MathsCBSE Class 10 · Coordinate Geometry

Distance Formula Class 10: Derivation, Uses and Solved Questions

Updated 14 October 2026 · Fact-checked

The distance formula gives the length of the straight line joining two points on the Cartesian plane. It comes from Pythagoras theorem: square the difference of the x-coordinates, square the difference of the y-coordinates, add them and take the square root. The board asks you to find lengths, unknown coordinates, equidistant points and triangle types with it.

Understand Distance Formula

Every point on the Cartesian plane has an address $(x, y)$ . Once points have addresses, you can measure how far apart they are without a ruler. The distance formula does this using only the coordinates.

Take $P (x_{1}, y_{1})$ and $Q (x_{2}, y_{2})$ . Draw a horizontal line through P and a vertical line through Q. They meet at $R (x_{2}, y_{1})$ . Triangle PRQ is right-angled at R. The horizontal side is $P R = ∣ x_{2} - x_{1} ∣$ and the vertical side is $QR = ∣ y_{2} - y_{1} ∣$ . By Pythagoras theorem, $P Q^{2} = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$ . Taking the positive root, $P Q = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$ .

Because each difference is squared, the order does not matter: $(x_{2} - x_{1})^{2} = (x_{1} - x_{2})^{2}$ . A negative difference also becomes positive. But you must still subtract carefully. For example, $5 - (- 3) = 8$ , not $2$ . Most lost marks in this topic come from sign slips, not from the formula.

The formula holds for any two points in any quadrants. It also covers the special cases. If one point is the origin, the distance is $x^{2} + y^{2}$ . If both points have the same y-coordinate, the segment is horizontal and the length is just $∣ x_{2} - x_{1} ∣$ . If they share the x-coordinate, it is $∣ y_{2} - y_{1} ∣$ . A distance is never negative, so you always take the positive square root.

The formula is the base of the whole chapter. With it you can find an unknown coordinate from a given length, find a point equidistant from two points, and decide whether a triangle is isosceles or right-angled using the converse of Pythagoras theorem. These uses are developed on the Applications page. The section formula, the other main tool of the chapter, can also be checked with distances.

Two working habits help. First, work with squared distances such as $A B^{2}$ when comparing lengths; this avoids surds until the end. Second, give the final answer in simplest surd form, like $52$ units, and always write the units.

Distance Formula: formulas to remember

Distance formula: $P Q = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$; Length between $P (x_{1}, y_{1})$ and $Q (x_{2}, y_{2})$ . Order of points does not matter.
Squared distance: $P Q^{2} = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$; Use this when comparing lengths or solving for an unknown; it avoids square roots.
Distance from the origin: $O P = x^{2} + y^{2}$; Distance of $P (x, y)$ from $O (0, 0)$ .
Horizontal and vertical segments: $P Q = ∣ x_{2} - x_{1} ∣ (same y), P Q = ∣ y_{2} - y_{1} ∣ (same x)$; Quick case when two points share a coordinate.
Equidistant condition: $P A = P B ⟺ P A^{2} = P B^{2}$; Squaring is safe because distances are never negative.
Right angle test (converse of Pythagoras theorem): $A B^{2} + B C^{2} = A C^{2} \Rightarrow ∠ B = 9 0^{\circ}$; The right angle is opposite the longest side.
Collinearity test: $A B + B C = A C \Rightarrow A, B, C are collinear$; Collinear: the sum of the two smaller distances equals the third. B then lies between A and C. If the sum is not equal, the points form a triangle.
Square: $A B = B C = C D = D A, A C = B D$; Four equal sides and equal diagonals.
Rhombus: $A B = B C = C D = D A, A C \neq = B D$; Four equal sides but unequal diagonals.
Rectangle: $A B = C D, B C = D A, A C = B D$; Opposite sides equal and equal diagonals.
Parallelogram: $A B = C D, B C = D A$; Opposite sides equal. If the diagonals are also unequal, it is not a rectangle.
Point equidistant from three points: $P A^{2} = P B^{2} and P A^{2} = P C^{2}$; Each condition simplifies to a linear equation in x and y; solve the pair to get P.
Types of triangle by sides: $Equilateral: A B = B C = C A; Isosceles: two sides equal; Right-angled at A: A B^{2} + A C^{2} = B C^{2}$; Find all three squared distances first, then compare them.

