MathsCBSE Class 10 · Coordinate Geometry

Applications of the Distance Formula for Class 10

Updated 14 October 2026 · Fact-checked

The distance formula finds the length between two points. Once you have lengths, you can prove three points are collinear, show a triangle is isosceles, equilateral or right-angled, prove four points form a square, rhombus or rectangle, and find a point that is equidistant from two points. Board papers test all of these in 2- to 5-mark questions.

Every application follows one pattern: find the lengths with the distance formula, compare them, then state the conclusion.

Illustration drawn for QuestionBuddy.

Understand Applications of the Distance Formula

The distance formula, $A B = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$ , gives you one number: a length. On its own that is simple. The real power comes when you find several lengths and compare them. Almost every application on this page follows one pattern: find lengths, compare them, state a conclusion with a reason.

Shapes. A figure is decided by its lengths. Two equal sides make a triangle isosceles, three equal sides make it equilateral. If $A B^{2} + B C^{2} = A C^{2}$ , the converse of Pythagoras theorem says $∠ B = 9 0^{\circ}$ . For a quadrilateral ABCD taken in order, you find all four sides and both diagonals. Equal sides with equal diagonals give a square. Equal sides with unequal diagonals give a rhombus. Equal opposite sides with equal diagonals give a rectangle.

Collinearity. In any real triangle, two sides together are longer than the third. So if the two shorter distances add up exactly to the longest one, the triangle has collapsed into a straight line. That is your test: if $A B + B C = A C$ , then A, B, C are collinear and B lies between A and C. Simplify surds first, because only like surds can be added.

Equidistant points and unknowns. A condition such as "P is equidistant from A and B" becomes the equation $P A = P B$ . Squaring gives $P A^{2} = P B^{2}$ , which is safe because lengths are never negative. When you expand, the squared unknowns cancel, so you get a linear equation. A condition such as "the distance is 10 units" becomes $P Q^{2} = 100$ , which usually gives two values. Keep both unless the question rules one out.

Work with squared lengths wherever you only need to compare. This avoids surds and arithmetic slips. Take square roots only when the question asks for an actual length, and then give the answer in simplest surd form with units. These ideas connect to the section formula too: when three points are collinear, the middle point divides the segment in the ratio of the two smaller distances.

Applications of the Distance Formula: formulas to remember

Distance formula: $A B = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$; The starting point of every application. Order of points does not matter.
Squared distance: $A B^{2} = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$; Use squared lengths to compare sides and test right angles without surds.
Collinearity test: $A B + B C = A C$; AC is the longest distance. If equality holds, A, B, C are collinear with B between A and C.
Isosceles and equilateral triangles: $A B = A C (isosceles), A B = B C = C A (equilateral)$; Compare the three side lengths.
Right-angled triangle (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.
Square: $A B = B C = C D = D A and A C = B D$; Vertices in order. Both conditions are needed.
Rhombus: $A B = B C = C D = D A and A C \neq = B D$; Equal sides, unequal diagonals.
Rectangle: $A B = C D, B C = D A and A C = B D$; Opposite sides equal and diagonals equal.
Parallelogram: $A B = C D and B C = D A$; Opposite sides equal only.
Equidistant condition: $P A = P B ⟺ P A^{2} = P B^{2}$; Square both sides; the squared unknowns cancel, leaving a linear equation.
Unknown point on an axis: $(x, 0) on the x-axis, (0, y) on the y-axis$; Write the point in this form first to reduce the unknowns to one.
Distance from the origin: $O P = x^{2} + y^{2}$; The special case of the distance formula when one point is $O (0, 0)$ .

How to solve Applications of the Distance Formula questions

Every question on this topic is a length-comparison question. Use this method whether you are proving a shape, testing collinearity or finding an unknown.

1Label the points and write their coordinates clearly. If a point lies on an axis, write it as $(x, 0)$ or $(0, y)$ .
2Decide which lengths you need: three sides for a triangle or collinearity, four sides and two diagonals for a quadrilateral, two distances for an equidistant condition.
3Find each length as a squared length first, showing the differences, for example $A B^{2} = (4)^{2} + (1)^{2} = 17$ .
4For collinearity or an actual length, take square roots and simplify surds, for example $45 = 35$ .
5For an unknown, turn the condition into an equation ( $P A^{2} = P B^{2}$ or $P Q^{2} = d^{2}$ ), expand and solve. Keep both roots unless a condition rejects one.
6Compare the lengths against the exact condition for the shape, line or point.
7Write a one-line conclusion with the reason, for example: all sides equal and diagonals equal, so ABCD is a square.
8Check: substitute your answer back, or confirm that a right angle satisfies Pythagoras.

Applications of the 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.

Equidistant points and unknowns

Imagine three villages want to share one new water tank. On a district map with 1 unit = 1 km, the villages are at $A (4, 7)$ , $B (- 3, 6)$ and $C (5, 0)$ . The panchayat wants the tank to be equally far from all three, so every pipeline has the same length.

Find where the tank should be placed and the total length of pipe if a separate straight pipeline runs from the tank to each village.

