Core practice ← Back to lesson

Partitioning a Segment: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

0 of 10 done

Progress saved in this browser.

Problem 1 of 10

Problem 1 Turn a ratio into a coordinate point

Let $A(-11,2)$ and $B(9,12)$ . Find internal P with $AP:PB=1:4$ . State the fraction of AB that AP occupies.
1. Hint 1
  Count the total number of ratio parts.
2. Hint 2
  Apply the first-part fraction to both signed coordinate changes.
Answer

$\begin{aligned}P&=(-7,4),\\AP/AB&=1/5.\end{aligned}$

Full solution

The ratio has five parts in total, so $AP/AB=1/5$ .

The changes from A to B are 20 and 10.
$\begin{aligned}x_P&=-11+\tfrac15(20)\\&=-7,\\y_P&=2+\tfrac15(10)\\&=4.\end{aligned}$

Both coordinates use the same fraction, and $0<1/5<1$ , so P is internal.

Answer

$\begin{aligned}P&=(-7,4),\\AP/AB&=1/5.\end{aligned}$

Key idea

The first ratio part is a fraction of the sum of both parts.
Problem 2 Preserve decreasing coordinate changes

Let $A(14,9)$ and $B(-6,-11)$ . Find internal P with $AP:PB=3:2$ and explain why both coordinates decrease.
1. Hint 1
  Read the ratio from A toward B.
2. Hint 2
  Find the signed whole changes before taking three fifths.
Answer

$P=(2,-3)$ ; both endpoint coordinates decrease from A to B.

Full solution

The first part is three fifths of AB.
Both whole coordinate changes are -20.

Take three fifths of each change.
$\begin{aligned}x_P&=14+\tfrac35(-20)\\&=2,\\y_P&=9+\tfrac35(-20)\\&=-3.\end{aligned}$

Both changes are negative because B is left of and below A.
The fraction $3/5$ is internal.

Answer

$P=(2,-3)$ ; both endpoint coordinates decrease from A to B.

Key idea

Ratios of lengths are positive while coordinate changes retain their direction.
Problem 3 Read an ordered ratio from a horizontal segment

The points $A(-9,6)$ , $P(3,6)$ , and $B(11,6)$ are on one horizontal segment. Find $AP:PB$ and $BP:PA$ . Explain the change in order.

The given point lies between the two endpoints on the horizontal segment.

Text description of this figure

A schematic internal division with endpoint order and positive ratio parts. Use the written coordinates, not a measurement of the picture.
1. Hint 1
  Use the horizontal gaps, since the second coordinate is fixed.
2. Hint 2
  Reduce the two positive lengths in each stated order.
Answer

$\begin{aligned}AP:PB&=3:2,\\BP:PA&=2:3.\end{aligned}$

Full solution

The horizontal lengths are
$\begin{aligned}AP&=3-(-9)\\&=12,\\PB&=11-3\\&=8.\end{aligned}$

The second coordinate agrees throughout, and $-9<3<11$ , so P is internal.
The ordered ratio $12:8$ reduces to $3:2$ .

Starting from B exchanges the first and last parts, giving $BP:PA=8:12$ , or $2:3$ .

Answer

$\begin{aligned}AP:PB&=3:2,\\BP:PA&=2:3.\end{aligned}$

Key idea

Reversing the endpoints reverses the ratio for the same point.
Problem 4 Partition a vertical segment

Let $A(7,-15)$ and $B(7,20)$ . Find internal P satisfying $AP:PB=4:3$ . Explain how you handle the coordinate that does not change.
1. Hint 1
  The first coordinates agree, so every point of AB has that first coordinate.
2. Hint 2
  Four of the seven ratio parts lie before P.
Answer

$P=(7,5)$ ; its first coordinate stays 7.

Full solution

The first coordinate remains 7 because the entire segment is vertical.
Its horizontal change is zero.

The vertical change is 35.
$\begin{aligned}y_P&=-15+\tfrac47(35)\\&=5.\end{aligned}$

The second-coordinate gaps are 20 and 15, whose ratio is $4:3$ .
Also $-15<5<20$ , checking internality.

