MathsCBSE Class 10 · Coordinate Geometry

Finding the Ratio and Unknowns with the Section Formula

Updated 13 October 2026 · Fact-checked

This topic uses the section formula backwards. Instead of finding the dividing point, you find the ratio in which a point or an axis divides a segment, or an unknown coordinate or vertex. The board asks 2- and 3-mark ratio questions using the k : 1 method, and longer questions on parallelogram vertices and missing end points.

Set the y-coordinate of P to 0 for the x-axis, solve for k, then write k : 1 as a whole-number ratio, here 1 : 2.

Illustration drawn for QuestionBuddy.

Understand Finding the Ratio and Unknowns with Section Formula

In the forward use of the section formula you know the two ends A and B and the ratio, and you find the dividing point P. Board papers often turn this around. You may know P only partly, or not know the ratio, or not know one end. The formula still links all of them, so you treat it as an equation and solve for whatever is missing.

The k : 1 idea. An unknown ratio $m_{1} : m_{2}$ has two unknowns, but only their ratio matters. Divide both by $m_{2}$ and it becomes $k : 1$ with $k = \frac{m _{1}}{m _{2}}$ . Now the point is $(\frac{k x _{2} + x _{1}}{k + 1}, \frac{k y _{2} + y _{1}}{k + 1})$ and you have only one unknown. One fact about P, such as a known coordinate, gives one equation for k.

What you set equal. If P lies on the x-axis, its y-coordinate is 0. If P lies on the y-axis, its x-coordinate is 0. If P is given as $(- 2, y)$ , its x-coordinate is $- 2$ . Solve for k, write the ratio in whole numbers (for example $k = \frac{1}{2}$ gives $1 : 2$ ), then put k into the other coordinate to finish P.

Why k must be positive. For internal division, P lies between A and B, so both parts AP and PB point the same way and k is positive. A negative k would mean P is outside the segment, which is external division, not in your syllabus. An axis meets segment AB internally only when A and B lie on opposite sides of it, so the relevant coordinates of A and B have opposite signs. That is a quick check before you start.

Unknown ends and vertices. If the ratio and the dividing point are known, equate each coordinate separately to find a missing end. The most common case is a parallelogram: its diagonals bisect each other, so the midpoint of AC equals the midpoint of BD. This gives $x_{A} + x_{C} = x_{B} + x_{D}$ and $y_{A} + y_{C} = y_{B} + y_{D}$ , which find a missing vertex in two lines. The vertices must be taken in order.

These questions often finish with a distance formula check, such as showing a side length or confirming the figure, so this topic ties the two main tools of the chapter together.

Finding the Ratio and Unknowns with Section Formula: formulas to remember

Section formula in k : 1 form: $P = (\frac{k x _{2} + x _{1}}{k + 1}, \frac{k y _{2} + y _{1}}{k + 1})$; P divides $A (x_{1}, y_{1})$ and $B (x_{2}, y_{2})$ internally with $A P : P B = k : 1$ . k pairs with B.
Ratio in which the x-axis divides AB: $\frac{k y _{2} + y _{1}}{k + 1} = 0 \Rightarrow k = - \frac{y _{1}}{y _{2}}$; Works only when $y_{1}$ and $y_{2}$ have opposite signs, so k is positive. Show the equation in the exam, not just the shortcut.
Ratio in which the y-axis divides AB: $\frac{k x _{2} + x _{1}}{k + 1} = 0 \Rightarrow k = - \frac{x _{1}}{x _{2}}$; Needs $x_{1}$ and $x_{2}$ of opposite signs.
Missing end from a known ratio: $\frac{m _{1} x _{2} + m _{2} x _{1}}{m _{1} + m _{2}} = x_{P}, \frac{m _{1} y _{2} + m _{2} y _{1}}{m _{1} + m _{2}} = y_{P}$; Equate each coordinate separately and solve for the unknown end.
Fourth vertex of parallelogram ABCD: $D = (x_{A} + x_{C} - x_{B}, y_{A} + y_{C} - y_{B})$; From midpoint of AC = midpoint of BD. Vertices in order, A opposite C.
Missing end from the midpoint: $B = (2 x_{M} - x_{A}, 2 y_{M} - y_{A})$; M is the midpoint of AB.

