chevron_backward

Points on a line segment

chevron_forward

visibility 103update 6 months agobookmarkshare

🎯 In this topic you will

Work out the coordinates of the midpoint of a line segment
Use coordinates to find points on a line segment

🧠 Key Words

line segment
midpoint

Show Definitions

line segment: A straight part of a line between two endpoints.
midpoint: The point that is exactly halfway between the two endpoints of a line segment.

📐 Midpoints of Line Segments

The diagram shows two line segments, AB and CD. The midpoint of AB is halfway between A and B.

You can see from the diagram that the midpoint of AB is $(3, 3)$. You can see from the diagram that the midpoint of CD is $(1, 0)$.

📘 Worked example

The diagram shows two line segments, LM and PQ.

a. Write the coordinates of the midpoint of LM.
b. Work out the coordinates of the midpoint of PQ.

Answer:

a.$(3, 3)$

b.$(1, 0)$

You can see that the $y$-coordinate of the midpoint is $3$, because all the points on the line LM have a $y$-coordinate of $3$.

You can see that the $x$-coordinate of the midpoint is $3$, because it is exactly halfway along the line LM.

To go from P to Q, you go $8$ squares across and $4$ squares up (shown by the red line).

To go from P to the midpoint, you do half of this, so you go $4$ squares across and $2$ squares up (shown by the blue line).

🧠 PROBLEM-SOLVING Strategy

Finding the Midpoint of a Line Segment

Use this method whenever you need the midpoint between two coordinates.

Label the endpoints of the line as $A(x_1, y_1)$ and $B(x_2, y_2)$.
Add the $x$-coordinates: $x_1 + x_2$.
Add the $y$-coordinates: $y_1 + y_2$.
Divide each sum by $2$ to find the midpoint values.
The midpoint is given by the formula: $M = \Big(\dfrac{x_1 + x_2}{2}, \; \dfrac{y_1 + y_2}{2}\Big)$.
Check your answer by plotting the points on a grid — the midpoint should be exactly halfway along the segment.

Endpoints	Midpoint Calculation	Result
$A(2,3)$, $B(6,7)$	$\Big(\dfrac{2+6}{2}, \dfrac{3+7}{2}\Big)$	$(4,5)$
$A(8,0)$, $B(12,6)$	$\Big(\dfrac{8+12}{2}, \dfrac{0+6}{2}\Big)$	$(10,3)$

❓ EXERCISES

1. Write the coordinates of the midpoint of each line segment.

👀 Show answer

For each line segment, the midpoint is found using the formula:

$\left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\right)$

Midpoint of $AB$: $(3, 1)$
Midpoint of $CD$: $(3, 4)$
Midpoint of $EF$: $(2, 3)$
Midpoint of $GH$: $(4, 1)$

❓ EXERCISES

2. Match each line segment with the correct midpoint.
An example is done for you.
Line segment $AB$ and iii.

i. $(1, -2)$
ii. $(-1, -6)$
iii. $(2, 3)$
iv. $(-5, 4)$
v. $\left(-3 \tfrac{1}{2}, -3\right)$
vi. $(2 \tfrac{1}{2}, 5)$
vii. $(-3, -1)$
viii. $(5, -2 \tfrac{1}{2})$

👀 Show answer

$AB \;\;\to\;\; iii.$$(2, 3)$
$CD \;\;\to\;\; iv.$$(-5, 4)$
$EF \;\;\to\;\; vii.$$(-3, -1)$
$GH \;\;\to\;\; i.$$(1, -2)$
$IJ \;\;\to\;\; vi.$$(2 \tfrac{1}{2}, 5)$
$KL \;\;\to\;\; viii.$$(5, -2 \tfrac{1}{2})$
$MN \;\;\to\;\; ii.$$(-1, -6)$
$PQ \;\;\to\;\; v.$$\left(-3 \tfrac{1}{2}, -3\right)$

🧠 Think like a Mathematician

Task: Zalika and Maha use different methods to find the midpoint of the line segment $AB$ where $A = (3,4)$ and $B = (11,4)$.