How to solve Distance Formula questions

This method works for any distance question, from a 1-mark MCQ to a 5-mark proof.

1Label the points clearly as $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ . If a point is on an axis, write it as $(x, 0)$ or $(0, y)$ first.
2Find the x-difference and y-difference separately, using brackets for negatives, e.g. $5 - (- 3) = 8$ .
3Square each difference and add them to get the squared distance.
4If an unknown is involved, set the squared distance equal to the square of the given length (or equal to another squared distance) and solve.
5Keep both roots of a square equation unless a condition such as a quadrant rules one out.
6For collinearity, find all three distances, then check whether the two smaller ones add up to the largest. If they do, the points are collinear; if not, they form a triangle.
7For a quadrilateral ABCD with vertices taken in order, find all four sides and both diagonals. Then match: square (four equal sides, equal diagonals), rhombus (four equal sides, unequal diagonals), rectangle (opposite sides equal, equal diagonals) or parallelogram (opposite sides equal).
8For a point equidistant from three points, write $P A^{2} = P B^{2}$ and $P A^{2} = P C^{2}$ , simplify each to a linear equation and solve the pair.
9Take the positive square root and simplify the surd, e.g. $50 = 52$ .
10Write units and, for proofs, a one-line conclusion with the reason.
11Check by substituting the answer back into the distance condition.

Distance 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.

The distance formula and why it works

Imagine your village panchayat marks a plan on a grid where 1 unit = 10 m. The main road runs along the x-axis. House A is at $(1, 4)$ and house B is at $(8, 3)$ . A water tank must stand on the main road so that the straight pipes to the two houses are equal in length.

Find the position of the tank on the road and the length of each pipe in metres. Suppose laying pipe costs Rs 40 per metre. Find the total cost of both pipes.

Try it first, then show the solutionSolution

The tank is on the x-axis, so let it be $T (x, 0)$
Equal pipes: $T A^{2} = T B^{2}$ , so $(x - 1)^{2} + 4^{2} = (x - 8)^{2} + 3^{2}$
$x^{2} - 2 x + 17 = x^{2} - 16 x + 73$
$14 x = 56$ , so $x = 4$ and the tank is at $T (4, 0)$
$T A^{2} = (4 - 1)^{2} + (0 - 4)^{2} = 9 + 16 = 25$ , so $T A = 5$ units $= 50$ m
Check: $T B^{2} = (4 - 8)^{2} + (0 - 3)^{2} = 16 + 9 = 25$ , so $T B = 5$ units $= 50$ m
Total pipe $= 50 + 50 = 100$ m
Cost $= 100 \times 40 =$ Rs 4000

Answer: Tank at $(4, 0)$ , that is 40 m along the road from the origin; each pipe is 50 m; total cost Rs 4000.

Memory hook: Tank on the road: write it as $(x, 0)$ ; equal pipes means equal squared distances.

Tricks to remember Distance Formula

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

Across, up, Pythagoras
Think of walking from one point to the other: go across by the x-difference, then up by the y-difference. The straight path is the hypotenuse.
Use it when: Recall the derivation and the formula in any exam question.
Square kills the sign, not the subtraction
Order does not matter because you square, but $3 - (- 5)$ is still 8. Bracket every negative coordinate.
Use it when: Avoid the commonest sign error.
Compare squares, root at the end
When checking equal sides or a right angle, compare $A B^{2}$ , $B C^{2}$ , $A C^{2}$ . Take roots only for the final length.
Use it when: Saves time and avoids surd mistakes.

Distance Formula: solved examples

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

Example 1: Distance with letters · 2 marks · Very short answerBoard level

Find the distance between the points $P (a + b, a - b)$ and $Q (a - b, - a - b)$ .

Try it first, then show the solutionSolution

x-difference: $(a + b) - (a - b) = 2 b$ ; y-difference: $(a - b) - (- a - b) = 2 a$ (½ mark)
$P Q^{2} = (2 b)^{2} + (2 a)^{2} = 4 a^{2} + 4 b^{2}$ (½ mark)
$P Q = 4 (a^{2} + b^{2}) = 2 a^{2} + b^{2}$ units (1 mark)

Answer: $P Q = 2 a^{2} + b^{2}$ units

Example 2: Unknown coordinate from a given distance · 3 marks · Short answerBoard level

The distance between $P (2, - 3)$ and $Q (10, y)$ is 10 units. Find the possible values of y. Which value places Q in the fourth quadrant?

