Coordinate Geometry, titled Orienting Yourself: The Use of Coordinates in the new NCERT book, shows how two perpendicular number lines let us name any point on a flat surface with two numbers. It builds from the axes, the origin and the four quadrants to distances along grid lines, the distance formula from the Baudhāyana-Pythagoras theorem, and the midpoint of a segment. These tools then decide whether points are collinear, whether a triangle is right-angled, and where a point sits relative to a circle. The same formulas return in Class 10.
Key Concepts
1. Why We Need a System of Coordinates
NCERT defines a system of coordinates as a structured framework, like the grid lines on a map or on graph paper, that lets us use numbers to describe the exact physical locations of points or objects. Two numbers, counted from a fixed starting point along two fixed directions, pin down a place exactly.
A short history from the chapter. Streets in the Sindhu-Sarasvatī Civilisation ran North-South and East-West about 10 metres apart, so a merchant could find a shop by counting units from the city centre: a coordinate system in practice. Baudhāyana used East-West and North-South lines in his constructions and developed the Baudhāyana-Pythagoras theorem. Brahmagupta (c. 628 CE) formalised zero and negative numbers; the origin is zero and the negative axes hold values below zero, so the four-quadrant plane depends on his work. René Descartes (1637 CE), following related work of Fermat, formalised the fact that any point in a plane is fixed by two numbers, its distances from two perpendicular axes. That is why it is called the Cartesian plane.
Reiaan’s room. Reiaan cannot see, and his family has moved to a new city. His sister Shalini maps his room on a rectangular grid, with pins for key points and thick wool joining the corners of objects so he can feel their positions. Her scale is 1 cm : 1 foot.
The map shows only the floor. A window sits some height above the floor, and a floor map cannot show height. The floor is a two-dimensional (2-D) space, where two numbers are enough; a point in the air of the room would need a third number.
2. Coordinate Axes, Origin and Sign Convention
The number line from earlier grades is one-dimensional: it locates points on a line only. The 2-D coordinate system uses two number lines at right angles.
- The horizontal line is called the x-axis.
- The vertical line is called the y-axis.
- The point where they intersect is called the origin O. Its coordinates are (0, 0).
- Distances from O are marked off in equal units on both axes.
Sign convention. Distances to the right of O or upwards from O are positive; distances to the left or downwards are negative.
Points on the axes. NCERT uses these three points to show the idea:
| Point | Where it is | Coordinates |
|---|---|---|
| B | On the x-axis, 4.5 units to the right of O | (4.5, 0) |
| G | On the y-axis, 4.5 units below O | (0, −4.5) |
| H | On the y-axis, 4 units above O | (0, 4) |
A point (x, 0) lies on the x-axis, right of O if x > 0 and left if x < 0. A point (0, y) lies on the y-axis, above O if y > 0 and below if y < 0. The origin is the only point on both axes.
Notation. B = (4.5, 0) is often written without the “=” sign as B (4.5, 0), especially on a graph.
3. Coordinates of a Point and the Four Quadrants
The plane of the axes is called the Cartesian plane, the coordinate plane or the xy-plane. The axes divide it into four quadrants: I is top-right, and the numbering goes anticlockwise, so II is top-left, III bottom-left and IV bottom-right.
Meaning of the coordinates. The coordinates of a point P are written (x, y).
- x is the perpendicular distance of P from the y-axis, measured along the x-axis. It is the x-coordinate.
- y is the perpendicular distance of P from the x-axis, measured along the y-axis. It is the y-coordinate.
Signs in each quadrant.
| Quadrant | Position | Sign of x | Sign of y | Pattern | Example |
|---|---|---|---|---|---|
| I | Top right | Positive | Positive | (+, +) | (2, 7) |
| II | Top left | Negative | Positive | (−, +) | Q (−5, 3) |
| III | Bottom left | Negative | Negative | (−, −) | (−4, −1) |
| IV | Bottom right | Positive | Negative | (+, −) | S (3, −5) |
NCERT’s examples: S (3, −5) is in Quadrant IV and Q (−5, 3) is in Quadrant II. To plot Q, move 5 units left of O along the x-axis, then 3 units up.