Try it first, then show the solutionSolution

Let the tank be at $P (x, y)$ , with $P A^{2} = P B^{2} = P C^{2}$ .
$P A^{2} = P B^{2}$ : $(x - 4)^{2} + (y - 7)^{2} = (x + 3)^{2} + (y - 6)^{2}$
$- 8 x - 14 y + 65 = 6 x - 12 y + 45$ , so $7 x + y = 10$ ... (1)
$P A^{2} = P C^{2}$ : $(x - 4)^{2} + (y - 7)^{2} = (x - 5)^{2} + y^{2}$
$- 8 x - 14 y + 65 = - 10 x + 25$ , so $x - 7 y = - 20$ ... (2)
From (1), $y = 10 - 7 x$ . Substituting in (2): $x - 70 + 49 x = - 20$ , so $x = 1$ and $y = 3$ .
$P A = (4 - 1)^{2} + (7 - 3)^{2} = 9 + 16 = 5$ km.
Check: $P B = 16 + 9 = 5$ km and $P C = 16 + 9 = 5$ km.
Total pipe $= 3 \times 5 = 15$ km.

Answer: The tank should be at $(1, 3)$ on the map, 5 km from each village; the total pipe length is 15 km.

Memory hook: One tank, three villages: two equal-distance equations, two unknowns.

Collinearity using the distance formula

Suppose your school is decorating for a festival and plans three food stalls in a straight line so one string of lights can run along them. On the plan, 1 unit = 2 m. The end stalls are at $A (- 2, 1)$ and $C (7, 13)$ , and the middle stall is at $B (1, k)$ , exactly 5 units from A. A power socket is at $D (4, 1)$ .

Find k so that the three stalls are collinear with B in the middle, find the length of light string from A to C, and find the length of cable from the socket D to the middle stall.

Try it first, then show the solutionSolution

$A B = 5$ : $(1 + 2)^{2} + (k - 1)^{2} = 25$ , so $(k - 1)^{2} = 16$ .
$k - 1 = \pm 4$ , so $k = 5$ or $k = - 3$ .
$A C = (7 + 2)^{2} + (13 - 1)^{2} = 81 + 144 = 15$ units.
For B in the middle we need $A B + B C = A C$ , so $B C = 10$ , that is $B C^{2} = 100$ .
$k = 5$ : $B C^{2} = (7 - 1)^{2} + (13 - 5)^{2} = 36 + 64 = 100$ . This works.
$k = - 3$ : $B C^{2} = (7 - 1)^{2} + (13 + 3)^{2} = 36 + 256 = 292 \neq = 100$ . Reject.
So B is $(1, 5)$ , and the light string is $15 \times 2 = 30$ m.
$D B = (1 - 4)^{2} + (5 - 1)^{2} = 9 + 16 = 5$ units $= 10$ m.

Answer: $k = 5$ ; the light string from A to C is 30 m and the cable from the socket to the middle stall is 10 m.

Memory hook: One straight string: the two short pieces add up to the whole.

Tricks to remember Applications of the Distance Formula

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

Square needs S + D
Sides equal plus Diagonals equal. Sides alone only say Rhombus.
Use it when: Proving a square or deciding between square and rhombus.
Short + short = long
For collinearity, add the two smaller distances and compare with the biggest. Equal means flat, so one line.
Use it when: Collinearity questions using distances.
Square first, root last
Compare $A B^{2}$ , $B C^{2}$ and so on. Take square roots only when an actual length is asked.
Use it when: Shape proofs and right-angle tests, to avoid surd slips.
Equal means squares cancel
In $P A^{2} = P B^{2}$ the $x^{2}$ or $y^{2}$ terms always cancel. If they do not, recheck your expansion.
Use it when: Equidistant-point questions.

Applications of the Distance Formula: solved examples

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

Example 1: Unknown coordinate from a given distance · 2 marks · Very short answerBoard level

The distance between the points $P (3, - 2)$ and $Q (11, y)$ is 10 units. Find the possible values of y.

Try it first, then show the solutionSolution

$P Q^{2} = 100$ , so $(11 - 3)^{2} + (y + 2)^{2} = 100$ . [½]
$64 + (y + 2)^{2} = 100$ , so $(y + 2)^{2} = 36$ . [½]
$y + 2 = 6$ or $y + 2 = - 6$ . [½]
$y = 4$ or $y = - 8$ . Check: $64 + 36 = 100$ in both cases. [½]

Answer: $y = 4$ or $y = - 8$ .

Example 2: Collinearity with surds and the middle point · 3 marks · Short answerBoard level

Show that the points $A (- 3, - 1)$ , $B (- 1, 3)$ and $C (2, 9)$ are collinear. Also find the ratio in which B divides AC.

Try it first, then show the solutionSolution

$A B = (- 1 + 3)^{2} + (3 + 1)^{2} = 4 + 16 = 20 = 25$ [½]
$B C = (2 + 1)^{2} + (9 - 3)^{2} = 9 + 36 = 45 = 35$ [½]
$A C = (2 + 3)^{2} + (9 + 1)^{2} = 25 + 100 = 125 = 55$ [½]
$A B + B C = 25 + 35 = 55 = A C$ , so A, B, C are collinear with B between A and C. [1]
B divides AC in the ratio $A B : B C = 25 : 35 = 2 : 3$ . [½]

Answer: The points are collinear and B divides AC internally in the ratio $2 : 3$ .

