘š + 𝑛 𝑦 = π‘š π‘š + 𝑛 π‘₯ + 𝑛 π‘š + 𝑛 π‘₯ , π‘š π‘š + 𝑛 𝑦 + 𝑛 π‘š + 𝑛 𝑦 = π‘š π‘₯ + 𝑛 π‘₯ π‘š + 𝑛 , π‘š 𝑦 + 𝑛 𝑦 π‘š + 𝑛 .

We make it at the post-obit formula.

Theorem: The Section Formula

If we accept distinct points 𝐴 ( π‘₯ , 𝑦 ) and 𝐡 ( π‘₯ , 𝑦 ) and the betoken 𝑃 ∈ 𝐴 𝐡 divides 𝐴 𝐡 such that 𝐴 𝑃 ∶ 𝑃 𝐡 = π‘š ∶ 𝑛 , and so 𝑃 has the coordinates 𝑃 = π‘š π‘₯ + 𝑛 π‘₯ π‘š + 𝑛 , π‘š 𝑦 + 𝑛 𝑦 π‘š + 𝑛 .

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

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

If the coordinates of 𝐴 and 𝐡 are ( 5 , v ) and ( − 1 , − 4 ) , respectively, find the coordinates of betoken 𝐢 that divides 𝐴 𝐡 internally past the ratio 2 ∶ 1 .

Answer

We can sketch this directed line segment 𝐴 𝐡 as shown.

We can apply the following formula to discover point 𝐢 that divides 𝐴 𝐡 internally in the ratio 2 ∶ ane . This means that 𝐢 ∈ 𝐴 𝐡 and the ratio will be given as 𝐴 𝐢 ∶ 𝐢 𝐡 = two ∶ 1 .

We tin 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 𝐴 𝐡 such that 𝐴 𝑃 ∶ 𝑃 𝐡 = π‘š ∶ 𝑛 , so 𝑃 has the coordinates 𝑃 = π‘š π‘₯ + 𝑛 π‘₯ π‘š + 𝑛 , π‘š 𝑦 + 𝑛 𝑦 π‘š + 𝑛 .

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

The ratio values ii ∶ 1 tin can be substituted for π‘š and 𝑛 respectively.

Therefore, we will have the coordinates of 𝐢 as 𝐢 = 2 ( − 1 ) + 1 ( v ) ii + 1 , ii ( − 4 ) + 1 ( five ) 2 + 1 .

Simplifying, we have 𝐢 = 3 3 , − 3 iii = ( 1 , − ane ) .

Thus, the coordinates of point 𝐢 that divides 𝐴 𝐡 internally by the ratio 2 ∶ one are ( one , − ane ) .

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

And then far, we take observed how to place the coordinates of a signal that divides a line segment in a given ratio. We refer to such problems as internal sectionalisation bug since the betoken that we are looking for lies within the line segment.

Let us at present consider a different type of problems, known as external sectionalization bug. In these problems, the point that divides the line segment does not lie inside but rather on an extension of the line segment equally shown on the diagram below.

Because the diagram above, we say that point 𝑃 divides 𝐴 𝐡 externally in the ratio π‘š ∶ 𝑛 , where π‘š > 𝑛 . We can solve such external division bug by slightly modifying our previous arroyo to internal partitioning problems. The master difference for this example is that 𝐴 𝑃 , with magnitude π‘š length units, is the larger vector compared to 𝐡 𝑃 , with magnitude 𝑛 length units. By subtraction, we tin can meet that the length of 𝐴 𝐡 is equal to π‘š − 𝑛 units. This leads to ‖ ‖ 𝐴 𝑃 ‖ ‖ = π‘š π‘š − 𝑛 ‖ ‖ 𝐴 𝐡 ‖ ‖ .

As in the previous context, 𝐴 𝑃 and 𝐴 𝐡 have the same management, and so we can write 𝐴 𝑃 = π‘š π‘š − 𝑛 𝐴 𝐡 .

As we have washed for the internal partitioning problem, nosotros can calculate the formula for the position vector of 𝑃 : 𝑂 𝑃 − 𝑂 𝐴 = π‘š π‘š − 𝑛 𝑂 𝐡 − 𝑂 𝐴 𝑂 𝑃 = π‘š π‘š − 𝑛 𝑂 𝐡 − 𝑂 𝐴 + 𝑂 𝐴 𝑂 𝑃 = π‘š π‘š − 𝑛 𝑂 𝐡 + − π‘š π‘š − 𝑛 + 1 𝑂 𝐴 𝑂 𝑃 = π‘š π‘š − 𝑛 𝑂 𝐡 − 𝑛 π‘š − 𝑛 𝑂 𝐴 .