Try it first, then show the solutionSolution

$P Q^{2} = (10 - 2)^{2} + (y + 3)^{2} = 100$ (½ mark)
$64 + (y + 3)^{2} = 100$ , so $(y + 3)^{2} = 36$ (1 mark)
$y + 3 = \pm 6$ , so $y = 3$ or $y = - 9$ (1 mark)
Fourth quadrant needs $x > 0, y < 0$ , so $y = - 9$ , giving $Q (10, - 9)$ (½ mark)
Check: $8^{2} + (- 6)^{2} = 100$ ✓

Answer: $y = 3$ or $y = - 9$ ; Q is in the fourth quadrant when $y = - 9$ .

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

Find the point on the x-axis which is equidistant from $A (2, - 5)$ and $B (- 2, 9)$ . Also find this common distance.

Try it first, then show the solutionSolution

Let the point be $P (x, 0)$ since it lies on the x-axis (½ mark)
$P A^{2} = P B^{2}$ : $(x - 2)^{2} + 25 = (x + 2)^{2} + 81$ (½ mark)
$x^{2} - 4 x + 29 = x^{2} + 4 x + 85$ , so $- 8 x = 56$ , $x = - 7$ (1 mark)
$P (- 7, 0)$ ; $P A^{2} = (- 9)^{2} + 5^{2} = 106$ , so $P A = 106$ units (½ mark)
Check: $P B^{2} = (- 5)^{2} + 9^{2} = 106$ ✓ (½ mark)

Answer: $P (- 7, 0)$ ; common distance $106$ units.

Example 4: Proving a quadrilateral is a square · 5 marks · Long answerHarder board level

Show that the points $A (- 1, - 2)$ , $B (1, 0)$ , $C (- 1, 2)$ and $D (- 3, 0)$ , taken in order, are the vertices of a square. Also find its perimeter and the length of each diagonal.

Try it first, then show the solutionSolution

$A B^{2} = (1 + 1)^{2} + (0 + 2)^{2} = 4 + 4 = 8$ , so $A B = 22$ units (½ mark)
$B C^{2} = (- 1 - 1)^{2} + (2 - 0)^{2} = 4 + 4 = 8$ , so $B C = 22$ units (½ mark)
$C D^{2} = (- 3 + 1)^{2} + (0 - 2)^{2} = 4 + 4 = 8$ , so $C D = 22$ units (½ mark)
$D A^{2} = (- 1 + 3)^{2} + (- 2 - 0)^{2} = 4 + 4 = 8$ , so $D A = 22$ units (½ mark)
Diagonal $A C^{2} = (- 1 + 1)^{2} + (2 + 2)^{2} = 16$ , so $A C = 4$ units (½ mark)
Diagonal $B D^{2} = (- 3 - 1)^{2} + (0 - 0)^{2} = 16$ , so $B D = 4$ units (½ mark)
All four sides are equal and the diagonals are equal, so ABCD is a square (1 mark)
Perimeter $= 4 \times 22 = 82$ units; each diagonal is 4 units (1 mark)
Check: $A B^{2} + B C^{2} = 8 + 8 = 16 = A C^{2}$ , so $∠ B = 9 0^{\circ}$ , as a square needs ✓

Answer: ABCD is a square with side $22$ units; perimeter $82$ units; each diagonal 4 units.

Example 5: Collinearity using distances · 3 marks · Short answerHarder board level

Show that the points $A (1, 5)$ , $B (2, 3)$ and $C (- 2, 11)$ are collinear. Which point lies between the other two?

Try it first, then show the solutionSolution

$A B^{2} = (2 - 1)^{2} + (3 - 5)^{2} = 1 + 4 = 5$ , so $A B = 5$ units (½ mark)
$A C^{2} = (- 2 - 1)^{2} + (11 - 5)^{2} = 9 + 36 = 45$ , so $A C = 35$ units (½ mark)
$B C^{2} = (- 2 - 2)^{2} + (11 - 3)^{2} = 16 + 64 = 80$ , so $B C = 45$ units (½ mark)
Sum of the two smaller distances: $A B + A C = 5 + 35 = 45 = B C$ (1 mark)
Hence A, B, C are collinear, and A lies between B and C (½ mark)

Answer: $A B + A C = B C = 45$ units, so the points are collinear with A between B and C.

