MathsCBSE Class 9 · Orienting Yourself: The Use of Coordinates

Distance Formula, Midpoint and Collinear Points: Class 9

Updated 17 October 2026 · Fact-checked

The distance between $A(x_1, y_1)$ and $B(x_2, y_2)$ is $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$, which is Pythagoras on the right triangle with legs $|x_2 - x_1|$ and $|y_2 - y_1|$. The midpoint is $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$. Three points are collinear if the largest of the three distances equals the sum of the other two. To classify a figure, compare side lengths and diagonals.

Join the two points, drop a right triangle, and the run and rise are the legs, so the distance is the hypotenuse, √(12² + 5²) = 13.

Illustration drawn for QuestionBuddy.

Understand Distance Formula, Midpoint and Collinear Points

Distance formula. Draw a right triangle with $A (x_{1}, y_{1})$ and $B (x_{2}, y_{2})$ at the ends of the hypotenuse: the horizontal leg is $∣ x_{2} - x_{1} ∣$ and the vertical leg is $∣ y_{2} - y_{1} ∣$ . By the Baudhāyana–Pythagoras theorem,
$ $A B = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} .$ $
The distance from $(- 5, - 2)$ to $(7, 3)$ is $1 2^{2} + 5^{2} = 13$ . The distance of $(x, y)$ from the origin is $x^{2} + y^{2}$ ; for $(- 8, 15)$ it is $17$ . The order of the points does not matter because the differences are squared.

Midpoint. The midpoint $M$ of $A B$ is $(\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})$ . The midpoint of $(- 6, 4)$ and $(10, - 2)$ is $(2, 1)$ . If $M$ and one end $A$ are known, the other end is $B = (2 x_{M} - x_{A}, 2 y_{M} - y_{A})$ : with $M (3, - 1)$ and $A (7, 5)$ , $B = (- 1, - 7)$ . The midpoint of a hypotenuse is equidistant from all three vertices of a right triangle.

Right-angle test. A triangle is right-angled if the squares of two sides add up to the square of the third (converse of the Baudhāyana–Pythagoras theorem). With coordinates, find the three side lengths squared: you do not need the square roots. For $A (- 1, 2)$ , $B (3, - 1)$ , $C (6, 3)$ : $A B^{2} = 25$ , $B C^{2} = 25$ , $A C^{2} = 50$ , so $A B^{2} + B C^{2} = A C^{2}$ and the triangle is right-angled and isosceles at $B$ .

Collinear points. Three points are collinear if they lie on one line. The test with distances: the largest distance equals the sum of the other two. For $A (- 2, 1)$ , $B (1, 4)$ , $C (4, 7)$ : $A B = B C = 32$ and $A C = 62$ , so $A B + B C = A C$ and $B$ is the midpoint of $A C$ . If the three distances do not satisfy this, the points form a triangle.

Identifying quadrilaterals. Compute all four sides and both diagonals:

all four sides equal: rhombus; if the diagonals are also equal: square;
opposite sides equal: parallelogram; with equal diagonals: rectangle;
the diagonals of a parallelogram bisect each other: equal midpoints of the two diagonals is a quick test.

For $A (0, 0)$ , $B (10, 0)$ , $C (16, 8)$ , $D (6, 8)$ : every side is $10$ and the diagonals are $85$ and $45$ , so it is a rhombus but not a square; the area is $\frac{1}{2} \times 85 \times 45 = 80$ .

Points on a circle. A point $P$ is on the circle with centre $C$ and radius $r$ if $C P = r$ . For $C (2, - 1)$ , $r = 5$ : $P (5, 3)$ has $C P = 9 + 16 = 5$ , so it lies on the circle, while $(5, 5)$ is at distance $45 > 5$ , outside. The other end of the diameter through $P$ is the reflection of $P$ through $C$ : $(- 1, - 5)$ .

Method. Square the differences, add, and keep the square root only at the end. Compare squared lengths when you only need to check equalities.