We notation that this formula closely resembles the ane 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 to a higher place by π‘š − 𝑛 .

On the right-hand side above, the expression π‘š + 𝑛 for internal partitioning is replaced by π‘š − 𝑛 for external segmentation. Let'southward derive the formula for the Cartesian coordinates of the division point, given the coordinates 𝐴 ( π‘₯ , 𝑦 ) and 𝐡 ( π‘₯ , 𝑦 ) : 𝑂 𝑃 = π‘š π‘š − 𝑛 ( π‘₯ , 𝑦 ) − 𝑛 π‘š − 𝑛 ( π‘₯ , 𝑦 ) = π‘š π‘š − 𝑛 π‘₯ , π‘š π‘š − 𝑛 𝑦 − 𝑛 π‘š − 𝑛 π‘₯ , 𝑛 π‘š − 𝑛 𝑦 = π‘š π‘š − 𝑛 π‘₯ − 𝑛 π‘š − 𝑛 π‘₯ , π‘š π‘š − 𝑛 𝑦 − 𝑛 π‘š − 𝑛 𝑦 = π‘š π‘₯ − 𝑛 π‘₯ π‘š − 𝑛 , π‘š 𝑦 − 𝑛 𝑦 π‘š − 𝑛 .

This leads to the post-obit formula.

Theorem: The Section Formula with External Division

If we have singled-out points 𝐴 ( π‘₯ , 𝑦 ) and 𝐡 ( π‘₯ , 𝑦 ) and point 𝑃 ∉ 𝐴 𝐡 divides 𝐴 𝐡 such that 𝐴 𝑃 ∶ 𝑃 𝐡 = π‘š ∶ 𝑛 , then 𝑃 has the coordinates 𝑃 = π‘š π‘₯ − 𝑛 π‘₯ π‘š − 𝑛 , π‘š 𝑦 − 𝑛 𝑦 π‘š − 𝑛 .

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

Instance four: Finding the Coordinates of a Point That Divides a Line Segment Externally into a Given Ratio

If 𝐴 ( iii , − 2 ) and 𝐡 ( − 2 , 4 ) , find in vector form the coordinates of point 𝐢 that divides 𝐴 𝐡 externally in the ratio iv ∶ 3 .

Answer

We tin brainstorm past sketching the points 𝐴 and 𝐡 and extending the directed line segment 𝐴 𝐡 to point 𝐢 that divides 𝐴 𝐡 externally. Nosotros can write that 𝐴 𝐢 ∶ 𝐢 𝐡 = 4 ∶ 3 .

We recall the section formula for external division.

If we have distinct points 𝐴 ( π‘₯ , 𝑦 ) and 𝐡 ( π‘₯ , 𝑦 ) and indicate 𝑃 ∉ 𝐴 𝐡 divides 𝐴 𝐡 such that 𝐴 𝑃 ∶ 𝑃 𝐡 = π‘š ∶ 𝑛 , then 𝑃 has the coordinates 𝑃 = π‘š π‘₯ − 𝑛 π‘₯ π‘š − 𝑛 , π‘š 𝑦 − 𝑛 𝑦 π‘š − 𝑛 .

We can substitute the values 𝐴 ( 3 , − 2 ) and 𝐡 ( − 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 united states 𝐢 = 4 ( − 2 ) − 3 ( three ) 4 − 3 , 4 ( 4 ) − 3 ( − 2 ) 4 − 3 = − viii − 9 i , one 6 + vi 1 = ( − i 7 , 2 2 ) .

As we are asked to give our answer in vector class, we can give the position vector of 𝐢 equally ( − 1 vii , 2 2 ) .

We will now see an case of how we tin utilize the section formula to find the ratio in which a line segment is divided.

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

Fill in the bare: Given that 𝐢 ( − three , 3 ) and 𝐷 ( 4 , − 2 ) , the π‘₯ -axis divides 𝐢 𝐷 in the ratio .

  1. three ∶ 5
  2. v ∶ 3
  3. two ∶ three
  4. 3 ∶ 2

Reply

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

In social club to notice how the π‘₯ -axis divides 𝐢 𝐷 , we commencement need to detect the point where 𝐢 𝐷 crosses the π‘₯ -axis. Given the coordinates of 𝐢 and 𝐷 , we can observe the equation of 𝐢 𝐷 , beginning by finding the slope of this line.

The slope, π‘š , of a line joining two points ( π‘₯ , 𝑦 ) and ( π‘₯ , 𝑦 ) can exist establish using π‘š = 𝑦 − 𝑦 π‘₯ − π‘₯ .