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 $A (7, 6)$ and $B (- 3, 4)$ . Also find this common distance.

Try it first, then show the solutionSolution

Let the point be $P (x, 0)$ . Since $P A = P B$ , $P A^{2} = P B^{2}$ . [½]
$(x - 7)^{2} + 36 = (x + 3)^{2} + 16$ , so $x^{2} - 14 x + 85 = x^{2} + 6 x + 25$ . [1]
$60 = 20 x$ , so $x = 3$ . [½]
The point is $P (3, 0)$ . [½]
$P A = (3 - 7)^{2} + 6^{2} = 52 = 213$ units; check $P B = 6^{2} + 4^{2} = 52$ . [½]

Answer: $P (3, 0)$ ; the common distance is $213$ units.

Example 4: Name the quadrilateral · 5 marks · Long answerBoard level

Show that the points $A (- 2, - 1)$ , $B (4, 2)$ , $C (2, 6)$ and $D (- 4, 3)$ , taken in order, are the vertices of a rectangle. Show that it is not a square, and find the length of a diagonal.

Try it first, then show the solutionSolution

$A B^{2} = (4 + 2)^{2} + (2 + 1)^{2} = 36 + 9 = 45$ [½]
$B C^{2} = (2 - 4)^{2} + (6 - 2)^{2} = 4 + 16 = 20$ [½]
$C D^{2} = (- 4 - 2)^{2} + (3 - 6)^{2} = 36 + 9 = 45$ [½]
$D A^{2} = (- 2 + 4)^{2} + (- 1 - 3)^{2} = 4 + 16 = 20$ [½]
So $A B = C D = 35$ and $B C = D A = 25$ : opposite sides are equal. [½]
$A C^{2} = (2 + 2)^{2} + (6 + 1)^{2} = 16 + 49 = 65$ [½]
$B D^{2} = (- 4 - 4)^{2} + (3 - 2)^{2} = 64 + 1 = 65$ , so $A C = B D$ . [½]
Opposite sides equal and diagonals equal, so ABCD is a rectangle. [1]
$A B \neq = B C$ , so the sides are not all equal and ABCD is not a square. Diagonal $= 65$ units (check: $45 + 20 = 65$ , so $∠ B = 9 0^{\circ}$ ). [½]

Answer: ABCD is a rectangle but not a square; each diagonal is $65$ units.

Example 5: Case study: lamp posts in a park · 4 marks · Case-basedBoard level

Imagine a park is drawn on a grid in metres. Three lamp posts stand at $A (3, 0)$ , $B (6, 4)$ and $C (- 1, 3)$ , and cables run along AB, AC and BC.
(i) Find the length of cable AB. (1 mark)
(ii) Find the length of cable AC. (1 mark)
(iii) Show that $△ A B C$ is a right-angled isosceles triangle and find the length of cable BC. (2 marks)

Try it first, then show the solutionSolution

(i) $A B = (6 - 3)^{2} + (4 - 0)^{2} = 9 + 16 = 5$ m. [1]
(ii) $A C = (- 1 - 3)^{2} + (3 - 0)^{2} = 16 + 9 = 5$ m. [1]
(iii) $B C^{2} = (- 1 - 6)^{2} + (3 - 4)^{2} = 49 + 1 = 50$ , so $B C = 52$ m. [½]
$A B = A C$ , so the triangle is isosceles. [½]
$A B^{2} + A C^{2} = 25 + 25 = 50 = B C^{2}$ , so by the converse of Pythagoras theorem $∠ A = 9 0^{\circ}$ . [1]

Answer: (i) 5 m (ii) 5 m (iii) $A B = A C$ and $A B^{2} + A C^{2} = B C^{2}$ , so the triangle is right-angled isosceles with $∠ A = 9 0^{\circ}$ ; $B C = 52$ m.

Example 6: Point equidistant from three points · 5 marks · Long answerHarder board level

Find the point P which is equidistant from $A (5, 5)$ , $B (- 2, 4)$ and $C (6, - 2)$ , and find this common distance. Hence show that $△ A B C$ is right-angled at A and that P is the midpoint of BC.