Distance Formula, Midpoint and Collinear Points: formulas to remember

Distance formula: $A B = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$; Pythagoras on the grid.
Distance from the origin: $x^{2} + y^{2}$; Special case.
Midpoint: $M = (\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})$; Average of the coordinates.
Other end from a midpoint: $B = (2 x_{M} - x_{A}, 2 y_{M} - y_{A})$; Reflection of $A$ in $M$ .
Right-angle test: $A B^{2} + B C^{2} = A C^{2}$; Right angle at $B$ .
Collinear test: $A B + B C = A C$; $B$ lies between $A$ and $C$ .
Point on a circle: $C P = r$; $C$ centre, $r$ radius.

How to solve Distance Formula, Midpoint and Collinear Points questions

Convert every geometric statement into a distance or midpoint calculation.

1Write the coordinates of each point clearly and label them.
2Compute the required distances using the formula; keep squared values when only comparing.
3For midpoints, average the x-coordinates and y-coordinates separately.
4For other-end questions, use $2 M - A$ coordinate by coordinate.
5For classification, compute all sides and both diagonals and use the properties of each quadrilateral.
6Verify with a second property, such as equal midpoints of the diagonals.

Distance Formula, Midpoint and Collinear Points 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.

Distance and midpoint

Suppose two villages are at (2, 3) and (8, 11) on a map where each unit is 1 km, and a hospital is to be built exactly midway between them.

How far apart are the villages along a straight road, and where should the hospital be?

Try it first, then show the solutionSolution

Distance $= (8 - 2)^{2} + (11 - 3)^{2} = 36 + 64 = 10$ km.
Midpoint $= (\frac{2 + 8}{2}, \frac{3 + 11}{2}) = (5, 7)$ .

Answer: The villages are 10 km apart and the hospital should be at (5, 7).

Memory hook: Pythagoras for distance, average for midpoint.

Tricks to remember Distance Formula, Midpoint and Collinear Points

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

Differences squared, then add
Subtract x from x and y from y, square both, add, root.
Use it when: Distance.
Average the coordinates
Midpoint is the average of the x-values and the average of the y-values.
Use it when: Midpoint questions.
Longest equals the sum
If the longest of three distances equals the sum of the two others, the points are on a line.
Use it when: Collinearity.

Distance Formula, Midpoint and Collinear Points: solved examples

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

Example 1: Distances · 3 marks · Short answerBoard level

Find the distance between $(- 5, - 2)$ and $(7, 3)$ , and the distance of $(- 8, 15)$ from the origin.

Try it first, then show the solutionSolution

Differences: $7 - (- 5) = 12$ and $3 - (- 2) = 5$ . (1)
Distance $= 1 2^{2} + 5^{2} = 169 = 13$ . (1)
From the origin: $(- 8)^{2} + 1 5^{2} = 289 = 17$ . (1)

Answer: 13 units and 17 units.

Example 2: Midpoint and other end · 3 marks · Short answerBoard level

Find the midpoint of the segment joining $(- 6, 4)$ and $(10, - 2)$ . If $M (3, - 1)$ is the midpoint of $A B$ and $A$ is $(7, 5)$ , find $B$ .

Try it first, then show the solutionSolution

Midpoint $= (\frac{- 6 + 10}{2}, \frac{4 + ( - 2 )}{2}) = (2, 1)$ . (1)
For $B$ : $\frac{7 + x}{2} = 3$ gives $x = - 1$ . (1)
$\frac{5 + y}{2} = - 1$ gives $y = - 7$ , so $B = (- 1, - 7)$ . (1)

Answer: The midpoint is $(2, 1)$ and $B = (- 1, - 7)$ .

Example 3: Show a rhombus · 5 marks · Long answerHarder board level

Show that $A (0, 0)$ , $B (10, 0)$ , $C (16, 8)$ and $D (6, 8)$ are the vertices of a rhombus and not a square. Find its area in two ways.

Try it first, then show the solutionSolution

$A B = 10$ ; $B C = 6^{2} + 8^{2} = 10$ ; $C D = ∣16 - 6∣ = 10$ ; $D A = 6^{2} + 8^{2} = 10$ . (1½)
All four sides are equal, so it is a rhombus. (½)
Diagonals: $A C = 1 6^{2} + 8^{2} = 320 = 85$ and $B D = 4^{2} + 8^{2} = 80 = 45$ ; they are unequal, so it is not a square. (1)
Area $= \frac{1}{2} \times 85 \times 45 = \frac{1}{2} \times 160 = 80$ square units. (1)
Check: base $A B = 10$ and the height is the vertical distance $8$ , so area $= 10 \times 8 = 80$ ; both diagonals have the midpoint $(8, 4)$ . (1)

Answer: It is a rhombus (not a square) with area 80 square units.