Therefore, the slope betwixt 𝐢 ( − 3 , 3 ) and 𝐷 ( four , − 2 ) is given by π‘š = − 2 − 3 four − ( − iii ) = − 5 vii = − five 7 .

Nosotros tin and so apply the point–slope form of a line such that, given a signal ( π‘₯ , 𝑦 ) and the slope, π‘š , we can write the equation of the line as 𝑦 − 𝑦 = π‘š ( π‘₯ − π‘₯ ) .

We tin can substitute the coordinates of either 𝐢 or 𝐷 into this class, and then using 𝐢 ( − 3 , 3 ) for the ( π‘₯ , 𝑦 ) values and π‘š = − 5 7 , we have 𝑦 − 3 = − 5 seven ( π‘₯ − ( − iii ) ) 𝑦 − 3 = − 5 7 ( π‘₯ + 3 ) .

We can then multiply both sides by 7 and expand the parentheses on the right-hand side, giving seven 𝑦 − 2 1 = − 5 ( π‘₯ + 3 ) 7 𝑦 − 2 1 = − 5 π‘₯ − ane 5 .

Rearranging to write this in the general grade of the equation of a line, π‘Ž π‘₯ + 𝑏 𝑦 + 𝑐 = 0 , we have 5 π‘₯ + 7 𝑦 − half-dozen = 0 .

We retrieve that a line crosses the π‘₯ -axis when 𝑦 = 0 , so substituting this into the equation five π‘₯ + 7 𝑦 − six = 0 and simplifying gives v π‘₯ + 7 ( 0 ) − six = 0 , 5 π‘₯ − 6 = 0 , 5 π‘₯ = half dozen , π‘₯ = half-dozen 5 .

Nosotros have now calculated that the line segment 𝐢 𝐷 crosses the π‘₯ -axis at the point half dozen 5 , 0 .

Nosotros now need to discover the ratio by which this coordinate, 6 five , 0 , divides 𝐢 𝐷 .

To do this, we tin can use the section formula for internal partitioning of a line segment. If we have singled-out points 𝐴 ( π‘₯ , 𝑦 ) and 𝐡 ( π‘₯ , 𝑦 ) and point 𝑃 ∈ 𝐴 𝐡 divides 𝐴 𝐡 such that 𝐴 𝑃 ∶ 𝑃 𝐡 = π‘š ∶ 𝑛 , then 𝑃 has the coordinates 𝑃 = π‘š π‘₯ + 𝑛 π‘₯ π‘š + 𝑛 , π‘š 𝑦 + 𝑛 𝑦 π‘š + 𝑛 .

In this question, we know points 𝐢 ( − 3 , 3 ) and 𝐷 ( 4 , − 2 ) and point 𝑃 6 5 , 0 that divides 𝐢 𝐷 . We need to calculate the ratio values of π‘š and 𝑛 .

Substituting 𝐢 ( − 3 , iii ) and 𝐷 ( iv , − ii ) for the ( π‘₯ , 𝑦 ) and ( π‘₯ , 𝑦 ) values, respectively, into the section formula, we take 𝑃 = π‘š ( 4 ) + 𝑛 ( − iii ) π‘š + 𝑛 , π‘š ( − two ) + 𝑛 ( 3 ) π‘š + 𝑛 = 4 π‘š − 3 𝑛 π‘š + 𝑛 , − two π‘š + iii 𝑛 π‘š + 𝑛 .

Nosotros know that 𝑃 has coordinates 6 5 , 0 , so nosotros can write half dozen 5 , 0 = iv π‘š − three 𝑛 π‘š + 𝑛 , − ii π‘š + 3 𝑛 π‘š + 𝑛 .

Evaluating the π‘₯ -coordinates, we have six 5 = 4 π‘š − 3 𝑛 π‘š + 𝑛 .

We can cross multiply and simplify to write an expression for π‘š in terms of 𝑛 as v ( 4 π‘š − 3 𝑛 ) = 6 ( π‘š + 𝑛 ) 2 0 π‘š − i 5 𝑛 = half-dozen π‘š + 6 𝑛 one 4 π‘š = 2 1 𝑛 π‘š 𝑛 = 2 1 ane four π‘š 𝑛 = 3 two .

The ratio of 𝐢 𝑃 ∶ 𝑃 𝐷 = π‘š ∶ 𝑛 , and then 𝐢 𝑃 ∶ 𝑃 𝐷 = iii ∶ two .