Example 6: Unknown vertex of a right-angled triangle · 5 marks · Long answerHarder board level

The points $A (1, 2)$ , $B (4, k)$ and $C (- 1, 5)$ form a triangle which is right-angled at A. Find the value of k. Show that the triangle is also isosceles and find its perimeter.

Try it first, then show the solutionSolution

$A B^{2} = (4 - 1)^{2} + (k - 2)^{2} = 9 + (k - 2)^{2}$ (½ mark)
$A C^{2} = (- 1 - 1)^{2} + (5 - 2)^{2} = 4 + 9 = 13$ (½ mark)
$B C^{2} = (- 1 - 4)^{2} + (5 - k)^{2} = 25 + (5 - k)^{2}$ (½ mark)
Right angle at A, so BC is the hypotenuse: $A B^{2} + A C^{2} = B C^{2}$ (½ mark)
$9 + k^{2} - 4 k + 4 + 13 = 25 + 25 - 10 k + k^{2}$ (½ mark)
$k^{2} - 4 k + 26 = k^{2} - 10 k + 50$ , so $6 k = 24$ , $k = 4$ (1 mark)
With $B (4, 4)$ : $A B^{2} = 9 + 4 = 13$ , $B C^{2} = 25 + 1 = 26$ , and $13 + 13 = 26$ ✓ (½ mark)
$A B = A C = 13$ units, so the triangle is isosceles (½ mark)
Perimeter $= 13 + 13 + 26 = (213 + 26)$ units (½ mark)

Answer: $k = 4$ ; $A B = A C = 13$ units, so the triangle is isosceles; perimeter $(213 + 26)$ units.

Example 7: Case study: placing a bus stop · 4 marks · Case-basedHOTS / competency

Suppose a town map is drawn on a grid where 1 unit = 100 m. The main road runs along the x-axis. A school S is at $(1, 7)$ and a hospital H is at $(7, 5)$ . The town wants a bus stop P on the main road that is equidistant from the school and the hospital. A library L is at $(6, 3)$ and a market M is at $(- 3, 4)$ .
(i) Find the position of the bus stop P. (1 mark)
(ii) Find the distance of the bus stop from the school in metres. (1 mark)
(iii) Of the school, the library and the market, which is nearest to the bus stop? Justify, and give that distance in metres. (2 marks)

Try it first, then show the solutionSolution

(i) P is on the x-axis, so $P (x, 0)$ ; $P S^{2} = P H^{2}$ : $(x - 1)^{2} + 49 = (x - 7)^{2} + 25$ (½ mark)
$x^{2} - 2 x + 50 = x^{2} - 14 x + 74$ , so $12 x = 24$ , $x = 2$ ; bus stop at $P (2, 0)$ (½ mark)
(ii) $P S^{2} = (2 - 1)^{2} + (0 - 7)^{2} = 50$ , so $P S = 52$ units $= 5002$ m; check $P H^{2} = 25 + 25 = 50$ ✓ (1 mark)
(iii) $P L^{2} = (6 - 2)^{2} + (3 - 0)^{2} = 16 + 9 = 25$ , so $P L = 5$ units (½ mark)
$P M^{2} = (- 3 - 2)^{2} + (4 - 0)^{2} = 25 + 16 = 41$ (½ mark)
Compare squares: $25 < 41 < 50$ , so the library is nearest, at $5 \times 100 = 500$ m (1 mark)

Answer: (i) $P (2, 0)$ (ii) $5002$ m (iii) The library, at 500 m.

