Class 9 Maths Ch 1: Coordinate Geometry Notes

1The Cartesian plane and ordered pairs

Draw a horizontal number line (the x-axis) and a vertical number line (the y-axis) that cross at their zeros. The crossing point is the origin O, written $(0, 0)$ . Together the two axes make the Cartesian plane, named after René Descartes.

Every point P is described by an ordered pair $(x, y)$ . The first number, the x-coordinate or abscissa, tells you how far to move right (positive) or left (negative) from the origin. The second, the y-coordinate or ordinate, tells you how far to move up (positive) or down (negative).

Order matters: $(2, 5)$ means 2 right and 5 up, while $(5, 2)$ means 5 right and 2 up — two different points. Think of a floor plan drawn on squared paper: once you fix one corner as the origin, every table, door and window gets an exact address.

Key formulas & definitions

Origin: $O(0, 0)$ , where the axes meet at right angles.
Point $P(x, y)$ : $x$ = signed distance from the y-axis, $y$ = signed distance from the x-axis.
$(a, b) = (c, d)$ only when $a = c$ and $b = d$ .

Worked example 1

A classroom floor plan uses the bottom-left corner as the origin and 1 unit = 1 m. The teacher's desk is 4 m from the left wall and 1 m from the front wall (the x-axis). A student's bench is 2 m from the left wall and 5 m from the front wall. Write their coordinates.

Distance from the left wall (the y-axis) gives the x-coordinate; distance from the front wall (the x-axis) gives the y-coordinate.
Desk: $x = 4$ , $y = 1$ , so $D(4, 1)$ .
Bench: $x = 2$ , $y = 5$ , so $B(2, 5)$ .

Answer: $D(4, 1)$ and $B(2, 5)$ .

Trick / memory aid

“Walk before you climb”: the x-coordinate (walk along the floor) always comes before the y-coordinate (climb the ladder).

Common mistakes

Swapping the coordinates — writing $(5, 2)$ for a point 2 right and 5 up.
Counting grid lines instead of units when the scale is not 1 unit per square.

2Quadrants and points on the axes

The axes cut the plane into four quadrants, numbered anticlockwise starting from the top right. The signs of the coordinates tell you the quadrant immediately, without plotting.

A point that lies on an axis belongs to no quadrant. Every point on the x-axis has $y = 0$ (it has not moved up or down), and every point on the y-axis has $x = 0$ .

The distance of $P(x, y)$ from the x-axis is $|y|$ and from the y-axis is $|x|$ — distances are never negative, so we take absolute values.

Key formulas & definitions

Quadrant I: $(+, +)$ · Quadrant II: $(-, +)$ · Quadrant III: $(-, -)$ · Quadrant IV: $(+, -)$
On the x-axis: $(a, 0)$ . On the y-axis: $(0, b)$ .
Distance of $(x, y)$ from the x-axis $= |y|$ ; from the y-axis $= |x|$ .

Worked example 1

State where each point lies: $A(-3, 4)$ , $B(0, -6)$ , $C(7, -2)$ , $D(-1, -1)$ .

$A(-3, 4)$ : $x$ negative, $y$ positive → Quadrant II.
$B(0, -6)$ : $x = 0$ → on the y-axis (below the origin).
$C(7, -2)$ : $x$ positive, $y$ negative → Quadrant IV.
$D(-1, -1)$ : both negative → Quadrant III.

Answer: A: II, B: y-axis, C: IV, D: III.

Trick / memory aid

Go round like a clock running backwards starting at the top right: I (+,+), II (−,+), III (−,−), IV (+,−). Only the x-sign flips first, then the y-sign, then the x-sign again.

Common mistakes

Saying $(0, 5)$ is in Quadrant I. Points on an axis are in no quadrant.
Giving the distance from the x-axis as $|x|$ . Distance from the x-axis is measured vertically, so it is $|y|$ .

3Distance between two points

To find the distance between $A(x_1, y_1)$ and $B(x_2, y_2)$ , draw a horizontal line from A and a vertical line from B. They meet at $C(x_2, y_1)$ , making a right-angled triangle ACB.