Follow-up Questions:

a. Write the advantages and disadvantages of Zalika’s method.

b. Explain how Maha’s method works.

c. Write the advantages and disadvantages of Maha’s method.

d. Whose method do you prefer? Explain why.

e. Can you think of a better method?

f. Summarize your answers clearly.

👀 Show Answers

a: Zalika’s method is simple and visual but less precise for larger coordinates or without graph paper.
b: Maha’s method uses arithmetic: average the coordinates to find the midpoint, ensuring accuracy.
c: Maha’s method is exact and doesn’t need a diagram, but requires calculation skills.
d: Maha’s method is usually preferable because it is both quicker and more accurate.
e: A better method is the general midpoint formula: $\left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\right)$.
f: Zalika’s method is good for visualization, but Maha’s is more efficient. The best is to use the general midpoint formula.

❓ EXERCISES

4. Work out the midpoint of the line segment joining each pair of points.
Write whether A, B or C is the correct answer.
Use your preferred method.

a.$(7,1)$ and $(7,7)$
A $(7,6)$ B $(7,3)$ C $(7,4)$

b.$(4,2)$ and $(10,2)$
A $(7,2)$ B $(6,2)$ C $(5,2)$

c.$(4,11)$ and $(4,2)$
A $(4,9)$ B $(4,6 \tfrac{1}{2})$ C $(4,4 \tfrac{1}{2})$

d.$(8,15)$ and $(15,15)$
A $(11 \tfrac{1}{2},15)$ B $(7,15)$ C $(12 \tfrac{1}{2},15)$

👀 Show answer

a. Midpoint: $(7,4)$ → Answer: C
b. Midpoint: $(7,2)$ → Answer: A
c. Midpoint: $(4,6 \tfrac{1}{2})$ → Answer: B
d. Midpoint: $(11 \tfrac{1}{2},15)$ → Answer: A

Method used: Apply the midpoint formula $\left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\right)$.

📘 Worked example

You can calculate the midpoint of a line segment by finding the means of the $x$-coordinates and the $y$-coordinates of the end points.

The diagram shows the line segment $PQ$.
Calculate the coordinates of the midpoint of $PQ$.

Answer:

$\dfrac{10 + (-4)}{2} = \dfrac{6}{2} = 3$
$\dfrac{4 + (-6)}{2} = \dfrac{-2}{2} = -1$

The midpoint of $PQ$ is $(3, -1)$.

Add the $x$-coordinates of P and Q and divide the result by 2.

Add the $y$-coordinates of P and Q and divide the result by 2.

❓ EXERCISES

5. Copy and complete the workings to calculate the midpoint of the line segment joining each pair of points.

a.$(2,3)$ and $(6,7)$
$\left(\dfrac{2+6}{2}, \dfrac{3+7}{2}\right) = \left(\dfrac{8}{2}, \dfrac{10}{2}\right) = (4, \, \square)$

b.$(8,0)$ and $(12,6)$
$\left(\dfrac{8+12}{2}, \dfrac{0+6}{2}\right) = \left(\dfrac{20}{2}, \dfrac{6}{2}\right) = (\square, \, \square)$

c.$(5,2)$ and $(8,10)$
$\left(\dfrac{5+8}{2}, \dfrac{2+10}{2}\right) = \left(\dfrac{13}{2}, \dfrac{12}{2}\right) = \left(6 \tfrac{1}{2}, \, \square\right)$

d.$(0,4)$ and $(7,11)$
$\left(\dfrac{0+7}{2}, \dfrac{4+11}{2}\right) = \left(\dfrac{\square}{2}, \dfrac{\square}{2}\right) = (\square, \, \square)$

👀 Show answer

a. $(4,5)$
b. $(10,3)$
c. $\left(6 \tfrac{1}{2}, 6\right)$
d. $\left(3 \tfrac{1}{2}, 7 \tfrac{1}{2}\right)$

Method: Apply the midpoint formula: $\left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\right)$.