Where students lose marks in Distance Formula

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

Mistake: Dropping the sign when subtracting a negative coordinate
Students write $x_{2} - x_{1}$ but ignore that $x_{1}$ is negative.
Wrong working
For $(- 2, 1)$ and $(4, 5)$ : $(4 - 2)^{2} + (5 - 1)^{2} = 4 + 16 = 20$
Right working
$(4 - (- 2))^{2} + (5 - 1)^{2} = 36 + 16 = 52$ , so distance $= 213$ units
Do this: Put every negative coordinate in brackets before subtracting.
Usually costs: Usually all marks after that step.
Mistake: Splitting the square root over a sum
Treating $a^{2} + b^{2}$ as $a + b$ .
Wrong working
$6^{2} + 8^{2} = 6 + 8 = 14$
Right working
$36 + 64 = 100 = 10$
Do this: Add the squares first, then take one square root.
Usually costs: The final answer mark and often method marks.
Mistake: Keeping only one root
Taking $36 = 6$ only when solving $(y + 3)^{2} = 36$ .
Wrong working
$y + 3 = 6$ , so $y = 3$
Right working
$y + 3 = \pm 6$ , so $y = 3$ or $y = - 9$
Do this: Write $\pm$ whenever you root both sides; reject a value only with a stated reason.
Usually costs: ½ to 1 mark.
Mistake: Unsimplified surd or missing units
Stopping at $50$ or forgetting units.
Wrong working
$A C = 50$
Right working
$A C = 52$ units
Do this: Pull out perfect squares and always end with units.
Usually costs: ½ mark.

Exam tips

Show the squared-distance line before the root; examiners give method marks for it.
For equidistant points on an axis, write the point as $(x, 0)$ or $(0, y)$ in the first line.
When a quadrant or condition is given, state clearly why you reject a root.
In proofs about triangles, end with a sentence naming the result used, such as the converse of Pythagoras theorem.
In case studies, find each distance separately and give metres or km as the question uses.

