Partitioning a Segment

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Learning goals

Interpret an ordered internal ratio as a fraction measured from its first endpoint
Derive partition coordinates from similar triangles and explain the reversed endpoint weights
Locate internal division points on oblique, horizontal, and vertical segments
Recover a missing endpoint from a partition point and an ordered ratio
Check both coordinates and internality when recovering an ordered division ratio

Read the ratio from the first endpoint

An internal partition point lies strictly between two distinct endpoints. Suppose $AP:PB=2:3$ . There are five equal ratio parts in the whole segment, and $P$ is two of them from $A$ .

Therefore $AP/AB=2/5$ , not $2/3$ .

For example, take $A(-4,1)$ and $B(11,11)$ . The whole horizontal change is $15$ , and the whole vertical change is $10$ . Two fifths of those changes are $6$ and $4$ . Starting at $A$ puts $P$ at $(2,5)$ .

The small and whole right triangles have corresponding legs in the same ratio as their hypotenuses.

In the earlier distance and midpoint lesson, equal parts gave the midpoint. Here the parts can differ, so the fraction from the starting endpoint need not be one half.

For an ordered ratio $AP:PB=u:v$ , with positive $u,v$ , the fraction from $A$ toward $B$ is

\frac{AP}{AB}=\frac{u}{u+v}.

The notation keeps $AP$ first and $PB$ second. Reversing the endpoint order turns the same physical division into $BP:PA=v:u$ .

Check your understanding

An internal point satisfies $AP:PB=5:7$ . What fraction of the journey from A to B reaches P?

5/7

7/12

5/12

12/5

Why both coordinates use one fraction

Similar right triangles explain why the same fraction controls both coordinate changes. In the picture above, the smaller triangle has hypotenuse $AP$ and the larger has hypotenuse $AB$ . Their right angles agree, and they share the angle at $A$ , so AA similarity gives the same fraction on corresponding legs.

Why an internal ratio gives weighted coordinates#

For another numerical division, take $A(-6,-2)$ and $B(14,13)$ with $AP:PB=3:2$ . Similarity makes the horizontal and vertical leg fractions both $3/5$ . The whole legs are $20$ and $15$ , so the smaller legs are $12$ and $9$ . Adding those changes to $A$ produces

\begin{aligned} x_P&=-6+\tfrac35(20),\\ &=6,\\ y_P&=-2+\tfrac35(15),\\ &=7. \end{aligned}

Now let $A(x_A,y_A)$ and $B(x_B,y_B)$ be distinct, and let $P$ be internal with $AP:PB=u:v$ , where $u,v$ are positive. When both coordinate changes are nonzero, the same AA argument makes each leg change the fraction $u/(u+v)$ of its whole change. If a coordinate decreases, its small and whole changes have the same negative sign, so the same positive fraction still applies.

If the segment is horizontal, the vertical change is zero throughout. If it is vertical, the horizontal change is zero throughout. Apply the fraction along the changing coordinate in either case; no division by a zero coordinate change is needed.

Thus start at $A$ and add the required fraction of each signed change.

\begin{aligned} x_P&=x_A+\frac{u}{u+v}(x_B-x_A),\\ y_P&=y_A+\frac{u}{u+v}(y_B-y_A). \end{aligned}

Combining terms in each coordinate gives the weighted-coordinate formula.

\begin{aligned} x_P&=\frac{v x_A+u x_B}{u+v},\\ y_P&=\frac{v y_A+u y_B}{u+v}. \end{aligned}

The argument used no special property of the example’s numbers. The weight on $A$ is $v$ , and the weight on $B$ is $u$ . A larger first part $AP$ moves $P$ farther toward $B$ , which is why $B$ receives the weight $u$ .

Worked example 1 Move left and up with one fraction

Problem. Let $A(8,-6)$ and $B(-7,4)$ . Find the internal point $P$ satisfying $AP:PB=2:3$ .

Solution. The required fraction from $A$ is $2/5$ . The signed coordinate changes from $A$ to $B$ are $-15$ and $10$ .

\begin{aligned} x_P&=8+\tfrac25(-15),\\ &=2,\\ y_P&=-6+\tfrac25(10),\\ &=-2. \end{aligned}

So $P=(2,-2)$ . Both coordinates are between the corresponding endpoint coordinates, and both changes use the same fraction $2/5$ .

The whole signed changes are -15 and 10; the smaller changes are -6 and 4. Both have the same fraction 2/5.

Check your understanding

For $AP:PB=4:9$ , which expression gives the first coordinate of P?

(4x_A+9x_B)/13

(9x_A+4x_B)/13

(4x_A+9x_B)/5

(x_A+x_B)/2

Keep the endpoints in order

Changing only the ratio usually changes the point. Changing both the endpoint order and the ratio order preserves the original point.

Worked example 2 Compare the two ordered divisions

Problem. Let $A(-3,-2)$ and $B(9,6)$ .

(a) Find $P$ with $AP:PB=1:3$ .