❓ EXERCISES

6. E is the point $(6,0)$, F is the point $(14,8)$ and G is the point $(3,15)$.

a. Work out the midpoint of the line segments:

i. EF
ii. EG
iii. FG

b. Draw a coordinate grid. Plot the points E, F and G. Check your answers to part a by finding the midpoints on your diagram.

👀 Show answer

i. EF: $\left(\dfrac{6+14}{2}, \dfrac{0+8}{2}\right) = (10,4)$
ii. EG: $\left(\dfrac{6+3}{2}, \dfrac{0+15}{2}\right) = \left(4 \tfrac{1}{2}, 7 \tfrac{1}{2}\right)$
iii. FG: $\left(\dfrac{14+3}{2}, \dfrac{8+15}{2}\right) = \left(8 \tfrac{1}{2}, 11 \tfrac{1}{2}\right)$

Check: Plotting the points on a coordinate grid confirms that these are the correct midpoints.

🧠 Think like a Mathematician

Task: Shen and Hassan calculate the midpoint of the line segment joining the points $(-5,-8)$ and $(-1,9)$. Compare their methods and decide who is correct.

Follow-up questions (answer on your own first):

a. Who has the correct midpoint? Explain the other student’s mistake.

b. Shen added the coordinates of $(-5,-8)$ to $(-1,9)$. Hassan added the coordinates of $(-1,9)$ to $(-5,-8)$. Does the order matter? Explain.

c. Summarize what this tells you about calculating midpoints.

👀 Show Answers

a.Hassan is correct: midpoint $(-3,\,\tfrac{1}{2})$. Shen’s error is arithmetic: $-5 + (-1) = -6$, not $-4$; the wrong sum led to $x=-2$ instead of $x=-3$.
b. Order does not matter. Addition is commutative, so $\dfrac{x_1+x_2}{2}=\dfrac{x_2+x_1}{2}$ and likewise for $y$. Both orders give the same midpoint when the arithmetic is done correctly.
c. The reliable method is to average the coordinates: $\text{Midpoint}=\left(\dfrac{x_1+x_2}{2},\;\dfrac{y_1+y_2}{2}\right)$. Check simple sums carefully—small sign mistakes change the answer.

❓ EXERCISES

8. Calculate the midpoint of the line segment between

a.$(5,-2)$ and $(2,-6)$ b.$(-4,5)$ and $(3,0)$ c.$(-7,5)$ and $(-10,10)$

👀 Show answer

a.$\left(\dfrac{5+2}{2},\dfrac{-2+(-6)}{2}\right)=\left(\dfrac{7}{2},-4\right)=\left(3\tfrac{1}{2},-4\right)$
b.$\left(\dfrac{-4+3}{2},\dfrac{5+0}{2}\right)=\left(-\dfrac{1}{2},\dfrac{5}{2}\right)$
c.$\left(\dfrac{-7+(-10)}{2},\dfrac{5+10}{2}\right)=\left(-\dfrac{17}{2},\dfrac{15}{2}\right)$

Rule:$\text{Midpoint}=\left(\dfrac{x_1+x_2}{2},\;\dfrac{y_1+y_2}{2}\right)$.

9. A parallelogram has vertices at $P(2,5)$, $Q(-2,3)$, $R(2,-1)$ and $S(6,1)$. The diagonals are $PR$ and $QS$. Show that the diagonals have the same midpoint.

👀 Show answer

$PR$ midpoint: $\left(\dfrac{2+2}{2},\dfrac{5+(-1)}{2}\right)=(2,2)$.

$QS$ midpoint: $\left(\dfrac{-2+6}{2},\dfrac{3+1}{2}\right)=(2,2)$.

Both midpoints are $(2,2)$, so the diagonals have the same midpoint.

10. Calculate the coordinates of the midpoint of each side of this triangle.

👀 Show answer

The exact coordinates of the triangle’s vertices are not stated in the text; they must be read from the diagram. Use $\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)$ on each pair of endpoints to find each side’s midpoint. (If precise vertex coordinates are provided, the numeric midpoints can be computed immediately.)