Thus, we tin can give the reply that the π‘₯ -axis divides 𝐢 𝐷 in the ratio 3 ∶ 2 .

Equally a check of our reply, nosotros could discover the distance of 𝐢 𝑃 and the distance of 𝑃 𝐷 and find the ratio of 𝐢 𝑃 ∶ 𝑃 𝐷 directly.

We recall that the distance formula for finding the distance, 𝑑 , betwixt 2 points ( π‘₯ , 𝑦 ) and ( π‘₯ , 𝑦 ) is given by 𝑑 = ( π‘₯ − π‘₯ ) + ( 𝑦 − 𝑦 ) .

To find the length of 𝐢 𝑃 , ‖ ‖ 𝐢 𝑃 ‖ ‖ , we substitute the values of 𝐢 ( − three , 3 ) and 𝑃 6 5 , 0 for the ( π‘₯ , 𝑦 ) and ( π‘₯ , 𝑦 ) values to give ‖ ‖ 𝐢 𝑃 ‖ ‖ = half dozen v − ( − 3 ) + ( 0 − iii ) = 2 1 5 + 3 = 6 6 six 2 v = iii √ 7 4 5 .

To find the length of 𝑃 𝐷 , ‖ ‖ 𝑃 𝐷 ‖ ‖ , we substitute 𝑃 six 5 , 0 and 𝐷 ( 4 , − 2 ) for the ( π‘₯ , 𝑦 ) and ( π‘₯ , 𝑦 ) values, which gives ‖ ‖ 𝑃 𝐷 ‖ ‖ = 4 − half dozen 5 + ( − 2 − 0 ) = 1 4 5 + ( − 2 ) = 2 nine 6 ii 5 = ii √ vii 4 v .

We tin then write the ratio of 𝐢 𝑃 ∶ 𝑃 𝐷 as 𝐢 𝑃 ∶ 𝑃 𝐷 = 3 √ 7 four 5 ∶ 2 √ 7 4 5 .

Multiplying both sides of the ratio by 5, and then dividing by √ 7 4 , gives 𝐢 𝑃 ∶ 𝑃 𝐷 = iii √ 7 4 ∶ ii √ seven 4 = 3 ∶ ii .

Thus, we have confirmed our answer, that the π‘₯ -axis divides 𝐢 𝐷 by the ratio 3 ∶ 2 .

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

Example vi: Solving a Word Problem past Dividing a Line Segment

A bus is traveling from urban center 𝐴 ( 1 0 , − ane 0 ) to city 𝐡 ( − 8 , eight ) . Its first stop is at 𝐢 , which is halfway between the cities. Its second stop is at 𝐷 , which is 2-thirds of the fashion from 𝐴 to 𝐡 . What are the coordinates of 𝐢 and 𝐷 ?

Answer

We are given that urban center 𝐴 has coordinates ( one 0 , − 1 0 ) and city 𝐡 has coordinates ( − 8 , 8 ) . Firstly, we need to discover the coordinates of city 𝐢 , halfway betwixt these.

Nosotros tin can make utilise of the formula for the midpoint of a line. To find the midpoint, 𝑀 , of a line segment betwixt two points ( π‘₯ , 𝑦 ) and ( π‘₯ , 𝑦 ) nosotros can utilize 𝑀 = π‘₯ + π‘₯ two , 𝑦 + 𝑦 2 .

Substituting ( π‘₯ , 𝑦 ) = ( 1 0 , − 1 0 ) and ( π‘₯ , 𝑦 ) = ( − 8 , 8 ) into this formula gives the midpoint, 𝐢 , as 𝐢 = 1 0 + ( − 8 ) 2 , − 1 0 + 8 2 = ii two , − two 2 = ( 1 , − one ) .

Adjacent, we need to find the coordinates of 𝐷 , which is 2-thirds of the fashion from 𝐴 to 𝐡 . The direction, 𝐴 to 𝐡 , is important every bit it indicates the position of 𝐷 . 𝐷 will be closer to 𝐡 than 𝐴 . Nosotros can think of the position past dividing 𝐴 𝐡 into 3 equal pieces. We can write the ratio 𝐴 𝐷 ∶ 𝐷 𝐡 as 2 ∶ 1 .

We can apply the formula to division a line segment in a given ratio. If 𝐴 ( π‘₯ , 𝑦 ) and 𝐡 ( π‘₯ , 𝑦 ) and point 𝑃 divides 𝐴 𝐡 such that 𝐴 𝑃 ∶ 𝑃 𝐡 = π‘š ∶ 𝑛 , and so 𝑃 has the coordinates 𝑃 = π‘š π‘₯ + 𝑛 π‘₯ π‘š + 𝑛 , π‘š 𝑦 + 𝑛 𝑦 π‘š + 𝑛 .