Try it first, then show the solutionSolution

Let $P (x, y)$ . Since $P A = P B = P C$ , we use $P A^{2} = P B^{2}$ and $P A^{2} = P C^{2}$ . [½]
$P A^{2} = P B^{2}$ : $(x - 5)^{2} + (y - 5)^{2} = (x + 2)^{2} + (y - 4)^{2}$ gives $- 10 x - 10 y + 50 = 4 x - 8 y + 20$ , so $7 x + y = 15$ ... (1) [1]
$P A^{2} = P C^{2}$ : $(x - 5)^{2} + (y - 5)^{2} = (x - 6)^{2} + (y + 2)^{2}$ gives $- 10 x - 10 y + 50 = - 12 x + 4 y + 40$ , so $x - 7 y = - 5$ ... (2) [1]
From (1), $y = 15 - 7 x$ . In (2): $x - 105 + 49 x = - 5$ , so $50 x = 100$ , $x = 2$ and $y = 1$ . P is $(2, 1)$ . [1]
$P A = 9 + 16 = 5$ units; check $P B = 16 + 9 = 5$ and $P C = 16 + 9 = 5$ units. [½]
$A B^{2} = 49 + 1 = 50$ , $A C^{2} = 1 + 49 = 50$ , $B C^{2} = 64 + 36 = 100$ . [½]
$A B^{2} + A C^{2} = 100 = B C^{2}$ , so $∠ A = 9 0^{\circ}$ (converse of Pythagoras theorem). Midpoint of BC $= (\frac{- 2 + 6}{2}, \frac{4 - 2}{2}) = (2, 1) = P$ . [½]

Answer: $P (2, 1)$ , common distance 5 units; $∠ A = 9 0^{\circ}$ and P is the midpoint of the hypotenuse BC.

Example 7: Right-angled triangle with an unknown k · 3 marks · Short answerHarder board level

The points $A (k, 4)$ , $B (2, 1)$ and $C (5, 5)$ form a triangle right-angled at B. Find k. Hence show that the triangle is also isosceles.

Try it first, then show the solutionSolution

$A B^{2} = (2 - k)^{2} + (1 - 4)^{2} = (k - 2)^{2} + 9$ [½]
$B C^{2} = (5 - 2)^{2} + (5 - 1)^{2} = 9 + 16 = 25$ [½]
$A C^{2} = (5 - k)^{2} + (5 - 4)^{2} = (k - 5)^{2} + 1$ [½]
Right angle at B, so $A B^{2} + B C^{2} = A C^{2}$ (converse of Pythagoras theorem): $k^{2} - 4 k + 4 + 9 + 25 = k^{2} - 10 k + 25 + 1$ [½]
$- 4 k + 38 = - 10 k + 26$ , so $6 k = - 12$ and $k = - 2$ . [½]
Check: $A B^{2} = 16 + 9 = 25$ and $A C^{2} = 49 + 1 = 50 = 25 + 25$ . Since $A B = B C = 5$ units, the triangle is isosceles. [½]

Answer: $k = - 2$ ; $A B = B C = 5$ units and $A C = 52$ units, so the triangle is right-angled and isosceles.

Example 8: Equilateral triangle with an unknown vertex · 3 marks · Short answerHarder board level

Two vertices of an equilateral triangle are $A (1, 1)$ and $B (5, 1)$ . The third vertex C lies in the first quadrant. Find the coordinates of C.

Try it first, then show the solutionSolution

Let $C (x, y)$ . $C A = C B$ , so $(x - 1)^{2} + (y - 1)^{2} = (x - 5)^{2} + (y - 1)^{2}$ , giving $- 2 x + 1 = - 10 x + 25$ , so $x = 3$ . [1]
$A B = (5 - 1)^{2} + 0^{2} = 4$ units. [½]
$C A = A B$ : $(3 - 1)^{2} + (y - 1)^{2} = 16$ , so $(y - 1)^{2} = 12$ and $y - 1 = \pm 23$ . [1]
$1 - 23 < 0$ would put C in the fourth quadrant, so reject it. $C = (3, 1 + 23)$ . Check: $C B^{2} = 4 + 12 = 16$ . [½]

Answer: $C (3, 1 + 23)$ .

Example 9: Rhombus with unequal diagonals · 5 marks · Long answerBoard level

Show that the points $A (3, 3)$ , $B (0, 2)$ , $C (- 1, - 1)$ and $D (2, 0)$ , taken in order, are the vertices of a rhombus but not a square. Also show that its diagonals bisect each other.

Try it first, then show the solutionSolution

$A B^{2} = (0 - 3)^{2} + (2 - 3)^{2} = 9 + 1 = 10$ [½]
$B C^{2} = (- 1 - 0)^{2} + (- 1 - 2)^{2} = 1 + 9 = 10$ [½]
$C D^{2} = (2 + 1)^{2} + (0 + 1)^{2} = 9 + 1 = 10$ [½]
$D A^{2} = (3 - 2)^{2} + (3 - 0)^{2} = 1 + 9 = 10$ [½]
So $A B = B C = C D = D A = 10$ units. [½]
$A C^{2} = (- 1 - 3)^{2} + (- 1 - 3)^{2} = 32$ and $B D^{2} = (2 - 0)^{2} + (0 - 2)^{2} = 8$ , so $A C = 42$ and $B D = 22$ units. [½]
All sides equal but diagonals unequal, so ABCD is a rhombus and not a square. [1]
Midpoint of AC $= (\frac{3 - 1}{2}, \frac{3 - 1}{2}) = (1, 1)$ and midpoint of BD $= (\frac{0 + 2}{2}, \frac{2 + 0}{2}) = (1, 1)$ . Same point, so the diagonals bisect each other. [1]

Answer: ABCD is a rhombus (sides $10$ units, diagonals $42$ and $22$ units), not a square; both diagonals have midpoint $(1, 1)$ .