11. A quadrilateral has vertices at $(-2,1)$, $(0,4)$, $(5,2)$ and $(1,-1)$. Do the diagonals have the same midpoint? Justify your answer.

👀 Show answer

Label the points $A(-2,1)$, $B(0,4)$, $C(5,2)$, $D(1,-1)$. Midpoint of diagonal $AC$: $\left(\dfrac{-2+5}{2},\dfrac{1+2}{2}\right)=\left(\dfrac{3}{2},\dfrac{3}{2}\right)$. Midpoint of diagonal $BD$: $\left(\dfrac{0+1}{2},\dfrac{4+(-1)}{2}\right)=\left(\dfrac{1}{2},\dfrac{3}{2}\right)$. Since the midpoints are different, the diagonals do **not** have the same midpoint.

🧠 Think like a Mathematician

The midpoint of a line segment is $(4,1)$. One end of the line segment is $(2,5)$.

a. Work out the coordinates of the other end of the line segment.

b. Reflect on the method you used to solve part a. Could you have used a different method? Compare the advantages and disadvantages of different approaches.

c. Decide which method is the best to use for this type of question and explain why.

👀 Show Answers

a. Let the other endpoint be $(x,y)$. Midpoint formula: $\left(\dfrac{2+x}{2},\dfrac{5+y}{2}\right)=(4,1)$. So $\dfrac{2+x}{2}=4 \Rightarrow x=6$, $\dfrac{5+y}{2}=1 \Rightarrow y=-3$. The other endpoint is $(6,-3)$.
b. Methods: – Solve equations from the midpoint formula (algebraic, precise). – Use geometric reasoning (average shifts in coordinates). Both give the same answer, but the algebraic method avoids miscounting.
c. The algebraic midpoint formula is the most reliable because it generalises easily and reduces errors compared to visual or trial methods.

❓ EXERCISES

13. The midpoint of a line segment is $(7,2)$. One end of the line segment is $(-1,6)$.
Work out the coordinates of the other end of the line segment.

👀 Show answer

Let the other endpoint be $(x,y)$. Midpoint formula: $\left(\dfrac{-1+x}{2},\dfrac{6+y}{2}\right)=(7,2)$. $\dfrac{-1+x}{2}=7 \;\Rightarrow\; x=15$. $\dfrac{6+y}{2}=2 \;\Rightarrow\; y=-2$. The other endpoint is $(15,-2)$.

14. Here are six cards showing the coordinates of the points A to F.

Three line segments are made using the six cards. The midpoint of all three line segments is $(-1,1)$. What are the three line segments? Show how you worked out your answers.

👀 Show answer

To have midpoint $(-1,1)$, the sums of the coordinates of each pair must be $(-2,2)$.

A $(2,0)$ and C $(-7,5)$ → midpoint $(-1,1)$
B $(-3,-2)$ and E $(5,-3)$ → midpoint $(-1,1)$
D $(1,4)$ and F $(-4,2)$ → midpoint $(-1,1)$

The three line segments are AC, BE, and DF.

📍 Midpoints and Beyond

In Stage $8$ you learned how to find the midpoint of a line segment. The diagram shows a line segment $AB$. You can see from the diagram that the midpoint of $AB$ is $(3,3)$.

In this section you will find the coordinates of different points along a line segment. For example, how can you work out the coordinates of the point that lies one third, two thirds, one quarter or three quarters of the way along line segment $AB$?

📘 Worked example

The diagram shows two line segments LM and PQ.
Write the coordinates of the points that lie:

a. one third of the way along LM
b. one quarter of the way along PQ

Answer:

a. $(2,5)$

x-coordinate: L is 0 and M is 6, so $\tfrac{1}{3}$ of the way across is $6 \div 3 = 2$.
Line LM is horizontal, so the y-coordinate is 5.
Therefore, the point is $(2,5)$.

b. $(2,1)$

x-coordinate: P is 0 and Q is 8, so $\tfrac{1}{4}$ of the way across is $8 \div 4 = 2$.
y-coordinate: P is 0 and Q is 4, so $\tfrac{1}{4}$ of the way up is $4 \div 4 = 1$.
Therefore, the point is $(2,1)$.

