MathsICSE Class 9 · Coordinate Geometry

The Distance Formula: Lengths, Unknown Coordinates and Identifying Triangles and Squares (ICSE Class 9)

Updated 17 October 2026 · Fact-checked

The distance between P(x₁, y₁) and Q(x₂, y₂) is PQ = √((x₂ − x₁)² + (y₂ − y₁)²), from Pythagoras. Use it to find lengths, solve for an unknown coordinate, find a point equidistant from two points, test collinearity, and show that a figure is isosceles, right-angled, a rhombus, a rectangle or a square.

The gaps in x and y are the legs of a right triangle, so PQ is its hypotenuse: square, add, then take the root.

Illustration drawn for QuestionBuddy.

Understand The Distance Formula and Identifying Figures from Coordinates

Formula. For $P (x_{1}, y_{1})$ and $Q (x_{2}, y_{2})$ the horizontal and vertical gaps $(x_{2} - x_{1})$ and $(y_{2} - y_{1})$ are the legs of a right triangle with hypotenuse $P Q$ , so
$ $P Q = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} .$ $
The distance from the origin is $x^{2} + y^{2}$ . Because the differences are squared, the order of the points does not matter.

Unknown coordinate. If $P Q$ is given, square both sides and solve; a quadratic typically gives two values (the two positions of the point on a line). Example: the distance from $(k, 3)$ to $(2, 7)$ is $5$ gives $(k - 2)^{2} + 16 = 25$ , so $k = 5$ or $k = - 1$ .

Equidistant points. For a point $(x, 0)$ on the x-axis equidistant from $A$ and $B$ , equate the squares of the distances $P A^{2} = P B^{2}$ ; the $x^{2}$ terms cancel and a linear equation remains. The same works for a point $(0, y)$ on the y-axis.

Identifying figures. Find all relevant distances and compare.

Collinear: the two shorter distances add to the longest.
Isosceles / equilateral triangle: two / three equal sides.
Right-angled triangle: longest $^{2} =$ sum of the squares of the other two (converse of Pythagoras).
Rhombus: four equal sides. Square: four equal sides and equal diagonals. Rectangle: opposite sides equal and equal diagonals. Parallelogram: opposite sides equal (and the diagonals bisect each other).

Write-up. Show every distance as a surd in simplest form (for example $18 = 32$ ), compare squares if surds are messy, then state the conclusion with its reason.

Level. Class IX is a school-exam class. Expect a 3 mark distance question and a 4 mark proof about a figure. Proofs of squares and of circumcentre-type problems are among the hardest items on this page.

The Distance Formula and Identifying Figures from Coordinates: formulas to remember

Distance: $P Q = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$; Any order of the two points.
From the origin: $O P = x^{2} + y^{2}$; Special case.
Collinear points: $P Q + QR = P R$; Q lies between P and R.
Right angle (converse of Pythagoras): $A B^{2} + B C^{2} = A C^{2}$; Right angle at B.
Square test: $4 equal sides and A C = B D$; Rhombus plus equal diagonals.

How to solve The Distance Formula and Identifying Figures from Coordinates questions

Subtract coordinates in the same order, square, add, root.

1Write the two points and label $x_{1}, y_{1}, x_{2}, y_{2}$ .
2Compute $(x_{2} - x_{1})^{2}$ and $(y_{2} - y_{1})^{2}$ separately, watching signs.
3Add, then take the square root; simplify the surd.
4For an unknown, equate the squares and solve; give all roots.
5For a figure, find all sides (and diagonals) and state which property they prove.
6Write the conclusion in words.

The Distance Formula and Identifying Figures from Coordinates 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.

Walking distance on a map

Suppose a town map has a library at $(1, - 2)$ and a stadium at $(9, 4)$ , with 1 unit = 100 m.

Find the straight-line distance between them.

Try it first, then show the solutionSolution

$d = (9 - 1)^{2} + (4 + 2)^{2} = 64 + 36 = 10$ units.
$10 \times 100 = 1000$ m.

Answer: 10 units, that is 1000 m.

Memory hook: A 6-8-10 triangle on the map.

A water tank for three houses

Imagine houses at $P (0, 0)$ , $Q (6, 0)$ and $R (0, 8)$ sharing a tank at the mid-point of $QR$ .