Example 10: Point on a given line equidistant from two points · 3 marks · Short answerHarder board level

Find the point on the line $y = 2 x - 3$ which is equidistant from $A (2, 3)$ and $B (6, - 1)$ . Also find the common distance.

Try it first, then show the solutionSolution

Any point on the line is $P (x, 2 x - 3)$ . [½]
$P A^{2} = (x - 2)^{2} + (2 x - 6)^{2} = x^{2} - 4 x + 4 + 4 x^{2} - 24 x + 36 = 5 x^{2} - 28 x + 40$ [½]
$P B^{2} = (x - 6)^{2} + (2 x - 2)^{2} = x^{2} - 12 x + 36 + 4 x^{2} - 8 x + 4 = 5 x^{2} - 20 x + 40$ [½]
$P A^{2} = P B^{2}$ gives $- 28 x = - 20 x$ , so $x = 0$ and $y = 2 (0) - 3 = - 3$ . P is $(0, - 3)$ . [1]
$P A = 4 + 36 = 40 = 210$ units; check $P B = 36 + 4 = 210$ units. [½]

Answer: $P (0, - 3)$ ; the common distance is $210$ units.

Example 11: Case study: planning a cable loop · 4 marks · Case-basedHOTS / competency

Imagine a village plans electric poles on a map grid where 1 unit = 100 m. Four poles stand at $A (1, 2)$ , $B (5, 0)$ , $C (7, 4)$ and $D (3, 6)$ . A straight road runs along the line $y = 7$ .
(i) Show that the cable loop A to B to C to D and back to A has four equal sides. (1 mark)
(ii) An engineer says the loop is only a rhombus, not a square. Decide whether he is right, with reason. (1 mark)
(iii) A junction box J is to be placed on the road so that it is equidistant from poles A and B. Find J. Two ways to run a cable from J to A are proposed: straight from J to A, or from J to D and then D to A. Which is shorter? Justify without a calculator. (2 marks)

Try it first, then show the solutionSolution

(i) $A B^{2} = 16 + 4 = 20$ , $B C^{2} = 4 + 16 = 20$ , $C D^{2} = 16 + 4 = 20$ , $D A^{2} = 4 + 16 = 20$ , so each side is $25$ units $= 2005$ m. [1]
(ii) $A C^{2} = (7 - 1)^{2} + (4 - 2)^{2} = 40$ and $B D^{2} = (3 - 5)^{2} + (6 - 0)^{2} = 40$ , so $A C = B D$ . Equal sides and equal diagonals make it a square, so the engineer is wrong. [1]
(iii) Let $J (x, 7)$ . $J A^{2} = J B^{2}$ : $(x - 1)^{2} + 25 = (x - 5)^{2} + 49$ . [½]
$- 2 x + 26 = - 10 x + 74$ , so $8 x = 48$ , $x = 6$ , and J is $(6, 7)$ . [½]
Direct: $J A = 25 + 25 = 52$ units. Via D: $J D = 9 + 1 = 10$ and $D A = 25$ units. [½]
J, D and A are not on one line, so by the triangle inequality in $△ J D A$ , $J A < J D + D A$ : the direct cable is shorter without any calculation. Check: $J A = 52$ units and $(10 + 25)^{2} = 30 + 202 > 50 = (52)^{2}$ . So the direct cable, $5002$ m, is shorter. [½]

Answer: (i) Each side is $2005$ m. (ii) Diagonals are equal, so it is a square; the engineer is wrong. (iii) $J (6, 7)$ ; the direct cable JA ( $5002$ m) is shorter than the route through D.

Example 12: Assertion-reason: collinearity · 1 mark · Assertion-reasonBoard level

Assertion (A): The points $A (1, 1)$ , $B (2, 3)$ and $C (3, 5)$ are collinear.
Reason (R): If three points A, B, C satisfy $A B + B C = A C$ , then they lie on one straight line.
(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.

Try it first, then show the solutionSolution

$A B = 1 + 4 = 5$ , $B C = 1 + 4 = 5$ , $A C = 4 + 16 = 25$ , so $A B + B C = A C$ and A is true. [½]
R is the distance test for collinearity, which is true, and it is exactly what proves A. [½]

Answer: (a) Both A and R are true and R is the correct explanation of A.