To find a point at a fraction of the way along a line segment, calculate the change in $x$ and $y$, then multiply each by the fraction. Add these to the starting coordinates.

Example: For PQ, the change in $x$ is $+8$ and the change in $y$ is $+4$. One quarter of the way is $(\tfrac{1}{4}\times8, \tfrac{1}{4}\times4) = (2,1)$.

🧠 PROBLEM-SOLVING Strategy

Finding Coordinates Along a Line Segment

Use this method whenever a point lies a fraction of the way between two coordinates.

Identify the coordinates of the two endpoints. Call them $A(x_1, y_1)$ and $B(x_2, y_2)$.
Work out the vector from $A$ to $B$: $(x_2 - x_1,\; y_2 - y_1)$.
Multiply this vector by the required fraction (e.g., $\tfrac{1}{3}$, $\tfrac{3}{4}$, $\tfrac{2}{5}$).
Add the result to the starting point $A$ to find the new coordinates.
If the point lies between $O(0,0)$ and some point $M(x,y)$, simply multiply the coordinates of $M$ by the fraction.
Always check your result makes sense by plotting it on a diagram — the point should lie on the line and between the two endpoints.

Step	Calculation
$\overrightarrow{AB}$	$(x_2-x_1,\;y_2-y_1)$
Fractional vector	$f \cdot (x_2-x_1,\; y_2-y_1)$
Point on line	$A + f \cdot \overrightarrow{AB}$

❓ EXERCISES

1. Make a copy of this diagram.

a. Write the coordinates of the points A and B.

b. Work out the coordinates of the point that lies one third $\left(\tfrac{1}{3}\right)$ of the way along AB.
Mark this point on your diagram and label it (b).

c. Work out the coordinates of the point that lies two thirds $\left(\tfrac{2}{3}\right)$ of the way along AB.
Mark this point on your diagram and label it (c).

d. Write the coordinates of the points C and D.

e. Work out the coordinates of the point that lies one quarter $\left(\tfrac{1}{4}\right)$ of the way along CD.
Mark this point on your diagram and label it (e).

f. Work out the coordinates of the point that lies three quarters $\left(\tfrac{3}{4}\right)$ of the way along CD.
Mark this point on your diagram and label it (f).

👀 Show answer

a. $A(0,2),\ B(2,2)$
b. $(\tfrac{2}{3},2)$
c. $(\tfrac{4}{3},2)$
d. $C(4,0),\ D(4,9)$
e. $(4,\tfrac{9}{4})$
f. $(4,\tfrac{27}{4})$

2. The diagram shows the line segment PQ.
Cards A to F show a fraction of the way along PQ.
Cards i to vi show coordinates.
Match each card A to F with the correct card i to vi.
The first one has been done for you: A and v.

Card	Fraction	Coordinates
A	$\tfrac{1}{6}$	$(2,2)$
B	$\tfrac{1}{4}$	$(3,3)$
C	$\tfrac{1}{3}$	$(4,4)$
D	$\tfrac{3}{4}$	$(9,9)$
E	$\tfrac{2}{3}$	$(8,8)$
F	$\tfrac{5}{6}$	$(10,10)$

👀 Show answer

A → v $(2,2)$
B → iii $(3,3)$
C → vi $(4,4)$
D → ii $(9,9)$
E → iv $(8,8)$
F → i $(10,10)$

🧠 Think like a Mathematician

Task: Chesa and Tefo use different methods to find the coordinates of the point that lies $\tfrac{2}{5}$ of the way along the line segment ST. S is at (0, 0) and T is at (10, 5).