Answer

$P=(7,5)$ ; its first coordinate stays 7.

Key idea

A zero change stays zero; use the changing coordinate to verify the ratio.
Problem 5 Compare two endpoint orders

Let $A(-2,-9)$ and $B(13,6)$ .

(a) Find internal P satisfying $AP:PB=2:3$ .

(b) Find internal Q satisfying $BQ:QA=2:3$ . Are P and Q the same point?
1. Hint 1
  Part (b) starts at B rather than A.
2. Hint 2
  Use two fifths of the change from the named starting endpoint in each part.
Answer

(a) $P=(4,-3)$ . (b) $Q=(7,0)$ . They are different points.

Full solution

(a) Starting from A, two fifths of each change 15 gives a change of 6.
Thus $P=(4,-3)$ .

(b) Starting from B, two fifths of each change -15 gives a change of -6.
Thus $Q=(7,0)$ .

For the original point P, the ratio from B would be $BP:PA=3:2$ .
Retaining the numerical ratio $2:3$ while reversing the endpoints specifies a different point.

The two starting endpoints place the points at different positions.

Answer

(a) $P=(4,-3)$ . (b) $Q=(7,0)$ . They are different points.

Key idea

A ratio identifies a point only after its endpoint order is fixed.
Problem 6 Recover the first endpoint

The endpoint is $B(17,-4)$ and the internal point is $P(5,8)$ . Given $AP:PB=3:2$ , find A and check the ratio in both coordinates.
1. Hint 1
  The change from P to B is two ratio parts.
2. Hint 2
  Find one part, then move three such parts backward from P.
Answer

$A=(-13,26)$ ; both coordinates give the internal fraction $3/5$ .

Full solution

From P to B the changes are 12 and -12, representing two parts.
One part therefore changes the coordinates by 6 and -6.

Move backward three parts from P.
$\begin{aligned}x_A&=5-3(6)\\&=-13,\\y_A&=8-3(-6)\\&=26.\end{aligned}$

The full changes from A to B are 30 and -30.
The changes to P are 18 and -18, so both fractions are $3/5$ .
This lies between zero and one, and its leftover is $2/5$ , checking the ordered ratio $3:2$ .

PB accounts for two parts. Moving three of those parts back from P recovers A.

Answer

$A=(-13,26)$ ; both coordinates give the internal fraction $3/5$ .

Key idea

When the known endpoint is B, the known gap PB represents the second ratio part.
Problem 7 Place an anchor on a straight cable

A straight cable runs from $A(-10,4)$ to $B(18,25)$ , with coordinates measured in meters. An internal anchor P must satisfy $AP:PB=5:2$ . Find its coordinates and the two cable lengths AP and PB.

The cable is straight and the anchor is internal. The coordinates and ratio are givens.

Text description of this figure

A schematic internal division with endpoint order and positive ratio parts. Use the written coordinates, not a measurement of the picture.
1. Hint 1
  Locate P by the fraction of the whole cable before it.
2. Hint 2
  Find AB from the endpoint differences, then divide that length in the same ratio.
Answer

$\begin{aligned}P&=(10,19),\\AP&=25\text{ m},\\PB&=10\text{ m}.\end{aligned}$

Full solution

The fraction from A is $5/7$ .
The coordinate changes are 28 and 21.
$\begin{aligned}x_P&=-10+\tfrac57(28)\\&=10,\\y_P&=4+\tfrac57(21)\\&=19.\end{aligned}$

The cable length comes from the earlier distance formula.
$\begin{aligned}AB&=\sqrt{28^2+21^2}\\&=35\text{ m}.\end{aligned}$

The two parts are
$\begin{aligned}AP&=\tfrac57(35)\\&=25\text{ m},\\PB&=\tfrac27(35)\\&=10\text{ m}.\end{aligned}$
Their sum is 35 m and their ratio is $5:2$ .