Where students lose marks in Applications of the Distance Formula

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

Mistake: Calling a figure a square after checking only the sides
Four equal sides also describe a rhombus. A square needs equal diagonals as well.
Wrong working
$A B = B C = C D = D A$ , hence ABCD is a square.
Right working
$A B = B C = C D = D A$ and $A C = B D$ , hence ABCD is a square.
Do this: For any quadrilateral, always find both diagonals before naming it.
Usually costs: The conclusion mark and often a step mark in a 4- or 5-mark question.
Mistake: Adding unsimplified surds in a collinearity check
$a + b$ is not $a + b$ . Only like surds can be added.
Wrong working
$8 + 18 = 26$ , which is not $50$ , so not collinear.
Right working
$22 + 32 = 52 = 50$ , so collinear.
Do this: Simplify every distance to simplest surd form before adding.
Usually costs: The whole conclusion, since the answer becomes wrong.
Mistake: Keeping only one value of the unknown
An equation like $(y + 3)^{2} = 36$ has two roots.
Wrong working
$(y + 3)^{2} = 36$ , so $y + 3 = 6$ , $y = 3$ .
Right working
$y + 3 = \pm 6$ , so $y = 3$ or $y = - 9$ .
Do this: Write $\pm$ when you take the square root, then reject a value only if the question gives a reason.
Usually costs: Half to one mark.
Mistake: Sign slips when subtracting negative coordinates
Squaring hides the sign of the difference but not a wrong size.
Wrong working
For $(- 3, 4)$ and $(5, 1)$ : $(5 - 3)^{2} + (1 - 4)^{2} = 4 + 9 = 13$ .
Right working
$(5 - (- 3))^{2} + (1 - 4)^{2} = 64 + 9 = 73$ .
Do this: Put negative coordinates in brackets before subtracting.
Usually costs: Every later mark that depends on that length.
Mistake: Writing a point on the x-axis as $(0, x)$
On the x-axis the y-coordinate is 0, so the point is $(x, 0)$ .
Wrong working
Point on the x-axis equidistant from A and B: let $P = (0, k)$ .
Right working
Let $P = (x, 0)$ , then set $P A^{2} = P B^{2}$ .
Do this: Say it aloud: on the x-axis, y is zero.
Usually costs: The full question, as all working is on the wrong point.

Exam tips

In shape proofs, write each squared length on its own line and end with a clear conclusion sentence giving the reason; the conclusion carries a mark.
In collinearity questions, say which point lies between the other two, since that is what $A B + B C = A C$ shows.
In equidistant-point questions, mention that you square both sides because lengths are non-negative; examiners like the reason.
When a question asks you to "name" the quadrilateral, test the diagonals even if the sides look different; it may be a rectangle.
Give final lengths in simplest surd form with units, and check one answer by substitution if time allows.