Follow-up Questions:

a. Critique Chesa’s method.

b. Critique Tefo’s method.

c. Whose method do you prefer? Explain why.

d. Can you think of a better method?

e. Will both Chesa’s and Tefo’s methods work when S is not at the point (0, 0)?

f. Reflect on which approach is most efficient for future problems.

Show Answers

a: Chesa’s method is clear and visual but relies on drawing accurately; it can be time-consuming for complex coordinates.
b: Tefo’s method is algebraic, quick, and avoids drawing errors; it works well for any coordinates.
c: Tefo’s method is usually preferable because it generalises better.
d: A better method is to use the general formula $(x_1+k(x_2-x_1),\ y_1+k(y_2-y_1))$ where $k$ is the fraction along the line.
e: Tefo’s formula works for any coordinates; Chesa’s diagram method only works easily when S is at the origin or simple points.
f: The formula method is most efficient for algebraic work, while the diagram method is useful for building understanding.

❓ EXERCISES

4. $O$ is at the point $(0,0)$, $M$ is at $(16,12)$ and $N$ is at $(10,15)$. Write whether A, B or C is the correct answer. Use your favourite method.

a. $\tfrac{1}{4}$ of the way along $OM$ is A $(3,4)$ B $(4,3)$ C $(4,4)$

👀 Show answer

$\tfrac{1}{4}(16,12) = (4,3)$ so the answer is B.

b. $\tfrac{3}{4}$ of the way along $OM$ is A $(12,9)$ B $(12,8)$ C $(9,12)$

👀 Show answer

$\tfrac{3}{4}(16,12) = (12,9)$ so the answer is A.

c. $\tfrac{1}{5}$ of the way along $ON$ is A $(5,3)$ B $(2,5)$ C $(2,3)$

👀 Show answer

$\tfrac{1}{5}(10,15) = (2,3)$ so the answer is C.

d. $\tfrac{4}{5}$ of the way along $ON$ is A $(12,8)$ B $(8,12)$ C $(8,10)$

👀 Show answer

$\tfrac{4}{5}(10,15) = (8,12)$ so the answer is B.

❓ EXERCISES

5. $\Omega$ (omega) is the point $(0, 0)$ and $A$ is the point $(2, 3)$.

a. Points $A$ and $B$ are equally spaced along the same line such that the distance $\Omega A$ is equal to the distance $AB$. What are the coordinates of point $B$?

Tip

You could draw a diagram to help you.

👀 Show answer

Vector $\overrightarrow{\Omega A}=(2,3)$. Equal spacing gives $B=A+\overrightarrow{\Omega A}=(2,3)+(2,3)=(4,6)$. So $B$ is $(4,6)$.

b. $C$ is the next point along the same line such that the distance $BC$ is equal to distances $\Omega A$ and $AB$. What are the coordinates of point $C$?

👀 Show answer

Step again by $(2,3)$: $C=B+(2,3)=(4,6)+(2,3)=(6,9)$.

c. The points continue along the line, equally spaced. Each point is labelled with a letter of the alphabet, in order from $A$ to $Z$. Show that point $J$ has coordinates $(20, 30)$.

Tip

$J$ is the $10$th letter in the alphabet.

👀 Show answer

Each step adds $(2,3)$. $J$ is the $10$th point, so $J=10\cdot(2,3)=(20,30)$.

d. What are the coordinates of point $P$?
Show how you worked out your answer.

Tip

$P$ is the $16$th letter in the alphabet.

👀 Show answer

$P=16\cdot(2,3)=(32,48)$.

e. What are the coordinates of the point labelled with the $20$th letter in the alphabet?
Show how you worked out your answer.

👀 Show answer

The $20$th letter is $T$. So the point is $20\cdot(2,3)=(40,60)$.

f. Write an expression for the coordinates of the point along the same line labelled with the $n$th letter of the alphabet.

Tip

You could write your expression as: $n$th letter is $(\_,\_)$.

👀 Show answer

Each step adds $(2,3)$, with $A$ as the first. Therefore the $n$th point is $(2n,\,3n)$.