The horizontal leg AC has length $|x_2 - x_1|$ and the vertical leg CB has length $|y_2 - y_1|$ . By the Baudhayana–Pythagoras theorem, $AB^2 = AC^2 + CB^2$ . Squaring removes the need for absolute values, which gives the distance formula.

Special case: the distance of $P(x, y)$ from the origin is $\sqrt{x^2 + y^2}$ .

Key formulas & definitions

$AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
Distance from the origin: $OP = \sqrt{x^2 + y^2}$ .
Points on the same horizontal line: distance $= |x_2 - x_1|$ ; on the same vertical line: $|y_2 - y_1|$ .

Worked example 1

Find the distance between $P(-2, 3)$ and $Q(4, -5)$ .

$x_2 - x_1 = 4 - (-2) = 6$ and $y_2 - y_1 = -5 - 3 = -8$ .
$PQ^2 = 6^2 + (-8)^2 = 36 + 64 = 100$ .
$PQ = \sqrt{100} = 10$ .

Answer: 10 units.

Worked example 2

Find $k$ if the point $(k, 2)$ is 5 units from $(1, -2)$ .

$(k - 1)^2 + (2 - (-2))^2 = 5^2$ , so $(k - 1)^2 + 16 = 25$ .
$(k - 1)^2 = 9$ , so $k - 1 = 3$ or $k - 1 = -3$ .
$k = 4$ or $k = -2$ . Both points work — they sit on either side.

Answer: $k = 4$ or $k = -2$ .

Trick / memory aid

Spot Pythagorean triples to skip the square root: legs 3 & 4 → 5, 6 & 8 → 10, 5 & 12 → 13, 8 & 15 → 17.

Common mistakes

Writing $(x_2 - x_1) + (y_2 - y_1)$ without squaring — you must square each difference, add, then take the root.
Sign slips with negatives: $4 - (-2) = 6$ , not 2.
Throwing away one value when solving for an unknown coordinate — a square usually gives two answers.

4Midpoint and section formula

The midpoint of a segment sits exactly halfway, so each of its coordinates is the average of the end-point coordinates.

More generally, a point P that divides AB in the ratio $m : n$ (so $AP : PB = m : n$ ) is found with the section formula. Each coordinate is a weighted average, and the weights are swapped: the coordinate of B is multiplied by $m$ , and that of A by $n$ . That makes sense — if P is closer to A, A's coordinates should count for more.

With $m = n = 1$ the section formula becomes the midpoint formula, so you only need to remember one idea.

Key formulas & definitions

Midpoint of $A(x_1, y_1)$ , $B(x_2, y_2)$ : $\left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right)$
Section formula ( $AP : PB = m : n$ ): $P = \left(\frac{m x_2 + n x_1}{m + n},\ \frac{m y_2 + n y_1}{m + n}\right)$
Centroid of a triangle (a preview of Chapter 12): $\left(\dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3}\right)$

Worked example 1

The midpoint of AB is $M(2, -1)$ and $A = (-3, 4)$ . Find B.

Let $B = (x, y)$ . Then $\dfrac{-3 + x}{2} = 2$ and $\dfrac{4 + y}{2} = -1$ .
$-3 + x = 4 \Rightarrow x = 7$ ; $4 + y = -2 \Rightarrow y = -6$ .

Answer: $B(7, -6)$ .

Worked example 2

Find the point that divides the segment joining $A(1, 2)$ and $B(7, 8)$ in the ratio $1 : 2$ .

$m = 1$ , $n = 2$ . $x = \dfrac{1 \cdot 7 + 2 \cdot 1}{3} = \dfrac{9}{3} = 3$ .
$y = \dfrac{1 \cdot 8 + 2 \cdot 2}{3} = \dfrac{12}{3} = 4$ .
Check: P is one-third of the way from A — moving 2 of the 6 units in x and 2 of the 6 units in y. ✓

Answer: $(3, 4)$ .

Trick / memory aid

For the section formula, cross the weights: the point near A is pulled by A, so A's coordinate is multiplied by the part of the ratio next to B ( $n$ ).