Answer

$\begin{aligned}P&=(10,19),\\AP&=25\text{ m},\\PB&=10\text{ m}.\end{aligned}$

Key idea

A straight segment has the same partition fraction in coordinates and in length.
Problem 8 Use one coordinate to recover the other

Let $A(1,10)$ and $B(16,-5)$ . An internal point P has coordinates $(k,4)$ . Find k and the ordered ratio $AP:PB$ . Check both coordinates.
1. Hint 1
  Recover the journey fraction from the second coordinates first.
2. Hint 2
  Apply that fraction to the horizontal change, then compare used and leftover parts.
Answer

$\begin{aligned}k&=7,\\AP:PB&=2:3.\end{aligned}$

Full solution

The vertical fraction is
$\frac{4-10}{-5-10}=\frac25.$

Use the same fraction horizontally.
$\begin{aligned}k&=1+\tfrac25(16-1)\\&=7.\end{aligned}$

The horizontal fraction is $6/15=2/5$ , agreeing with the vertical fraction.
The common fraction is strictly between zero and one, so P is internal and $AP:PB=2:3$ .

Answer

$\begin{aligned}k&=7,\\AP:PB&=2:3.\end{aligned}$

Key idea

A known changing coordinate determines the fraction for every other coordinate.
Problem 9 Decide whether endpoint data can coexist

A record lists $A(-7,-6)$ , $P(2,3)$ , and $B(b,14)$ . It requires P to be internal and $AP:PB=3:2$ . Is there a real value of b satisfying all the data? Justify your decision using both coordinate requirements.
1. Hint 1
  The ratio fixes the fraction from A before b is determined.
2. Hint 2
  Check the second coordinate, which does not contain b.
Answer

No real value of b satisfies all the data.

Full solution

The required fraction is $3/5$ .
The first-coordinate equation alone gives
$\begin{aligned}2&=-7+\tfrac35(b+7),\\b&=8.\end{aligned}$

The second coordinate required by the same fraction is
$\begin{aligned}y_P&=-6+\tfrac35(14+6)\\&=6.\end{aligned}$

The record instead gives 3.
Its actual vertical fraction is $9/20$ , not $3/5$ .
Since b cannot change that vertical fraction, no real b satisfies the complete record.

Answer

No real value of b satisfies all the data.

Key idea

An endpoint coordinate cannot repair an inconsistency in the other coordinate.
Problem 10 Recover a segment from two internal points

The points A, P, Q, B lie in that order on one straight segment. Their coordinates are $A(-12,9)$ , $P(-4,3)$ , $Q(4,-3)$ , and $B(b,-12)$ .

Find b, $AP:PB$ , and $AQ:QB$ . Verify that both P and Q have the required common fraction in the two coordinates and are internal.

The givens fix two internal positions; the missing endpoint coordinate is not supplied.

Text description of this figure

The points occur in the order A, P, Q, B on one segment. A has second coordinate nine, P three, Q negative three and B negative twelve. The first coordinate of B is unknown.
1. Hint 1
  The second coordinates determine the fractions from A for P and Q.
2. Hint 2
  Use either fraction to recover b, and then check the other point independently.
3. Hint 3
  Compare each used fraction with its own leftover fraction.
Answer

$\begin{aligned}b&=16,\\AP:PB&=2:5,\\AQ:QB&=4:3.\end{aligned}$
Both points satisfy both coordinate fractions and are internal.

Full solution

The full vertical change is -21.
The changes to P and Q are -6 and -12, respectively, so their fractions are $2/7$ and $4/7$ .

P has horizontal change 8.
Its required fraction determines the endpoint.
$\begin{aligned}8&=\tfrac27(b+12),\\b+12&=28,\\b&=16.\end{aligned}$

Now the full horizontal change is 28.
For P, $8/28=2/7$ matches the vertical fraction.
For Q, $16/28=4/7$ also matches its vertical fraction.
Thus both points lie on the recovered segment.

The two fractions satisfy $0<2/7<4/7<1$ , so both are internal and the stated order is correct.
Their leftover fractions are $5/7$ and $3/7$ , giving $AP:PB=2:5$ and $AQ:QB=4:3$ .

Answer

$\begin{aligned}b&=16,\\AP:PB&=2:5,\\AQ:QB&=4:3.\end{aligned}$
Both points satisfy both coordinate fractions and are internal.

Key idea

Recover the data, then verify every point with both coordinates and an internal fraction.

← Back to lesson