6. $O$ is the point $(0, 0)$ and $D$ is the point $(3, 7)$.

$D$ lies $\tfrac{1}{4}$ of the way along the line segment $OE$.

Sofia says: I think $E$ is the point $(12, 28)$.

Marcus says: I think the ratio of the lengths $OD:DE$ is $1:4$.

a. Is Sofia correct? Justify your answer.

👀 Show answer

Yes. Since $D$ is $\tfrac{1}{4}$ of the way from $O$ to $E$ and $O=(0,0)$, we have $D=\tfrac{1}{4}E$. Thus $E=4D=4\cdot(3,7)=(12,28)$.

b. Is Marcus correct? Justify your answer.

👀 Show answer

No. With $D$ at $\tfrac{1}{4}$ of $OE$, $OD:DE=\tfrac{1}{4}:\tfrac{3}{4}=1:3$, not $1:4$.

7. $O$ is the point $(0, 0)$ and $T$ is the point $(20, 25)$. The points $P$, $Q$, $R$ and $S$ are equally spaced along the line $OT$. Work out the coordinates of $R$.

👀 Show answer

Points $P,Q,R,S$ split $OT$ into $5$ equal parts: $O\!-\!P\!-\!Q\!-\!R\!-\!S\!-\!T$. Therefore $R$ is $\tfrac{3}{5}$ of the way from $O$ to $T$: $$R=\tfrac{3}{5}\,(20,25)=(12,15).$$

🧠 Think like a Mathematician

Task: Work through the following problem and questions step by step.

Follow-up Questions:

a. Only by looking at the diagram, explain why Briana’s coordinates for $C$ must be incorrect.

b. Look carefully at Briana’s solution. Explain the extra steps Briana must do to get the correct answer.

c. Work out the correct coordinates for $C$. Use a diagram to check that your coordinates are correct.

d. Write the ratio of the length of

$AC:AB$
$AC:BC$

e. Reflect on your answers to parts a–d.

👀 show answer

a: From the diagram, $C$ should lie between $A$ and $B$, not to the left of $A$. Briana’s coordinates $(1,2)$ are outside the segment.
b: After finding the differences, Briana should add the fractional amounts to the coordinates of $A$. She only calculated the differences but forgot to shift from $A$.
c: Move $\tfrac{1}{5}$ of the way from $A(3,2)$ towards $B(8,12)$. The vector $\overrightarrow{AB}=(5,10)$. One-fifth is $(1,2)$. So $C=(3+1,2+2)=(4,4)$.
d.i: $AC:AB=1:5$
d.ii: $AC:BC=1:4$
e: The method shows the importance of adding the fraction of the vector to the starting point. This ensures the new point lies along the correct line segment.

❓ EXERCISES

9. $F$ is the point $(3,4)$ and $G$ is the point $(9,13)$. $H$ is the point that lies $\tfrac{2}{3}$ of the way along $\overline{FG}$. Show that $H$ has coordinates $(7,10)$.

👀 Show answer

$\overrightarrow{FG}=(9-3,\,13-4)=(6,9)$. $\tfrac{2}{3}$ of this vector is $(\tfrac{2}{3}\cdot 6,\tfrac{2}{3}\cdot 9)=(4,6)$. So $H=F+(4,6)=(3+4,\,4+6)=(7,10)$.

10. $J$ is the point $(1,5)$ and $K$ is the point $(13,13)$. $L$ is the point that lies $\tfrac{3}{4}$ of the way along $\overline{JK}$.

a. Work out the coordinates of $L$.

b. Use a diagram to show that your answer to part a is correct.

👀 Show answer

$\overrightarrow{JK}=(13-1,\,13-5)=(12,8)$. $\tfrac{3}{4}$ of this is $(\tfrac{3}{4}\cdot 12,\tfrac{3}{4}\cdot 8)=(9,6)$. Therefore $L=J+(9,6)=(1+9,\,5+6)=(10,11)$. For a diagram check, plot $J(1,5)$ and $K(13,13)$; $L(10,11)$ lies on the line segment and divides it in the ratio $3:1$ from $J$.