Common mistakes

Subtracting instead of adding in the midpoint formula — it is the average, $(x_1 + x_2)/2$ .
Putting $m$ with $x_1$ instead of $x_2$ in the section formula.

5Using coordinates to test shapes

Once points have coordinates, you can prove geometric facts with arithmetic. Collinearity: three points A, B, C lie on one line exactly when the two shorter distances add up to the longest, e.g. $AB + BC = AC$ .

Right angle: a triangle has a right angle when the squares of its sides satisfy $a^2 + b^2 = c^2$ (the converse of the Pythagoras theorem, which you meet again in Chapter 9).

Type of quadrilateral: compute all four sides and both diagonals. Four equal sides and equal diagonals → square; four equal sides only → rhombus; opposite sides equal and equal diagonals → rectangle; diagonals with the same midpoint → parallelogram.

Key formulas & definitions

Collinear: $AB + BC = AC$ (sum of the two smaller distances equals the largest).
Right-angled at the vertex opposite the longest side if $(\text{longest})^2 =$ sum of squares of the other two.
Parallelogram test: the midpoints of the two diagonals coincide.

Worked example 1

Show that $A(1, 1)$ , $B(4, 5)$ and $C(7, 9)$ are collinear.

$AB = \sqrt{3^2 + 4^2} = 5$ ; $BC = \sqrt{3^2 + 4^2} = 5$ ; $AC = \sqrt{6^2 + 8^2} = 10$ .
$AB + BC = 5 + 5 = 10 = AC$ .

Answer: Collinear, with B the midpoint of AC.

Worked example 2

Is the triangle with vertices $P(0, 0)$ , $Q(4, 0)$ , $R(0, 3)$ right-angled?

$PQ = 4$ , $PR = 3$ , $QR = \sqrt{16 + 9} = 5$ .
$3^2 + 4^2 = 25 = 5^2$ , so the angle opposite QR (at P) is $90^\circ$ .

Answer: Yes, right-angled at P.

Common mistakes

Calling a quadrilateral a square just because all four sides are equal — a rhombus also has that; check the diagonals.
Comparing squared distances with plain distances in the same equation.

Important questions

Try each question on paper first, then tap it to check the answer.

Basic Recall and direct use

In which quadrant or on which axis do

(-5, 0)

and

(3, -8)

lie?Answer

(-5, 0)

lies on the x-axis (no quadrant).

(3, -8)

has

x > 0

y < 0

, so it lies in Quadrant IV.

Find the distance of the point

(-6, 8)

from the origin.Answer

\sqrt{(-6)^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10

units.

Find the midpoint of the segment joining

(2, -4)

and

(-6, 10)

.Answer

\left(\dfrac{2 + (-6)}{2}, \dfrac{-4 + 10}{2}\right) = (-2, 3)

What is the distance of

(-4, 7)

from the x-axis and from the y-axis?Answer

From the x-axis:

|7| = 7

units. From the y-axis:

|-4| = 4

units.

Medium Two or three steps

Find the point on the x-axis that is equidistant from

A(2, -5)

and

B(-2, 9)

.Answer

Let the point be

(x, 0)

(x - 2)^2 + 25 = (x + 2)^2 + 81

. Expanding:

x^2 - 4x + 29 = x^2 + 4x + 85

, so

-8x = 56

x = -7

. The point is

(-7, 0)

In what ratio does the point

(-1, 6)

divide the segment joining

(-3, 10)

and

(6, -8)

?Answer

Let the ratio be

k : 1

. Then

\dfrac{6k - 3}{k + 1} = -1 \Rightarrow 6k - 3 = -k - 1 \Rightarrow k = \dfrac{2}{7}

. Ratio

2 : 7

. Check with y:

\dfrac{2(-8) + 7(10)}{9} = \dfrac{54}{9} = 6

✓.

Show that

(1, 7)

(4, 2)

(-1, -1)

and

(-4, 4)

are the vertices of a square.Answer

Sides: each is

\sqrt{3^2 + 5^2} = \sqrt{34}

(check all four:

(1,7)\to(4,2)

(4,2)\to(-1,-1)

(-1,-1)\to(-4,4)

(-4,4)\to(1,7)

). Diagonals:

(1,7)\to(-1,-1)

\sqrt{4 + 64} = \sqrt{68}

and

(4,2)\to(-4,4)

\sqrt{64 + 4} = \sqrt{68}

. Equal sides and equal diagonals, so it is a square.