How to solve Finding the Ratio and Unknowns with Section Formula questions

Use this method for any ratio or unknown-coordinate question. Write each step, because the board gives marks for the set-up as well as the answer.

1Write the given points clearly as $A (x_{1}, y_{1})$ and $B (x_{2}, y_{2})$ , keeping the order the question uses.
2Decide what is unknown: the ratio, a coordinate of P, or an end point or vertex.
3If the ratio is unknown, take it as $k : 1$ and write P using the k : 1 form of the section formula.
4Use the known fact about P: 0 for a point on an axis, or the given coordinate. Form one equation in k and solve it.
5Check k is positive, then write the ratio in whole numbers, such as $\frac{2}{3}$ giving $2 : 3$ .
6Substitute k into the other coordinate to find the complete point P.
7For a missing vertex of a parallelogram, equate the midpoints of the diagonals; for a missing end, equate coordinates in the section formula.
8Verify: P must lie between A and B, and when both coordinates of P are given, both must give the same k.

Finding the Ratio and Unknowns with Section Formula in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Finding the ratio of division: the k : 1 method

Suppose a plumber lays a straight pipe from the water tank at $T (- 4, 7)$ to a kitchen tap at $K (8, - 2)$ on a house plan drawn in metres. A wall of the house runs along the y-axis, and the pipe must pass through a hole drilled in that wall.

Find the ratio in which the wall divides the pipe TK, the position of the hole, and the length of pipe on each side of the wall.

Try it first, then show the solutionSolution

Let the hole H divide TK in the ratio $k : 1$ .
x-coordinate of H: $\frac{8 k - 4}{k + 1} = 0$ because H is on the y-axis.
$8 k - 4 = 0$ , so $k = \frac{1}{2}$ ; the ratio is $1 : 2$ .
y-coordinate of H: $\frac{1 ( - 2 ) + 2 ( 7 )}{1 + 2} = \frac{12}{3} = 4$ , so H is $(0, 4)$ .
Length $T K = (8 + 4)^{2} + (- 2 - 7)^{2} = 144 + 81 = 225 = 15$ m.
Tank side $= \frac{1}{3} \times 15 = 5$ m, tap side $= \frac{2}{3} \times 15 = 10$ m.
Check: $T H = 16 + 9 = 5$ m.

Answer: Ratio $1 : 2$ ; the hole is at $(0, 4)$ ; 5 m of pipe on the tank side and 10 m on the tap side.

Memory hook: The wall is the y-axis, so the hole's x is zero: set x = 0 and k falls out.

Tricks to remember Finding the Ratio and Unknowns with Section Formula

Short memory aids. Say each one aloud once, then use it on a question the same day.

Axis means zero
Crossing the x-axis? Kill the y. Crossing the y-axis? Kill the x. Set that coordinate to 0 in the k : 1 form.
Use it when: Every 'in what ratio does the axis divide' question.
k sits with the far end
In $k : 1$ , k is the part AP next to A, but it multiplies B's coordinates: 'k goes across'.
Use it when: Writing the section formula correctly before solving.
Opposite signs, then it crosses
An axis cuts AB internally only if the two points sit on opposite sides, so the matching coordinates have opposite signs.
Use it when: A 10-second check before you start, and a check that k is positive.
Diagonal sums match
In parallelogram ABCD: A + C = B + D, coordinate by coordinate.
Use it when: Finding a fourth vertex in two lines.

Finding the Ratio and Unknowns with Section Formula: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Ratio in which the y-axis divides a segment · 2 marks · Very short answerBoard level

Find the ratio in which the y-axis divides the line segment joining $A (- 4, 3)$ and $B (2, - 6)$ . Also find the point of intersection.

Try it first, then show the solutionSolution

Let the y-axis divide AB at P in the ratio $k : 1$ . P is on the y-axis, so its x-coordinate is 0.
$\frac{k ( 2 ) + ( - 4 )}{k + 1} = 0 \Rightarrow 2 k - 4 = 0 \Rightarrow k = 2$ , so the ratio is $2 : 1$ . (1 mark)
y-coordinate of P: $\frac{2 ( - 6 ) + 1 ( 3 )}{2 + 1} = \frac{- 9}{3} = - 3$ . (½ mark)
So P is $(0, - 3)$ . Check: x-coordinates $- 4$ and $2$ have opposite signs, and k is positive. (½ mark)

Answer: Ratio $2 : 1$ ; point of intersection $(0, - 3)$ .