Practice questions on Distance Formula

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
If the distance between the points $(4, p)$ and $(1, 0)$ is 5 units, then the value of p is:
(a) $\pm 4$
(b) 4 only
(c) $- 4$ only
(d) 0
Hint
Square both sides; a square equal to a number has two roots.
Answer
(a) $\pm 4$ . $9 + p^{2} = 25$ , so $p^{2} = 16$ , $p = \pm 4$ .
Q2 · 3 marks · Short answerHarder board level
Find k so that the point $A (3, k)$ is equidistant from $B (- 1, 2)$ and $C (5, 4)$ . For this value, show that $△ A B C$ is isosceles and right-angled, and name the right angle.
Hint
Equate $A B^{2}$ and $A C^{2}$ ; then compare $A B^{2} + A C^{2}$ with $B C^{2}$ .
Answer
$16 + (k - 2)^{2} = 4 + (k - 4)^{2}$ gives $20 - 4 k = 20 - 8 k$ , so $k = 0$ and $A (3, 0)$ . Then $A B^{2} = 16 + 4 = 20$ , $A C^{2} = 4 + 16 = 20$ and $B C^{2} = 36 + 4 = 40$ . $A B = A C = 25$ units, so the triangle is isosceles, and $A B^{2} + A C^{2} = 40 = B C^{2}$ , so it is right-angled at A.
Q3 · 4 marks · Long answerHarder board level
Show that $A (3, 0)$ , $B (6, 4)$ and $C (- 1, 3)$ form an isosceles right-angled triangle. Then find the point D such that ABDC is a square, and verify that all its sides are equal.
Hint
The right angle is at the vertex where the two equal sides meet; in a square the diagonals have the same midpoint, so $D = B + C - A$ .
Answer
$A B^{2} = 9 + 16 = 25$ , $A C^{2} = 16 + 9 = 25$ , $B C^{2} = 49 + 1 = 50$ . So $A B = A C = 5$ units and $A B^{2} + A C^{2} = B C^{2}$ : isosceles and right-angled at A. (2) For the square ABDC, the midpoint of AD equals the midpoint of BC, $(\frac{5}{2}, \frac{7}{2})$ , so $D = (6 - 1 - 3, 4 + 3 - 0) = (2, 7)$ . (1) Check: $B D = 16 + 9 = 5$ and $C D = 9 + 16 = 5$ , so all four sides are 5 units and the diagonals $B C = A D = 52$ units. (1)
Q4 · 3 marks · Short answerHarder board level
Find a relation between x and y such that the point $P (x, y)$ is equidistant from $A (3, 6)$ and $B (- 3, 4)$ .
Hint
Expand $P A^{2} = P B^{2}$ ; the $x^{2}$ and $y^{2}$ terms cancel.
Answer
$(x - 3)^{2} + (y - 6)^{2} = (x + 3)^{2} + (y - 4)^{2}$ gives $- 6 x - 12 y + 45 = 6 x - 8 y + 25$ , so $12 x + 4 y = 20$ , i.e. $3 x + y = 5$ . Check: midpoint $(0, 5)$ satisfies it.
Q5 · 2 marks · Very short answerBoard level
A student found the distance between $A (- 3, 4)$ and $B (5, - 2)$ as $(5 - 3)^{2} + (- 2 - 4)^{2} = 40 = 210$ units. Find the error and give the correct distance.
Hint
Look at how the student subtracted the negative x-coordinate.
Answer
The x-difference should be $5 - (- 3) = 8$ , not $5 - 3$ . Correct: $8^{2} + (- 6)^{2} = 100 = 10$ units.
Q6 · 1 mark · MCQBoard level
The distance of the point $(a cos θ, a sin θ)$ from the origin, where $a > 0$ , is:
(a) $a$
(b) $2 a$
(c) $a^{2}$
(d) 1
Hint
Use $sin^{2} θ + cos^{2} θ = 1$ .
Answer
(a) $a$ . Distance $= a^{2} cos^{2} θ + a^{2} sin^{2} θ = a^{2} = a$ .
Q7 · 1 mark · MCQBoard level
The points on the y-axis which are at a distance of 10 units from the point $(6, - 3)$ are:
(a) $(0, 5)$ and $(0, - 11)$
(b) $(0, - 5)$ and $(0, 11)$
(c) $(0, 7)$ and $(0, - 13)$
(d) $(5, 0)$ and $(- 11, 0)$
Hint
A point on the y-axis is $(0, y)$ .
Answer
(a) $(0, 5)$ and $(0, - 11)$ . $36 + (y + 3)^{2} = 100$ , so $(y + 3)^{2} = 64$ , $y + 3 = \pm 8$ , $y = 5$ or $y = - 11$ .
Q8 · 1 mark · Assertion-reasonBoard level
Assertion (A): The points $O (0, 0)$ , $A (3, 4)$ and $B (6, 8)$ are collinear.
Reason (R): $O A + A B = O B$ .
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Find OA, AB and OB.
Answer
(a). $O A = 9 + 16 = 5$ , $A B = 9 + 16 = 5$ , $O B = 36 + 64 = 10$ . So $O A + A B = O B$ , which is exactly why the points are collinear.
Q9 · 2 marks · Very short answerBoard level
The distance between $P (2, y)$ and $Q (- 4, 3)$ is 10 units, and P lies in the fourth quadrant. Find y.
Hint
Get both roots first, then use the quadrant to reject one.
Answer
$36 + (y - 3)^{2} = 100$ , so $(y - 3)^{2} = 64$ , $y = 11$ or $y = - 5$ . Fourth quadrant needs $y < 0$ , so $y = - 5$ . Check: $6^{2} + 8^{2} = 100$ .
Q10 · 3 marks · Short answerHarder board level
Find the points on the x-axis which are at a distance of 13 units from $A (1, 5)$ . Then find the perimeter of the triangle formed by A and these two points.
Hint
Take the point as $(x, 0)$ ; the two points found are each 13 units from A.