Advanced Reasoning and HOTS

Three vertices of a parallelogram ABCD are

A(1, 2)

B(4, 3)

and

C(6, 6)

. Find D.Answer

Diagonals of a parallelogram bisect each other, so the midpoint of AC equals the midpoint of BD. Midpoint of AC

= (3.5, 4)

. If

D = (x, y)

\dfrac{4 + x}{2} = 3.5 \Rightarrow x = 3

and

\dfrac{3 + y}{2} = 4 \Rightarrow y = 5

. So

D(3, 5)

A circle has centre

(2, -3)

and one end of a diameter at

(-1, 1)

. Find the other end and the radius.Answer

The centre is the midpoint of the diameter. Other end

(x, y)

\dfrac{-1 + x}{2} = 2 \Rightarrow x = 5

;

\dfrac{1 + y}{2} = -3 \Rightarrow y = -7

. Other end

(5, -7)

. Radius

= \sqrt{3^2 + 4^2} = 5

units.

Prove that the midpoint of the hypotenuse of a right triangle is equidistant from all three vertices, using coordinates.Answer

Place the right angle at the origin:

O(0, 0)

A(2a, 0)

B(0, 2b)

. Midpoint of AB is

M(a, b)

MO = \sqrt{a^2 + b^2}

MA = \sqrt{(2a - a)^2 + b^2} = \sqrt{a^2 + b^2}

MB = \sqrt{a^2 + (2b - b)^2} = \sqrt{a^2 + b^2}

. All equal. (Choosing

2a

2b

avoids fractions.)

Self-check quiz

Five quick questions. Tap an option to see if you are right.

The point

(0, -4)

lies

The distance between

(1, 2)

and

(4, 6)

The midpoint of

(-2, 5)

and

(6, -1)

A point with negative x-coordinate and positive y-coordinate is in

Points A, B, C are collinear if

FAQ

Why is the distance formula just the Pythagoras theorem?

The horizontal and vertical gaps between two points form the two legs of a right triangle, and the straight-line distance is the hypotenuse. Squaring the legs, adding and taking the square root is exactly $c = \sqrt{a^2 + b^2}$ .

Does it matter which point I call

(x_1, y_1)

No. In the distance formula the differences are squared, so the order does not change the answer. In the section formula it does matter — the ratio $m : n$ is measured from the first point A.

Which formulas from this chapter must I memorise for the exam?

The distance formula, the midpoint formula and the quadrant signs. The section formula is easy to rebuild from the idea of a weighted average, but learning it saves time.

How do I check whether four points make a parallelogram quickly?

Find the midpoints of both diagonals. If they are the same point, the diagonals bisect each other and the figure is a parallelogram.

Book references

NCERT Ganita Manjari Part I, Chapter 1: Orienting Yourself: The Use of CoordinatesRead the chapter and do every exercise set first.
Mathematics for Class 9 — R.D. Sharma (Dhanpat Rai)Extra graded practice; its chapter numbers follow the old syllabus, so match by topic.
Mathematics for Class 9 — R.S. Aggarwal (Bharati Bhawan)Good for drill on basics; match by topic.
Exemplar Problems in Mathematics, Class IX — NCERTHarder, concept-testing questions (written for the old book).

These notes, examples and questions are written by PrepRung in our own words to help you study the NCERT Ganita Manjari book. NCERT and CBSE are not affiliated with PrepRung. Always study from your textbook too.

Class 9 Maths Chapter 1: Orienting Yourself: The Use of Coordinates — Notes, Examples & Practice

1The Cartesian plane and ordered pairs

2Quadrants and points on the axes

3Distance between two points

4Midpoint and section formula

5Using coordinates to test shapes

Important questions

Basic Recall and direct use

Medium Two or three steps

Advanced Reasoning and HOTS

Self-check quiz

FAQ

Book references