Example 2: Ratio when one coordinate of the point is given · 3 marks · Short answerBoard level

Find the ratio in which the point $P (- 2, y)$ divides the line segment joining $A (- 5, 8)$ and $B (4, - 1)$ . Hence find y.

Try it first, then show the solutionSolution

Let $A P : P B = k : 1$ . Then $x_{P} = \frac{4 k - 5}{k + 1}$ . (½ mark)
Given $x_{P} = - 2$ : $4 k - 5 = - 2 k - 2$ . (½ mark)
$6 k = 3 \Rightarrow k = \frac{1}{2}$ , so the ratio is $1 : 2$ . (1 mark)
$y = \frac{1 ( - 1 ) + 2 ( 8 )}{1 + 2} = \frac{15}{3} = 5$ . (½ mark)
Check: P $(- 2, 5)$ ; step A to P is $(3, - 3)$ , step P to B is $(6, - 6)$ , ratio $1 : 2$ . (½ mark)

Answer: Ratio $1 : 2$ ; $y = 5$ , so P is $(- 2, 5)$ .

Example 3: Finding an unknown end point and a coordinate · 3 marks · Short answerBoard level

The point $P (2, b)$ divides the line segment joining $A (a, 7)$ and $B (5, - 5)$ internally in the ratio $1 : 3$ . Find a and b.

Try it first, then show the solutionSolution

Here $m_{1} = 1$ , $m_{2} = 3$ . x-coordinate: $\frac{1 ( 5 ) + 3 a}{1 + 3} = 2$ . (½ mark)
$5 + 3 a = 8 \Rightarrow 3 a = 3 \Rightarrow a = 1$ . (1 mark)
y-coordinate: $b = \frac{1 ( - 5 ) + 3 ( 7 )}{4} = \frac{16}{4} = 4$ . (1 mark)
Check: A $(1, 7)$ , P $(2, 4)$ , B $(5, - 5)$ ; steps $(1, - 3)$ and $(3, - 9)$ are in ratio $1 : 3$ . (½ mark)

Answer: $a = 1$ , $b = 4$ .

Example 4: Ratio in which a given line divides a segment · 3 marks · Short answerHarder board level

Find the ratio in which the line $x - y - 2 = 0$ divides the line segment joining $A (3, - 1)$ and $B (8, 9)$ . Also find the point of division.

Try it first, then show the solutionSolution

Let the line divide AB at P in the ratio $k : 1$ . Then $P = (\frac{8 k + 3}{k + 1}, \frac{9 k - 1}{k + 1})$ . (½ mark)
P lies on the line, so its coordinates satisfy it: $\frac{8 k + 3}{k + 1} - \frac{9 k - 1}{k + 1} - 2 = 0$ . (½ mark)
Multiply by $k + 1$ : $8 k + 3 - 9 k + 1 - 2 k - 2 = 0 \Rightarrow - 3 k + 2 = 0 \Rightarrow k = \frac{2}{3}$ , so the ratio is $2 : 3$ . (1 mark)
$P = (\frac{2 ( 8 ) + 3 ( 3 )}{5}, \frac{2 ( 9 ) + 3 ( - 1 )}{5}) = (\frac{25}{5}, \frac{15}{5}) = (5, 3)$ . (½ mark)
Check: $5 - 3 - 2 = 0$ , so P is on the line, and k is positive, so the division is internal. (½ mark)

Answer: Ratio $2 : 3$ ; point of division $(5, 3)$ .

Example 5: Parallelogram vertex, then the ratio from an axis · 5 marks · Long answerHarder board level

Three vertices of parallelogram ABCD, taken in order, are $A (- 2, - 1)$ , $B (4, 0)$ and $C (3, 3)$ .
(i) Find the fourth vertex D.
(ii) Verify that $A D = B C$ .
(iii) Find the ratio in which the x-axis divides side AD, and the point of division.