Answer
$(x - 1)^{2} + 25 = 169$ , so $(x - 1)^{2} = 144$ , $x = 13$ or $x = - 11$ . Points $(13, 0)$ and $(- 11, 0)$ , which are $13 - (- 11) = 24$ units apart. Perimeter $= 13 + 13 + 24 = 50$ units.
Q11 · 3 marks · Short answerHOTS / competency
Riya says the quadrilateral with vertices $A (0, 0)$ , $B (5, 0)$ , $C (8, 4)$ and $D (3, 4)$ , taken in order, is a square because all four sides are equal. Is she right? Justify by naming the quadrilateral correctly.
Hint
Equal sides alone are not enough; check the diagonals.
Answer
$A B = 5$ , $B C = 9 + 16 = 5$ , $C D = 5$ , $D A = 9 + 16 = 5$ . Diagonals: $A C = 64 + 16 = 45$ units, $B D = 4 + 16 = 25$ units. The diagonals are unequal, so ABCD is a rhombus, not a square. Riya is wrong.
Q12 · 4 marks · Long answerHarder board level
Show that $A (1, 7)$ , $B (4, 2)$ , $C (- 1, - 1)$ and $D (- 4, 4)$ , taken in order, are the vertices of a square.
Hint
Find all four sides and both diagonals using squared distances.
Answer
$A B^{2} = 9 + 25 = 34$ , $B C^{2} = 25 + 9 = 34$ , $C D^{2} = 9 + 25 = 34$ , $D A^{2} = 25 + 9 = 34$ , so each side is $34$ units. $A C^{2} = 4 + 64 = 68$ and $B D^{2} = 64 + 4 = 68$ , so the diagonals are equal ( $217$ units). Four equal sides and equal diagonals, so ABCD is a square.
Q13 · 5 marks · Long answerHarder board level
Find the point P which is equidistant from $A (0, - 2)$ , $B (7, 5)$ and $C (- 2, 2)$ . Also find this equal distance.
Hint
Form two linear equations from $P A^{2} = P B^{2}$ and $P A^{2} = P C^{2}$ .
Answer
Let $P (x, y)$ . $P A^{2} = P B^{2}$ : $x^{2} + (y + 2)^{2} = (x - 7)^{2} + (y - 5)^{2}$ gives $14 x + 14 y = 70$ , so $x + y = 5$ . $P A^{2} = P C^{2}$ : $x^{2} + (y + 2)^{2} = (x + 2)^{2} + (y - 2)^{2}$ gives $8 y - 4 x = 4$ , so $x = 2 y - 1$ . Then $3 y - 1 = 5$ , $y = 2$ , $x = 3$ . $P (3, 2)$ ; $P A = 9 + 16 = 5$ units. Check: $P B = 16 + 9 = 5$ , $P C = 25 + 0 = 5$ .
Q14 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a garden is drawn on a grid where 1 unit = 1 m. A sprinkler at $S (2, 3)$ waters every point within 5 m of it, including points exactly 5 m away. Plants stand at $P (5, 7)$ , $Q (6, 1)$ and $R (- 1, 8)$ . A straight path runs along the x-axis.
(i) Does plant P get water? Justify. (1 mark)
(ii) Which of Q and R gets water? (1 mark)
(iii) Find the points on the path that are exactly 5 m from the sprinkler, and the length of the path that gets wet. (2 marks)
Hint
Compare each squared distance with $5^{2} = 25$ . For (iii), take the point as $(x, 0)$ .
Answer
(i) $S P^{2} = 9 + 16 = 25$ , so $S P = 5$ m; P just gets water (on the edge). (ii) $S Q^{2} = 16 + 4 = 20 < 25$ , so Q gets water; $S R^{2} = 9 + 25 = 34 > 25$ , so R does not. (iii) $(x - 2)^{2} + 9 = 25$ gives $x - 2 = \pm 4$ , so $(6, 0)$ and $(- 2, 0)$ ; wet length $= 6 - (- 2) = 8$ m.

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.Distance between $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ ?Show answer
$(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$
2.Distance of $(x, y)$ from the origin?Show answer
$x^{2} + y^{2}$
3.Distance of $(- 5, 12)$ from the origin?Show answer
13 units
4.Distance between $(3, 7)$ and $(- 4, 7)$ ?Show answer
Same y, so $∣3 - (- 4) ∣ = 7$ units
5.Which theorem is the distance formula based on?Show answer
Pythagoras theorem

Distance Formula: frequently asked questions

Does the order of the points matter in the distance formula?

No. Each difference is squared, so swapping the points gives the same result. Only be careful with signs while subtracting.

Why is the distance always positive?

A length cannot be negative, so you take only the positive square root. Negative roots may appear when solving for a coordinate, but not for the distance itself.

Should I leave answers as surds?

Yes, give them in simplest surd form such as 3 root 5 units unless the question asks for a decimal. Always write the units.

Is the area of a triangle using coordinates in the syllabus?

No, the area of a triangle from coordinates is not in the 2026-27 syllabus.

What kind of questions does Exercise 7.1 have?

It has questions on finding distances, finding unknown coordinates from distances, equidistant points and identifying shapes. Practise them with your own numbers using the steps on this page.