In this question, we have 𝐴 ( 1 0 , − one 0 ) , 𝐡 ( − 8 , 8 ) , and signal 𝐷 that divides 𝐴 𝐡 in the ratio 2 ∶ 1 . Substituting these into the formula to detect 𝐷 gives 𝐷 = 2 ( − viii ) + 1 ( i 0 ) 2 + one , 2 ( 8 ) + i ( − 1 0 ) 2 + ane = − 6 iii , 6 3 = ( − 2 , 2 ) .

Equally a useful check of our respond, nosotros can consider the lengths of 𝐴 𝐡 and 𝐴 𝐷 by applying the altitude formula. To find the distance, 𝑑 , between 2 points ( π‘₯ , 𝑦 ) and ( π‘₯ , 𝑦 ) , we calculate 𝑑 = ( π‘₯ − π‘₯ ) + ( 𝑦 − 𝑦 ) .

To summate the length of 𝐴 𝐷 , nosotros can substitute the coordinates 𝐴 ( one 0 , − 1 0 ) and 𝐷 ( − 2 , two ) into the formula to give 𝑑 = ( − 2 − 1 0 ) + ( ii − ( − 1 0 ) ) = ( − 1 2 ) + 1 two = √ 2 8 8 = one ii √ 2 .

To calculate the length of 𝐴 𝐡 , we substitute 𝐴 ( 1 0 , − ane 0 ) and 𝐡 ( − viii , 8 ) into the distance formula, giving 𝑑 = ( − viii − i 0 ) + ( 8 − ( − 1 0 ) ) = ( − 1 8 ) + 1 8 = √ vi 4 8 = 1 eight √ ii .

Thus, we can write the ratio of lengths 𝐴 𝐷 ∶ 𝐴 𝐡 as 𝐴 𝐷 ∶ 𝐴 𝐡 = 1 2 √ two ∶ ane 8 √ ii = 1 two ∶ 1 viii = 2 ∶ 3 .

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

We tin give the answer that the coordinates of 𝐢 and 𝐷 are ( 1 , − one ) ( − 2 , two ) . a n d

Key Points

  • The midpoint, 𝑀 , of the line segment between ( π‘₯ , 𝑦 ) and ( π‘₯ , 𝑦 ) is given by 𝑀 = π‘₯ + π‘₯ 2 , 𝑦 + 𝑦 2 .
  • If we take distinct points 𝐴 ( π‘₯ , 𝑦 ) and 𝐡 ( π‘₯ , 𝑦 ) and bespeak 𝑃 divides 𝐴 𝐡 internally such that 𝐴 𝑃 ∶ 𝑃 𝐡 = π‘š ∶ 𝑛 , so 𝑃 has the coordinates 𝑃 = π‘š π‘₯ + 𝑛 π‘₯ π‘š + 𝑛 , π‘š 𝑦 + 𝑛 𝑦 π‘š + 𝑛 .

  • When answering problems involving the partition of a line segment, nosotros need to be conscientious to establish the correct guild of the ratio. If 𝐴 𝐡 is split by point 𝑃 in the ratio π‘š ∶ 𝑛 , and so the ratio of 𝐴 𝑃 ∶ 𝑃 𝐡 will be π‘š ∶ 𝑛 . If 𝐡 𝐴 is split by point 𝑃 in the ratio π‘š ∶ 𝑛 , and then the ratio of 𝐴 𝑃 ∶ 𝑃 𝐡 will instead be 𝑛 ∶ π‘š .
  • If we have singled-out points 𝐴 ( π‘₯ , 𝑦 ) and 𝐡 ( π‘₯ , 𝑦 ) and point 𝑃 ∉ 𝐴 𝐡 divides 𝐴 𝐡 externally such that 𝐴 𝑃 ∶ 𝑃 𝐡 = π‘š ∶ 𝑛 , then 𝑃 has the coordinates 𝑃 = π‘š π‘₯ − 𝑛 π‘₯ π‘š − 𝑛 , π‘š 𝑦 − 𝑛 𝑦 π‘š − 𝑛 .

Partitioning A Line Segment Formula,

Source: https://www.nagwa.com/en/explainers/395106040365/#:~:text=Let%20%F0%9D%91%83%20be%20a%20point,%F0%9D%91%9A%20%2B%20%F0%9D%91%9B%20%EF%83%A0%20%F0%9D%91%82%20%F0%9D%90%B4%20.

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