Show that the tank at $(3, 4)$ is equally far from all three houses.

Try it first, then show the solutionSolution

$QR = 36 + 64 = 10$ , and the tank is at the mid-point $(3, 4)$ .
Distances: to $P$ : $9 + 16 = 5$ ; to $Q$ : $9 + 16 = 5$ ; to $R$ : $9 + 16 = 5$ .

Answer: Each distance is 5 units.

Memory hook: The mid-point of the hypotenuse is equidistant from the three vertices.

Tricks to remember The Distance Formula and Identifying Figures from Coordinates

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

Squares first
Handle signs by squaring each difference; negatives disappear.
Use it when: Avoiding sign errors.
Two roots
A quadratic from a distance gives two positions; check both.
Use it when: Unknown coordinate questions.
Longest = sum
For collinear points the biggest distance is the sum of the other two.
Use it when: Collinearity proofs.

The Distance Formula and Identifying Figures from Coordinates: solved examples

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

Example 1: Distance between two points · 1 mark · MCQBoard level

The distance between $(- 2, 3)$ and $(1, 7)$ is:
(a) 3
(b) 4
(c) 5
(d) 7

Try it first, then show the solutionSolution

$d = (1 + 2)^{2} + (7 - 3)^{2} = 9 + 16 = 5$ .

Answer: (c) 5

Example 2: Unknown coordinate · 3 marks · Short answerBoard level

Find the distance between $(1, - 2)$ and $(9, 4)$ . Also find $k$ if the distance between $(k, 3)$ and $(2, 7)$ is $5$ .

Try it first, then show the solutionSolution

First distance: $64 + 36 = 10$ .
$(k - 2)^{2} + 16 = 25$ , so $(k - 2)^{2} = 9$ and $k - 2 = \pm 3$ .
$k = 5$ or $k = - 1$ .

Answer: Distance 10; k = 5 or −1.

Example 3: Point on the x-axis equidistant from two points · 4 marks · Short answerHarder board level

Find the point on the x-axis equidistant from $A (0, 3)$ and $B (7, - 4)$ and the common distance.

Try it first, then show the solutionSolution

Let $P = (x, 0)$ . $P A^{2} = x^{2} + 9$ and $P B^{2} = (x - 7)^{2} + 16$ .
Equate: $x^{2} + 9 = x^{2} - 14 x + 65$ , so $14 x = 56$ and $x = 4$ .
$P = (4, 0)$ . $P A = 16 + 9 = 5$ , $P B = 9 + 16 = 5$ . ✓

Answer: $P = (4, 0)$ ; common distance $5$ .

Example 4: Right-angled isosceles triangle · 4 marks · Short answerHarder board level

Show that $A (7, 13)$ , $B (2, 1)$ and $C (- 10, 6)$ are the vertices of a right-angled isosceles triangle and name the right angle.

Try it first, then show the solutionSolution

$A B = 25 + 144 = 13$ , $B C = 144 + 25 = 13$ , $A C = 289 + 49 = 338$ .
$A B = B C$ , so the triangle is isosceles.
$A B^{2} + B C^{2} = 169 + 169 = 338 = A C^{2}$ , so it is right-angled at $B$ (converse of Pythagoras).

Answer: Right-angled isosceles triangle with the right angle at $B$ .

Example 5: Proving a square · 5 marks · Long answerHOTS / competency

Show that $A (- 1, 2)$ , $B (2, 5)$ , $C (5, 2)$ and $D (2, - 1)$ are the vertices of a square and find its area.

Try it first, then show the solutionSolution

$A B = 9 + 9 = 18$ , $B C = 9 + 9 = 18$ , $C D = 9 + 9 = 18$ , $D A = 9 + 9 = 18$ : four equal sides, so $A B C D$ is a rhombus.
$A C = 36 + 0 = 6$ and $B D = 0 + 36 = 6$ : equal diagonals.
A rhombus with equal diagonals is a square.
Area $= (18)^{2} = 18$ square units (also $\frac{1}{2} \times 6 \times 6 = 18$ ).

Answer: It is a square of area 18 square units.