Example 4: Collinear or not · 3 marks · Short answerHarder board level

Show that $A (- 2, 1)$ , $B (1, 4)$ and $C (4, 7)$ are collinear and that $B$ is the midpoint of $A C$ . Find $k$ so that $(7, k)$ lies on the same line, given that the line is $y = x + 3$ .

Try it first, then show the solutionSolution

$A B = 9 + 9 = 32$ , $B C = 9 + 9 = 32$ , $A C = 36 + 36 = 62$ . (1)
$A B + B C = 62 = A C$ , so the points are collinear. (1)
Midpoint of $A C = (\frac{- 2 + 4}{2}, \frac{1 + 7}{2}) = (1, 4) = B$ . On $y = x + 3$ , $k = 7 + 3 = 10$ . (1)

Answer: They are collinear, $B$ is the midpoint of $A C$ , and $k = 10$ .

Example 5: Fountain on a circle · 4 marks · Case-basedHOTS / competency

Case: a circular fountain has its centre at $C (2, - 1)$ on a map (units in metres) and a radius of 5 m. (i) Is a lamp at $P (5, 3)$ on the boundary? (ii) Is a bench at $T (5, 5)$ inside, on or outside the boundary? (iii) A diameter passes through $P$ ; find the other end $Q$ . (iv) Verify that $C$ is the midpoint of $P Q$ and that $Q$ is on the circle.

Try it first, then show the solutionSolution

(i) $C P = 3^{2} + 4^{2} = 5$ , equal to the radius, so $P$ is on the boundary. (1)
(ii) $C T = 3^{2} + 6^{2} = 45 \approx 6.7 > 5$ , so $T$ is outside. (1)
(iii) $Q = (2 \times 2 - 5, 2 \times (- 1) - 3) = (- 1, - 5)$ . (1)
(iv) Midpoint of $P Q = (\frac{5 + ( - 1 )}{2}, \frac{3 + ( - 5 )}{2}) = (2, - 1) = C$ ; $C Q = 3^{2} + 4^{2} = 5$ . (1)

Answer: (i) yes; (ii) outside; (iii) $Q = (- 1, - 5)$ ; (iv) verified.

Where students lose marks in Distance Formula, Midpoint and Collinear Points

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

Mistake: Forgetting to square the differences
Adding differences directly.
Wrong working
Distance $= 12 + 5 = 17$ .
Right working
$1 2^{2} + 5^{2} = 13$ .
Do this: Write the formula with squares each time.
Usually costs: The whole answer.
Mistake: Mixing up coordinates in the midpoint
Averaging x with y.
Wrong working
Midpoint of $(2, 3)$ and $(8, 11)$ is $(\frac{2 + 3}{2}, \frac{8 + 11}{2})$ .
Right working
$(\frac{2 + 8}{2}, \frac{3 + 11}{2}) = (5, 7)$ .
Do this: Add the two x-coordinates, then the two y-coordinates.
Usually costs: The whole answer.
Mistake: Concluding "square" from equal sides
Not checking the diagonals.
Wrong working
All sides are 5 so it is a square.
Right working
All sides equal gives a rhombus; the diagonals 8 and 6 are unequal, so it is not a square.
Do this: Compute diagonals before naming the shape.
Usually costs: 1 to 2 marks.

Exam tips

Write squared differences on separate lines, then add.
To check equalities, compare squares to avoid surds.
Name a shape only after checking all required properties (sides and diagonals).
For collinearity, say which point lies between the other two.