Try it first, then show the solutionSolution

Let D be $(x, y)$ . Diagonals of a parallelogram bisect each other, so midpoint of AC = midpoint of BD. (½ mark)
Midpoint of AC $= (\frac{- 2 + 3}{2}, \frac{- 1 + 3}{2}) = (\frac{1}{2}, 1)$ . (½ mark)
$\frac{4 + x}{2} = \frac{1}{2} \Rightarrow x = - 3$ ; $\frac{0 + y}{2} = 1 \Rightarrow y = 2$ . So D is $(- 3, 2)$ . (1 mark)
$A D = (- 3 + 2)^{2} + (2 + 1)^{2} = 10$ units. (½ mark)
$B C = (3 - 4)^{2} + (3 - 0)^{2} = 10$ units, so $A D = B C$ . (½ mark)
Let the x-axis divide AD in ratio $k : 1$ : $\frac{2 k + ( - 1 )}{k + 1} = 0 \Rightarrow k = \frac{1}{2}$ , ratio $1 : 2$ . (1 mark)
x-coordinate: $\frac{1 ( - 3 ) + 2 ( - 2 )}{3} = - \frac{7}{3}$ , so the point is $(- \frac{7}{3}, 0)$ . (1 mark)

Answer: (i) $D (- 3, 2)$ ; (ii) $A D = B C = 10$ units; (iii) ratio $1 : 2$ , point $(- \frac{7}{3}, 0)$ .

Example 6: Two unknowns from a ratio, then a parallelogram vertex · 5 marks · Long answerHarder board level

The point $P (2, 3)$ divides the line segment joining $A (p, 5)$ and $B (6, q)$ internally in the ratio $1 : 2$ .
(i) Find p and q.
(ii) ABCD is a parallelogram, with vertices taken in order, and $C$ is $(8, 3)$ . Find the fourth vertex D.
(iii) Use the distance formula to verify that both pairs of opposite sides of ABCD are equal.

Try it first, then show the solutionSolution

Here $m_{1} = 1$ , $m_{2} = 2$ . x-coordinate: $\frac{1 ( 6 ) + 2 p}{3} = 2$ . (½ mark)
$6 + 2 p = 6 \Rightarrow p = 0$ . (½ mark)
y-coordinate: $\frac{1 ( q ) + 2 ( 5 )}{3} = 3$ . (½ mark)
$q + 10 = 9 \Rightarrow q = - 1$ . So A is $(0, 5)$ and B is $(6, - 1)$ . (½ mark)
Diagonals of a parallelogram bisect each other, so midpoint of AC = midpoint of BD. Midpoint of AC $= (\frac{0 + 8}{2}, \frac{5 + 3}{2}) = (4, 4)$ . (½ mark)
Let D be $(x, y)$ : $\frac{6 + x}{2} = 4 \Rightarrow x = 2$ ; $\frac{- 1 + y}{2} = 4 \Rightarrow y = 9$ . So D is $(2, 9)$ . (1 mark)
$A B = (6 - 0)^{2} + (- 1 - 5)^{2} = 72 = 62$ ; $C D = (2 - 8)^{2} + (9 - 3)^{2} = 72 = 62$ . (½ mark)
$B C = (8 - 6)^{2} + (3 + 1)^{2} = 20 = 25$ ; $A D = (2 - 0)^{2} + (9 - 5)^{2} = 20 = 25$ . (½ mark)
So $A B = C D$ and $B C = A D$ , which agrees with ABCD being a parallelogram. Also check P: A to P is $(2, - 2)$ , P to B is $(4, - 4)$ , ratio $1 : 2$ . (½ mark)

Answer: (i) $p = 0$ , $q = - 1$ ; (ii) $D (2, 9)$ ; (iii) $A B = C D = 62$ units and $B C = A D = 25$ units.