Where students lose marks in The Distance Formula and Identifying Figures from Coordinates

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

Mistake: Sign slip in the difference
Subtracting a negative coordinate must add.
Wrong working
Distance from $(1, - 2)$ to $(9, 4)$ : $(4 - 2)^{2}$ .
Right working
$(4 - (- 2))^{2} = 36$ .
Do this: Put brackets around negative values.
Usually costs: Whole answer.
Mistake: Giving only one value for k
Squaring brings in a second root.
Wrong working
$(k - 2)^{2} = 9 \Rightarrow k = 5$ .
Right working
$k - 2 = \pm 3$ , so $k = 5$ or $- 1$ .
Do this: Always write ±.
Usually costs: One mark.
Mistake: Calling a rhombus a square
A square also needs equal diagonals.
Wrong working
Four equal sides, so a square.
Right working
Four equal sides and equal diagonals, so a square.
Do this: Check the diagonals.
Usually costs: Reason mark.

Exam tips

Class IX is examined in school papers; the proof-of-a-figure items are the hardest items in the chapter.
Compare squares of the sides instead of surds when numbers are not perfect squares.
In a figure proof, write every side before drawing any conclusion.
State the property you use: "converse of Pythagoras", "equal diagonals" and so on.

Practice questions on The Distance Formula and Identifying Figures from Coordinates

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

Q1 · 1 mark · MCQBoard level
The distance of the point $(- 5, 12)$ from the origin is:
(a) 7
(b) 13
(c) 17
(d) 169
Hint
$25 + 144$ .
Answer
(b) 13
Q2 · 3 marks · Short answerHarder board level
The distance between $(2, y)$ and $(10, - 3)$ is $10$ . Find the possible values of $y$ .
Hint
$(y + 3)^{2} = 36$ .
Answer
$64 + (y + 3)^{2} = 100$ , so $y + 3 = \pm 6$ : $y = 3$ or $y = - 9$ .
Q3 · 4 marks · Short answerHarder board level
Show that $P (- 2, - 1)$ , $Q (1, 2)$ and $R (4, 5)$ are collinear and that $Q$ is the mid-point of $P R$ .
Hint
Find PQ, QR, PR as surds.
Answer
$P Q = 9 + 9 = 32$ , $QR = 32$ , $P R = 36 + 36 = 62$ . Since $P Q + QR = P R$ , they are collinear, and $P Q = QR$ makes $Q$ the mid-point.
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The points $(0, 0)$ , $(3, 4)$ and $(6, 8)$ are collinear.
Reason (R): The distances are $5$ , $5$ and $10$ , and $5 + 5 = 10$ .
(a) Both (A) and (R) are true and (R) explains (A).
(b) Both are true but (R) does not explain (A).
(c) (A) is true, (R) is false.
(d) (A) is false, (R) is true.
Hint
Sum of the two shorter distances equals the longest.
Answer
(a)
Q5 · 4 marks · Case-basedHOTS / competency
Case study: Three sensors in a field are at $P (0, 0)$ , $Q (6, 0)$ and $R (0, 8)$ (units in metres). A relay is to be placed at the mid-point $O$ of $QR$ .
(a) Find $P Q$ , $P R$ and $QR$ and say what kind of triangle $P QR$ is.
(b) Find the coordinates of $O$ .
(c) Show that $O$ is equidistant from $P$ , $Q$ and $R$ and give the distance.
Hint
$P Q^{2} + P R^{2} = Q R^{2}$ .
Answer
(a) $6, 8, 10$ ; right-angled at $P$ (36 + 64 = 100). (b) $O = (3, 4)$ . (c) $O P = 9 + 16 = 5$ , $O Q = 9 + 16 = 5$ , $O R = 9 + 16 = 5$ .

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.Write the distance formula.Show answer
√((x₂ − x₁)² + (y₂ − y₁)²).
2.How do you test collinearity with distances?Show answer
The two shorter distances add up to the longest.
3.What makes a rhombus a square?Show answer
Equal diagonals.
4.Distance of (x, y) from the origin?Show answer
√(x² + y²).

The Distance Formula and Identifying Figures from Coordinates: frequently asked questions

Is the mid-point formula in Class IX?

The Class IX coordinate geometry topics are variables, plotting, graphs of linear equations and the distance formula; the mid-point formula appears in Class X.

Why can an unknown coordinate have two values?

Squaring removes the sign, so there are two points at the given distance.

Do I need the diagonals for a rectangle?

Yes: opposite sides equal and equal diagonals.