Practice questions on Applications of the 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 · 2 marks · Very short answerBoard level
For what values of p is the point $(p, 0)$ at a distance of 13 units from $(5, 12)$ ? Which of these points lies on the positive x-axis?
Hint
Square the distance and solve; then check the sign of p.
Answer
$(p - 5)^{2} + 144 = 169$ , so $(p - 5)^{2} = 25$ and $p = 10$ or $p = 0$ . (1) $(0, 0)$ is the origin, so only $(10, 0)$ lies on the positive x-axis. (1)
Q2 · 3 marks · Short answerBoard level
Find the point on the y-axis which is equidistant from $A (6, 5)$ and $B (- 4, 3)$ . Also find the common distance.
Hint
Take the point as $(0, y)$ and set $P A^{2} = P B^{2}$ .
Answer
$36 + (y - 5)^{2} = 16 + (y - 3)^{2}$ gives $61 - 10 y = 25 - 6 y$ , so $y = 9$ . Point $(0, 9)$ ; distance $36 + 16 = 52 = 213$ units.
Q3 · 3 marks · Short answerBoard level
A point $Q (1, 2)$ is at the same distance from $P (4, 6)$ as from $R (x, - 1)$ . Find the values of x, and find PR for each value.
Hint
Equate $Q P^{2}$ and $Q R^{2}$ .
Answer
$Q P^{2} = 9 + 16 = 25$ and $Q R^{2} = (x - 1)^{2} + 9$ , so $(x - 1)^{2} = 16$ and $x = 5$ or $x = - 3$ . For $R (5, - 1)$ : $P R = 1 + 49 = 52$ units. For $R (- 3, - 1)$ : $P R = 49 + 49 = 72$ units.
Q4 · 5 marks · Long answerHarder board level
Show that the points $A (2, 3)$ , $B (5, 1)$ , $C (7, 4)$ and $D (4, 6)$ , taken in order, are the vertices of a square. Find the length of its diagonal and its area.
Hint
Check all four sides and both diagonals; the area of a square is side squared.
Answer
$A B^{2} = 9 + 4 = 13$ , $B C^{2} = 4 + 9 = 13$ , $C D^{2} = 9 + 4 = 13$ , $D A^{2} = 4 + 9 = 13$ , so all sides are $13$ units. (2) $A C^{2} = 25 + 1 = 26$ and $B D^{2} = 1 + 25 = 26$ , so the diagonals are equal. A rhombus with equal diagonals is a square. (2) Diagonal $= 26$ units; area $= (13)^{2} = 13$ square units. (1)
Q5 · 3 marks · Short answerHOTS / competency
A student found that for $A (0, 0)$ , $B (3, 4)$ , $C (8, 4)$ and $D (5, 0)$ , taken in order, $A B = B C = C D = D A = 5$ units, and wrote "so ABCD is a square". Find the error, and name the quadrilateral correctly with reason.
Hint
Four equal sides are not enough. Check the diagonals.
Answer
The student did not check the diagonals. $A C^{2} = 64 + 16 = 80$ , so $A C = 45$ ; $B D^{2} = 4 + 16 = 20$ , so $B D = 25$ . Sides equal but diagonals unequal, so ABCD is a rhombus, not a square.
Q6 · 1 mark · MCQHarder board level
The point $P (x, 0)$ is equidistant from $A (2, 5)$ and $B (6, 3)$ . The common distance PA is
(a) 5 units
(b) $34$ units
(c) 4 units
(d) $41$ units
Hint
Find x first, then the distance.
Answer
(a) 5 units. $(x - 2)^{2} + 25 = (x - 6)^{2} + 9$ gives $8 x = 16$ , $x = 2$ ; then $P A = 0 + 25 = 5$ (and $P B = 16 + 9 = 5$ ).
Q7 · 1 mark · MCQHarder board level
The point $B (3, 4)$ is equidistant from $A (k, 2)$ and $C (5, 6)$ . Which of the following is a possible value of k?
(a) 1
(b) 3
(c) 7
(d) $- 1$
Hint
Solve $B A^{2} = B C^{2}$ ; there are two roots, check the options.
Answer
(a) 1. $(k - 3)^{2} + 4 = 4 + 4$ gives $(k - 3)^{2} = 4$ , so $k = 1$ or $k = 5$ ; only 1 is among the options.
Q8 · 1 mark · MCQBoard level
If the distance between $(4, k)$ and $(1, 0)$ is 5 units, then k is
(a) $4$ only
(b) $\pm 4$
(c) $\pm 3$
(d) $\pm 5$
Hint
Square the distance and remember both roots.
Answer
(b) $\pm 4$ . $9 + k^{2} = 25$ , so $k^{2} = 16$ and $k = \pm 4$ .
Q9 · 1 mark · Assertion-reasonBoard level
Assertion (A): The points $(3, 0)$ , $(0, 1)$ , $(- 3, 0)$ and $(0, - 1)$ , taken in order, are the vertices of a square.
Reason (R): All four sides of the quadrilateral formed by these points are equal to $10$ units.
(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 the sides, then the two diagonals.
Answer
(d) A is false but R is true. Each side is $9 + 1 = 10$ units, but the diagonals are 6 and 2 units, so the figure is a rhombus, not a square.
Q10 · 3 marks · Short answerHarder board level
Find the points P on the x-axis such that $△ P A B$ , with $A (1, 3)$ and $B (5, 1)$ , is right-angled at P.
Hint
Let P be $(x, 0)$ and use $P A^{2} + P B^{2} = A B^{2}$ .
Answer
$(x - 1)^{2} + 9 + (x - 5)^{2} + 1 = 16 + 4$ gives $2 x^{2} - 12 x + 36 = 20$ , so $x^{2} - 6 x + 8 = 0$ and $x = 2$ or $x = 4$ . P is $(2, 0)$ or $(4, 0)$ . Check for $(2, 0)$ : $10 + 10 = 20$ ; for $(4, 0)$ : $18 + 2 = 20$ .
Q11 · 3 marks · Short answerHarder board level
Find the point which is equidistant from $A (6, 3)$ , $B (- 1, 2)$ and $C (7, - 4)$ , and find the common distance.
Hint
Set $P A^{2} = P B^{2}$ and $P A^{2} = P C^{2}$ ; you get two linear equations.
Answer
$P A^{2} = P B^{2}$ gives $7 x + y = 20$ ; $P A^{2} = P C^{2}$ gives $x - 7 y = 10$ . Solving, $x = 3$ and $y = - 1$ , so the point is $(3, - 1)$ . Common distance $= 9 + 16 = 5$ units (also 5 units to B and to C).
Q12 · 3 marks · Short answerHarder board level
The points $A (1, 2)$ , $B (4, 6)$ and $C (k, 14)$ are such that $A C = 3 A B$ . Find the possible values of k, and decide which value makes A, B and C collinear.
Hint
Find AB first, then use $A C^{2} = 9 A B^{2}$ . Test each value with the distance test.
Answer
$A B = 9 + 16 = 5$ , so $A C = 15$ : $(k - 1)^{2} + 144 = 225$ , $(k - 1)^{2} = 81$ , $k = 10$ or $k = - 8$ . For $k = 10$ : $B C = 36 + 64 = 10$ and $A B + B C = 15 = A C$ , so collinear. For $k = - 8$ : $B C^{2} = 144 + 64 = 208$ , so $B C = 413$ , and no two of the distances add up to the third, so not collinear. So $k = 10$ makes them collinear.
Q13 · 3 marks · Short answerHOTS / competency
A student is asked to find a point on the line $y = x + 1$ that is equidistant from $A (2, 3)$ and $B (6, - 1)$ . Show that no such point exists. Then find the equation satisfied by every point $(x, y)$ equidistant from A and B, and use it to explain why.
Hint
Write the point as $(x, x + 1)$ and see what happens when you equate the squared distances.
Answer
For $P (x, x + 1)$ : $P A^{2} = (x - 2)^{2} + (x - 2)^{2} = 2 x^{2} - 8 x + 8$ and $P B^{2} = (x - 6)^{2} + (x + 2)^{2} = 2 x^{2} - 8 x + 40$ . Equating gives $8 = 40$ , which is impossible, so no such point exists. For a general point, $(x - 2)^{2} + (y - 3)^{2} = (x - 6)^{2} + (y + 1)^{2}$ simplifies to $y = x - 3$ . The pair $y = x - 3$ and $y = x + 1$ is inconsistent (the lines are parallel), so the given line never meets the set of equidistant points.
Q14 · 4 marks · Case-basedHOTS / competency
Suppose a PE teacher marks three cones on a ground grid where 1 unit = 5 m: $P (0, 0)$ , $Q (8, 0)$ and $R (4, 8)$ .
(i) Show that $△ P QR$ is isosceles. (1 mark)
(ii) A student claims the three cones form an equilateral triangle. Explain why the claim is wrong. (1 mark)
(iii) The teacher wants to stand at a point S equidistant from all three cones. Find S and her distance from each cone in metres. (2 marks)
Hint
In (iii), use $S P^{2} = S Q^{2}$ first; it fixes x.
Answer
(i) $P R^{2} = 16 + 64 = 80$ and $Q R^{2} = 16 + 64 = 80$ , so $P R = QR = 45$ units and the triangle is isosceles. (ii) $P Q = 8$ units, so $P Q^{2} = 64 \neq = 80$ ; the sides are not all equal. (iii) Let $S (x, y)$ . $S P^{2} = S Q^{2}$ gives $x^{2} = (x - 8)^{2}$ , so $x = 4$ . $S P^{2} = S R^{2}$ gives $16 + y^{2} = (8 - y)^{2}$ , so $16 = 64 - 16 y$ and $y = 3$ . S is $(4, 3)$ ; $S P = S Q = S R = 5$ units $= 25$ m.
Q15 · 3 marks · Short answerHarder board level
Show that the points $(a, a)$ , $(- a, - a)$ and $(- 3 a, 3 a)$ , where $a > 0$ , are the vertices of an equilateral triangle, and find its side.
Hint
Expand each squared distance; $(3)^{2} = 3$ .
Answer
Side 1: $(2 a)^{2} + (2 a)^{2} = 8 a^{2}$ . Side 2: $(- a + 3 a)^{2} + (- a - 3 a)^{2} = a^{2} (4 - 23) + a^{2} (4 + 23) = 8 a^{2}$ . Side 3: $(a + 3 a)^{2} + (a - 3 a)^{2} = a^{2} (4 + 23) + a^{2} (4 - 23) = 8 a^{2}$ . All three sides are $22 a$ , so the triangle is equilateral.
Q16 · 3 marks · Short answerHarder board level
Show that $A (1, 2)$ , $B (4, 3)$ , $C (6, 6)$ and $D (3, 5)$ , taken in order, form a parallelogram but not a rectangle, and verify that its diagonals bisect each other.
Hint
Opposite sides equal, diagonals unequal; then compare the midpoints of AC and BD.
Answer
$A B^{2} = 9 + 1 = 10 = C D^{2}$ and $B C^{2} = 4 + 9 = 13 = D A^{2}$ , so both pairs of opposite sides are equal: a parallelogram. $A C^{2} = 25 + 16 = 41$ and $B D^{2} = 1 + 4 = 5$ , so the diagonals are unequal and it is not a rectangle. Midpoint of AC $= (3.5, 4)$ = midpoint of BD, so the diagonals bisect each other.