Practice questions on Distance Formula, Midpoint and Collinear Points

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 $(- 6, 8)$ from the origin is
(a) 10
(b) 14
(c) 2
(d) 100
Hint
Use $x^{2} + y^{2}$ .
Answer
(a) 10. $36 + 64 = 10$ .
Q2 · 3 marks · Short answerBoard level
Find the midpoint of the segment joining $A (2, - 3)$ and $B (- 4, 7)$ . If $M (1, 2)$ is the midpoint of $P Q$ with $P (- 3, - 2)$ , find $Q$ .
Hint
Average for the midpoint; double and subtract for the end.
Answer
Midpoint of $A B = (\frac{2 - 4}{2}, \frac{- 3 + 7}{2}) = (- 1, 2)$ . $Q = (2 \times 1 - (- 3), 2 \times 2 - (- 2)) = (5, 6)$ . Check: $\frac{- 3 + 5}{2} = 1$ and $\frac{- 2 + 6}{2} = 2$ .
Q3 · 3 marks · Short answerHarder board level
Find the point on the x-axis that is equidistant from $A (2, - 5)$ and $B (- 2, 9)$ .
Hint
Let the point be $(x, 0)$ and equate the squared distances.
Answer
Let $P (x, 0)$ . $P A^{2} = (x - 2)^{2} + 25$ and $P B^{2} = (x + 2)^{2} + 81$ . Equating: $x^{2} - 4 x + 4 + 25 = x^{2} + 4 x + 4 + 81$ , so $- 8 x = 56$ and $x = - 7$ . $P (- 7, 0)$ ; check $P A^{2} = 81 + 25 = 106 = P B^{2} = 25 + 81$ .
Q4 · 5 marks · Long answerHarder board level
Show that $A (- 1, 2)$ , $B (3, - 1)$ and $C (6, 3)$ form a right-angled isosceles triangle. Find its area, and find the point $D$ such that $A B C D$ is a square. Verify with the midpoints of the diagonals.
Hint
Use squared lengths $A B^{2}$ , $B C^{2}$ , $A C^{2}$ ; for $D$ , use $D = A + C - B$ .
Answer
$A B^{2} = 16 + 9 = 25$ , $B C^{2} = 9 + 16 = 25$ , $A C^{2} = 49 + 1 = 50$ . So $A B = B C$ and $A B^{2} + B C^{2} = A C^{2}$ : a right-angled isosceles triangle with the right angle at $B$ . Area $= \frac{1}{2} \times 5 \times 5 = 12.5$ square units. In a square, the diagonals bisect each other, so $D = A + C - B = (- 1 + 6 - 3, 2 + 3 + 1) = (2, 6)$ . Midpoint of $A C = (2.5, 2.5)$ and midpoint of $B D = (\frac{3 + 2}{2}, \frac{- 1 + 6}{2}) = (2.5, 2.5)$ , and $A D^{2} = 9 + 16 = 25$ .
Q5 · 4 marks · Case-basedHOTS / competency
Case: three towns are at $A (0, 0)$ , $B (12, 5)$ and $C (- 5, 12)$ on a map (units in km). (i) Find $A B$ and $A C$ . (ii) Find $B C$ . (iii) Show that the triangle is right-angled and say where the right angle is. (iv) A hospital is built at the midpoint of $B C$ ; find its coordinates and its distance from $A$ .
Hint
$B C^{2} = 1 7^{2} + 7^{2}$ .
Answer
(i) $A B = 144 + 25 = 13$ , $A C = 25 + 144 = 13$ . (ii) $B C^{2} = 1 7^{2} + 7^{2} = 338$ , so $B C = 338 = 132$ . (iii) $A B^{2} + A C^{2} = 169 + 169 = 338 = B C^{2}$ , so the right angle is at $A$ (and the triangle is isosceles). (iv) The midpoint is $(3.5, 8.5)$ ; its distance from $A$ is $12.25 + 72.25 = 84.5 \approx 9.19$ km, which is half of $B C$ ( $132 /2$ ), as the midpoint of the hypotenuse of a right triangle is equidistant from all vertices.

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 formula?Show answer
$(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$ .
2.Midpoint formula?Show answer
$(\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})$ .
3.Distance of $(x, y)$ from the origin?Show answer
$x^{2} + y^{2}$ .
4.Collinear distance test?Show answer
The largest distance equals the sum of the other two.
5.Square versus rhombus?Show answer
A square has equal diagonals too.

Distance Formula, Midpoint and Collinear Points: frequently asked questions

Why is the distance formula the same as Pythagoras?

Because the horizontal and vertical differences form the legs of a right triangle whose hypotenuse is the segment.

Can the midpoint have fractional coordinates?

Yes. For example the midpoint of (1, 2) and (2, 4) is (1.5, 3).

Is a quick test for a parallelogram available?

Yes: a quadrilateral is a parallelogram if its diagonals have the same midpoint.