(b) Find $Q$ with $AQ:QB=3:1$ .

With fixed endpoints, exchanging the two ratio numbers moves the division to a different point.

Solution. In (a), move one fourth of the changes $12$ and $8$ from $A$ .

P=(0,0).

In (b), move three fourths of the same changes from $A$ .

Q=(6,4).

For (c), the original first and last parts exchange roles, so $BP:PA=3:1$ . Starting at $B$ and moving three fourths toward $A$ returns to $(0,0)$ .

When $u=v$ , the weights are equal and each coordinate is the average of its endpoint coordinates. This is the midpoint special case already established in the previous lesson.

Worked example 3 Divide vertical and horizontal segments

Problem. Solve each internal division.

(a) $A(4,9)$ and $B(4,-11)$ , with $AP:PB=3:2$ .

(b) $C(12,-5)$ and $D(-3,-5)$ , with $CQ:QD=1:4$ .

Solution. In (a), the first coordinate stays $4$ . Move three fifths of the vertical change $-20$ .

\begin{aligned} y_P&=9+\tfrac35(-20),\\ &=-3. \end{aligned}

Thus $P=(4,-3)$ .

Both panels have first coordinate 4 throughout. Endpoint order determines which vertical change to use.

The right panel starts at $B$ instead. Three fifths of the way from $B$ to $A$ gives $R=(4,1)$ , a different point.

In (b), the second coordinate stays $-5$ . Move one fifth of the horizontal change $-15$ .

\begin{aligned} x_Q&=12+\tfrac15(-15),\\ &=9. \end{aligned}

Thus $Q=(9,-5)$ . A fixed coordinate causes no difficulty because the change in that coordinate is zero.

Check your understanding

Let A be $(-2,14)$ and B be $(-2,-16)$ . Which internal point has $AP:PB=2:3$ ?

(-2,-4)

(2,2)

(-2,2)

(-2,-1)

Recover an endpoint

An internal point and a ratio tell us how much of the full change has already happened. We can undo that fraction to recover the endpoint.

Worked example 4 Extend the known part by the required amount

Problem. Let $A(-5,7)$ and $P(1,1)$ . The point $P$ is internal to $AB$ and $AP:PB=3:4$ . Find $B$ and verify the ratio.

P is given, while the endpoint B must be recovered. The schematic records the order and ratio.

Solution. Three of the seven ratio parts take $A$ to $P$ . The known changes are $6$ and $-6$ , so one part changes each coordinate by $2$ and $-2$ . Seven parts from $A$ give

\begin{aligned} x_B&=-5+7(2),\\ &=9,\\ y_B&=7+7(-2),\\ &=-7. \end{aligned}

The endpoint is $B=(9,-7)$ . The changes from $P$ to $B$ are $8$ and $-8$ , four copies of the same one-part changes. Therefore the straight subsegments have ratio $3:4$ , and $P$ is internal because $3/7$ is strictly between zero and one.

The formula gives the same endpoint by solving the two coordinate equations separately.

\begin{aligned} 7(1)&=4(-5)+3x_B,\\ 7(1)&=4(7)+3y_B. \end{aligned}

Check your understanding

A is $(2,-3)$ and internal P is $(6,5)$ . If $AP:PB=1:2$ , what is B?

(14,21)

(8,13)

(10,13)

(18,15)

Check the same division in both coordinates

A candidate can sit between the endpoint coordinates in each direction and still miss the segment. Both coordinates must describe the same fraction of the journey.

Worked example 5 Recover a ratio and reject inconsistent coordinates

Problem. Let $A(10,-8)$ and $B(-6,16)$ .

(a) Decide whether $P(4,1)$ is internal to $AB$ . If it is, find $AP:PB$ .

(b) Decide whether $Q(4,2)$ is internal to $AB$ .

Solution. In (a), use each changing coordinate to recover the fraction from $A$ .

\begin{aligned} \frac{4-10}{-6-10}&=\frac38,\\ \frac{1-(-8)}{16-(-8)}&=\frac38. \end{aligned}

Both agree, and $0<3/8<1$ , so $P$ is internal.

Three eighths lie before $P$ and five eighths after it, giving $AP:PB=3:5$ .

In (b), the first coordinate still requires $3/8$ of the journey. The second coordinate gives a different fraction.

\frac{2-(-8)}{16-(-8)}=\frac5{12}.

Thus $Q$ is not on $AB$ , even though each coordinate is within its endpoint range. For a vertical or horizontal segment, check that the fixed coordinate matches and use the other coordinate for the fraction. Fractions zero and one describe endpoints, not internal points.

Check your understanding

A is $(-8,12)$ , B is $(12,-3)$ , and P is $(0,6)$ . Which conclusion follows after checking both coordinates?

AP:PB=3:2

B P is not on AB C P is the midpoint D

AP:PB=2:3

Partitioning a Segment

Learning goals

Read the ratio from the first endpoint

Why both coordinates use one fraction

Keep the endpoints in order

Recover an endpoint

Check the same division in both coordinates

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice

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