Example 7: Case study: a relay line on the school ground · 4 marks · Case-basedHOTS / competency

Imagine your school marks a straight relay line on the ground plan, in metres, from $A (- 3, 9)$ to $B (9, - 3)$ . Flags are placed where the line crosses the two axes.
(i) Find the ratio in which the y-axis divides AB and the flag point on it. (1 mark)
(ii) Find the ratio in which the x-axis divides AB and the flag point on it. (1 mark)
(iii) Find the distance between the two flags and show it is half of AB. (2 marks)

Try it first, then show the solutionSolution

(i) Ratio $k : 1$ : $\frac{9 k - 3}{k + 1} = 0 \Rightarrow k = \frac{1}{3}$ , ratio $1 : 3$ ; y-coordinate $\frac{1 ( - 3 ) + 3 ( 9 )}{4} = 6$ , flag at $(0, 6)$ . (1 mark)
(ii) $\frac{- 3 k + 9}{k + 1} = 0 \Rightarrow k = 3$ , ratio $3 : 1$ ; x-coordinate $\frac{3 ( 9 ) + 1 ( - 3 )}{4} = 6$ , flag at $(6, 0)$ . (1 mark)
(iii) Distance between flags $= (6 - 0)^{2} + (0 - 6)^{2} = 72 = 62$ m. (1 mark)
$A B = 1 2^{2} + 1 2^{2} = 122$ m, and $62 = \frac{1}{2} \times 122$ . This fits the ratios: the flags are at the $\frac{1}{4}$ and $\frac{3}{4}$ marks of AB. (1 mark)

Answer: (i) $1 : 3$ , flag at $(0, 6)$ ; (ii) $3 : 1$ , flag at $(6, 0)$ ; (iii) $62$ m, which is half of $A B = 122$ m.

Where students lose marks in Finding the Ratio and Unknowns with Section Formula

The slips examiners see most often, and what to write instead.

Mistake: Pairing k with the wrong end point
Students think 'k is next to A, so it multiplies A'. In fact k (the part AP) multiplies B's coordinates.
Wrong working
For $A (- 4, 3)$ , $B (2, - 6)$ and the y-axis: $\frac{k ( - 4 ) + 2}{k + 1} = 0 \Rightarrow k = \frac{1}{2}$ , ratio $1 : 2$ .
Right working
$\frac{k ( 2 ) + ( - 4 )}{k + 1} = 0 \Rightarrow k = 2$ , ratio $2 : 1$ .
Do this: Write the formula $\frac{k x _{2} + x _{1}}{k + 1}$ first, then substitute.
Usually costs: The ratio and the point are both wrong; usually only ½ mark for the set-up survives.
Mistake: Setting the wrong coordinate to zero
Mixing up 'on the x-axis' with 'x = 0'.
Wrong working
x-axis divides AB, so $\frac{k x _{2} + x _{1}}{k + 1} = 0$ .
Right working
On the x-axis the y-coordinate is 0, so $\frac{k y _{2} + y _{1}}{k + 1} = 0$ .
Do this: Say it aloud: on the x-axis, y is zero.
Usually costs: The whole question, since every later step depends on k.
Mistake: Stopping after the ratio
The question says 'also find the point', but students forget to substitute k into the other coordinate.
Wrong working
Ratio $2 : 1$ . (End of answer.)
Right working
Ratio $2 : 1$ ; y-coordinate $\frac{2 ( - 6 ) + 3}{3} = - 3$ , so the point is $(0, - 3)$ .
Do this: Underline every 'find' in the question and tick each one off.
Usually costs: ½ to 1 mark.
Mistake: Taking parallelogram vertices out of order
Students equate midpoints of AB and CD instead of the diagonals AC and BD.
Wrong working
$A (- 2, - 1)$ , $B (4, 0)$ , $C (3, 3)$ : midpoint of AB = midpoint of CD gives D $= (- 1, - 4)$ .
Right working
Midpoint of AC = midpoint of BD gives D $= (- 2 + 3 - 4, - 1 + 3 - 0) = (- 3, 2)$ .
Do this: Diagonals join opposite vertices: A with C, B with D.
Usually costs: Usually 2 or more marks in a 5-mark question.