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.Condition for A, B, C to be collinear by distances?Show answer
The two smaller distances add to the largest, for example $A B + B C = A C$ .
2.What extra condition turns a rhombus into a square?Show answer
Equal diagonals: $A C = B D$ .
3.How do you test for a right angle at B?Show answer
Check $A B^{2} + B C^{2} = A C^{2}$ .
4.How is a point on the y-axis written?Show answer
$(0, y)$ .
5.Why square both sides of $P A = P B$ ?Show answer
To remove the roots; it is safe because lengths are never negative.
6.What do opposite sides equal and diagonals equal prove?Show answer
A rectangle.

Applications of the Distance Formula: frequently asked questions

Can I prove collinearity without the area of a triangle?

Yes. Find the three distances and check whether the two smaller ones add up to the largest. Area of a triangle from coordinates is not in the 2026-27 syllabus, so the distance method is the one to use.

Do I need to find the diagonals to prove a square?

Yes. Four equal sides only prove a rhombus. You must also show the two diagonals are equal to conclude it is a square.

Should I leave lengths as surds?

Yes, write them in simplest surd form, such as 3 root 5 units, unless the question asks for a decimal. For comparing sides, work with squared lengths and you can often avoid surds completely.

Why do I sometimes get two answers for an unknown coordinate?

A given distance leads to a square equal to a number, which has two roots. Both points are at that distance, so keep both unless the question names a quadrant or another condition that rules one out.

How do I show a triangle is isosceles using the distance formula?

Find all three sides with the distance formula. If any two are equal, the triangle is isosceles. Also check whether it is right-angled, as board questions often ask for both.