11. A kite has vertices at $A(1,1)$, $B(2,5)$, $C(5,5)$ and $D(5,2)$.

a. Draw a diagram of kite $ABCD$ on a coordinate grid.

b. On your diagram, draw the diagonals $AC$ and $BD$.

c. Line segments $AC$ and $BD$ cross at point $E$. Write the coordinates of $E$.

d. Show, using calculations, that $E$ is the midpoint of $BD$.

e. Show, using calculations, that $E$ lies $\tfrac{5}{8}$ of the way along $AC$.

👀 Show answer

c. $AC$ has equation $y=x$ (from $(1,1)$ to $(5,5)$). $BD$ has slope $-1$; through $B(2,5)$ its equation is $y=-x+7$. Intersection solves $x=-x+7\Rightarrow 2x=7\Rightarrow x=\tfrac{7}{2}$, $y=\tfrac{7}{2}$. Hence $E\!=\!\big(\tfrac{7}{2},\tfrac{7}{2}\big)=(3.5,3.5)$.

d. Midpoint of $BD$ is $\big(\tfrac{2+5}{2},\tfrac{5+2}{2}\big)=\big(\tfrac{7}{2},\tfrac{7}{2}\big)=E$ ✅.

e. $\overrightarrow{AC}=(4,4)$. $\tfrac{5}{8}$ of this from $A$ gives $A+\tfrac{5}{8}\overrightarrow{AC}=(1,1)+\big(\tfrac{20}{8},\tfrac{20}{8}\big)=(3.5,3.5)=E$.

12. $F$ is the point $(5,1)$ and $L$ is the point $(17,19)$. Points $G$, $H$, $I$, $J$, $K$ and $L$ are equally spaced along the line $FL$.

Which of the points $G$, $H$, $I$, $J$, $K$ and $L$ is the only point to have the same $x$ and $y$ coordinate?
Show all your working.

👀 Show answer

The sequence $F,G,H,I,J,K,L$ makes $6$ equal steps from $F$ to $L$. Step vector $=\dfrac{L-F}{6}=\left(\dfrac{17-5}{6},\,\dfrac{19-1}{6}\right)=(2,3)$. Coordinates: $G=(7,4)$, $H=(9,7)$, $I=(11,10)$, $J=(13,13)$, $K=(15,16)$, $L=(17,19)$. Only $J$ has equal coordinates, so the answer is $J(13,13)$.

📘 What we've learned

A point can lie at a fraction of the way along a line segment between two coordinates.
To find such a point, use vector methods: $P = A + f \cdot (B - A)$, where $f$ is the fraction along $AB$.
If the line starts at the origin $(0,0)$, simply multiply the endpoint coordinates by the fraction.
The midpoint formula is a special case: $M = \Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big)$.
When the midpoint and one endpoint are given, the other endpoint can be found by rearranging the midpoint formula.
We practiced applying these rules to work out coordinates of points such as quarters, thirds, or other fractional positions along a segment.

Points on a line segment

🎯 In this topic you will

🧠 Key Words

📐 Midpoints of Line Segments

🧠 PROBLEM-SOLVING Strategy

Finding the Midpoint of a Line Segment

❓ EXERCISES

❓ EXERCISES

🧠 Think like a Mathematician

❓ EXERCISES

❓ EXERCISES

❓ EXERCISES

🧠 Think like a Mathematician

❓ EXERCISES

🧠 Think like a Mathematician

❓ EXERCISES

📍 Midpoints and Beyond

🧠 PROBLEM-SOLVING Strategy

Finding Coordinates Along a Line Segment

❓ EXERCISES

🧠 Think like a Mathematician

❓ EXERCISES

❓ EXERCISES

🧠 Think like a Mathematician

❓ EXERCISES

📘 What we've learned

Related Past Papers

Related Tutorials

Blog