Exam tips

Always start a ratio question with 'Let the required ratio be $k : 1$ ' and the general point. Examiners give marks for this set-up line.
Convert a fractional k to a whole-number ratio in the final answer, such as $1 : 2$ , not $\frac{1}{2} : 1$ .
When both coordinates of the point are given, solve with one and verify with the other; this catches arithmetic slips.
In parallelogram questions, write the reason 'diagonals of a parallelogram bisect each other' before equating midpoints.
Do a sense check at the end: the dividing point must lie between the two ends, so each coordinate lies between theirs.

Practice questions on Finding the Ratio and Unknowns with Section Formula

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 2 marks · Very short answerBoard level
Find the ratio in which the x-axis divides the line segment joining $A (3, - 6)$ and $B (- 5, 2)$ . Also find the point of division.
Hint
On the x-axis the y-coordinate is 0. Take the ratio as $k : 1$ .
Answer
$\frac{2 k - 6}{k + 1} = 0 \Rightarrow k = 3$ , ratio $3 : 1$ . x-coordinate $= \frac{3 ( - 5 ) + 3}{4} = - 3$ . Point $(- 3, 0)$ .
Q2 · 3 marks · Short answerBoard level
In what ratio does the point $P (x, 2)$ divide the line segment joining $A (- 4, 6)$ and $B (8, - 3)$ ? Find x.
Hint
Use the y-coordinate first, since it is known.
Answer
$\frac{- 3 k + 6}{k + 1} = 2 \Rightarrow 5 k = 4 \Rightarrow k = \frac{4}{5}$ , ratio $4 : 5$ . $x = \frac{4 ( 8 ) + 5 ( - 4 )}{9} = \frac{12}{9} = \frac{4}{3}$ .
Q3 · 3 marks · Short answerBoard level
The point $P (2 a - 1, b + 3)$ divides the line segment joining $A (- 3, 6)$ and $B (7, - 4)$ internally in the ratio $2 : 3$ . Find the values of a and b, and write the coordinates of P.
Hint
First find P from the section formula with $m_{1} = 2$ , $m_{2} = 3$ . Then equate its coordinates to $2 a - 1$ and $b + 3$ .
Answer
$x_{P} = \frac{2 ( 7 ) + 3 ( - 3 )}{5} = \frac{5}{5} = 1$ and $y_{P} = \frac{2 ( - 4 ) + 3 ( 6 )}{5} = \frac{10}{5} = 2$ , so P is $(1, 2)$ . Then $2 a - 1 = 1 \Rightarrow a = 1$ and $b + 3 = 2 \Rightarrow b = - 1$ . Check: A to P is $(4, - 4)$ , P to B is $(6, - 6)$ , ratio $2 : 3$ .
Q4 · 3 marks · Short answerHarder board level
The point P divides the line segment joining $A (- 1, 7)$ and $B (5, - 5)$ internally in the ratio $1 : 2$ , and Q is the midpoint of PB. Find P and Q. Then use the distance formula to show that $A P = P Q = QB$ , so P and Q are the points of trisection of AB.
Hint
Find P first; Q depends on it. Then compute three distances.
Answer
$P = (\frac{1 ( 5 ) + 2 ( - 1 )}{3}, \frac{1 ( - 5 ) + 2 ( 7 )}{3}) = (1, 3)$ . $Q = (\frac{1 + 5}{2}, \frac{3 - 5}{2}) = (3, - 1)$ . $A P = 2^{2} + 4^{2} = 25$ , $P Q = 2^{2} + 4^{2} = 25$ , $QB = 2^{2} + 4^{2} = 25$ units. All three parts are equal, so P and Q trisect AB.
Q5 · 5 marks · Long answerHarder board level
$A (1, 1)$ , $B (5, 2)$ and $C (6, 6)$ are three vertices of parallelogram ABCD, taken in order. Find D. Show that ABCD is a rhombus and find its area using its diagonals.
Hint
$D = A + C - B$ coordinate by coordinate. Then compare AB and BC. Area of a rhombus $= \frac{1}{2} d_{1} d_{2}$ .
Answer
D $= (1 + 6 - 5, 1 + 6 - 2) = (2, 5)$ . $A B = 16 + 1 = 17$ , $B C = 1 + 16 = 17$ ; adjacent sides of a parallelogram are equal, so it is a rhombus (also $C D = D A = 17$ ). $A C = 25 + 25 = 52$ , $B D = 9 + 9 = 32$ . Area $= \frac{1}{2} \times 52 \times 32 = 15$ square units.
Q6 · 3 marks · Short answerHOTS / competency
A student finds the ratio in which the x-axis divides the segment joining $A (1, - 2)$ and $B (4, 6)$ . The student writes $\frac{k ( - 2 ) + 6}{k + 1} = 0$ , gets $k = 3$ and says the ratio is $3 : 1$ . Find the error, give the correct ratio and point, and explain a quick check that shows $3 : 1$ cannot be right.
Hint
Which point's coordinate should k multiply? Which end is the axis closer to?
Answer
The student paired k with A instead of B. Correct: $\frac{6 k - 2}{k + 1} = 0 \Rightarrow k = \frac{1}{3}$ , ratio $1 : 3$ . x-coordinate $= \frac{1 ( 4 ) + 3 ( 1 )}{4} = \frac{7}{4}$ , point $(\frac{7}{4}, 0)$ . Check: A is 2 units below the x-axis and B is 6 units above, so the crossing is nearer A and AP must be the smaller part.
Q7 · 4 marks · Case-basedHOTS / competency
Case study: Imagine a city map drawn with 1 unit = 1 km. A straight metro line is planned from station $A (- 4, - 2)$ to station $B (8, 7)$ . A highway runs exactly along the x-axis of the map.
(i) Find the length of the metro line AB. (1 mark)
(ii) A new station S is to be built on AB so that $A S : S B = 1 : 2$ . Find the coordinates of S. (1 mark)
(iii) The metro line crosses the highway at H. Find the ratio $A H : H B$ , the coordinates of H, and the distance AH. (2 marks)
Hint
For (iii), H is on the x-axis, so its y-coordinate is 0. Once you know the ratio, AH is that fraction of AB.
Answer
(i) $A B = (8 + 4)^{2} + (7 + 2)^{2} = 144 + 81 = 225 = 15$ km. (ii) $S = (\frac{1 ( 8 ) + 2 ( - 4 )}{3}, \frac{1 ( 7 ) + 2 ( - 2 )}{3}) = (0, 1)$ . (iii) Let $A H : H B = k : 1$ : $\frac{7 k - 2}{k + 1} = 0 \Rightarrow k = \frac{2}{7}$ , ratio $2 : 7$ . x-coordinate $= \frac{2 ( 8 ) + 7 ( - 4 )}{9} = - \frac{12}{9} = - \frac{4}{3}$ , so H is $(- \frac{4}{3}, 0)$ . $A H = \frac{2}{9} \times 15 = \frac{10}{3}$ km. Check: $A H = (\frac{8}{3})^{2} + 2^{2} = \frac{100}{9} = \frac{10}{3}$ km.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Why take an unknown ratio as $k : 1$ ?Show answer
It reduces two unknowns to one, so a single condition finds the ratio.
2.Which coordinate is 0 when a point lies on the x-axis?Show answer
The y-coordinate.
3.What must be true about k for internal division?Show answer
k must be positive.
4.Fourth vertex D of parallelogram ABCD?Show answer
$D = (x_{A} + x_{C} - x_{B}, y_{A} + y_{C} - y_{B})$ .
5.If $k = \frac{3}{4}$ , what is the ratio?Show answer
$3 : 4$ .

Finding the Ratio and Unknowns with Section Formula: frequently asked questions

How do I find the ratio in which the x-axis divides a line segment?

Take the ratio as k : 1 and write the y-coordinate of the dividing point using the section formula. Set it equal to 0, because every point on the x-axis has y-coordinate 0, and solve for k. Then put k into the x-coordinate to get the point.

What if I get a negative value of k?

A negative k means the point does not lie between the two ends, so there is no internal division. Recheck your working first, especially whether k multiplies the second point. In board questions on internal division, k comes out positive.

Can I use m1 : m2 instead of k : 1?

Yes, but you will have two unknowns and only their ratio can be found, so it takes longer. The k : 1 form is quicker and is the standard method examiners expect.

Is external division or area of a triangle from coordinates in this topic?

No. Both were removed and are not in the 2026-27 syllabus. Only internal division is tested.