Points on the axes belong to no quadrant. (−3, 0) and (0, 6) lie on the boundaries between quadrants.
Order matters. (2, 5) is 2 right and 5 up; (5, 2) is 5 right and 2 up. The chapter summary states:
- If x = y, then (x, y) = (y, x).
- If x ≠ y, then (x, y) ≠ (y, x).
So Q (y, x) coincides with P (x, y) only when x = y, for points such as (3, 3) or (−2, −2).
Why negative numbers are needed. With only zero and positive numbers we could name points in Quadrant I and on the positive axes. Everything left of the y-axis or below the x-axis, three quarters of the plane, would have no name.
4. Distance Along Grid Lines
Same y-coordinate (horizontal segment). The distance between (x1, y) and (x2, y) is the absolute value of the difference of the x-coordinates:
distance = |x2 − x1|
Same x-coordinate (vertical segment). The distance between (x, y1) and (x, y2) is:
distance = |y2 − y1|
The absolute value keeps a distance positive whichever point you start from. For example:
- (3, 0) and (−5, 0): |−5 − 3| = |−8| = 8 units.
- (0, 1.5) and (0, 4): |4 − 1.5| = 2.5 units. These are the ends B1 and B2 of Reiaan’s bathroom door, so the door is 2.5 units wide.
- (−5, −2) and (−5, 2): |2 − (−2)| = 4 units.
Mirror images in an axis. Reflecting in the y-axis changes (x, y) to (−x, y); reflecting in the x-axis changes (x, y) to (x, −y). So (−5, −2) and (−5, 2) are mirror images in the x-axis.
5. The Distance Formula
What if the segment joining two points is parallel to neither axis? NCERT answers this with the Baudhāyana-Pythagoras theorem from Grade 8: in a right-angled triangle, the square of the hypotenuse equals the sum of the squares of the other two sides.
NCERT’s triangle ADM. Take A (3, 4), D (7, 1) and M (9, 6), all in the first quadrant. To find AD, mark C (3, 1), directly below A and level with D. Triangle ACD has a right angle at C.
- Distance moved along the x-axis: CD = x-coordinate of D − x-coordinate of A = 7 − 3 = 4.
- Distance moved along the y-axis: AC = y-coordinate of A − y-coordinate of D = 4 − 1 = 3.
- By the Baudhāyana-Pythagoras theorem: AD = √(42 + 32) = √25 = 5 units.
- DM: shifts 9 − 7 = 2 and 6 − 1 = 5, so DM = √(22 + 52) = √29 units.
- MA: shifts 9 − 3 = 6 and 6 − 4 = 2, so MA = √(62 + 22) = √40 units (which simplifies to 2√10).
The general formula. For any two points (x1, y1) and (x2, y2), draw the horizontal and vertical lines through them to form a right triangle with its right angle at F (x1, y2). One leg is the shift along the x-axis, x2 − x1; the other is the shift along the y-axis, y2 − y1. The segment itself is the hypotenuse, so
distance = √[(x2 − x1)2 + (y2 − y1)2]
It makes no difference whether the shifts are positive or negative, because each is squared: (−4)2 and 42 are both 16. So you may subtract in either order.
Distance from the origin. Put (x1, y1) = (0, 0): the distance of P (x, y) from O is √(x2 + y2). For example, (−6, 8) is √(36 + 64) = √100 = 10 units from O.
Negative coordinates and reflection. NCERT reflects triangle ADM in the y-axis, giving A′ (−3, 4), D′ (−7, 1), M′ (−9, 6). Now C′D′ = −3 − (−7) = 4 and A′C′ = 4 − 1 = 3, so A′D′ = 5. Likewise D′M′ = √[(−2)2 + 52] = √29 and M′A′ = √[(−6)2 + 22] = √40.
Reflection preserves the lengths of the sides. Only the position changes: the triangle moves from Quadrant I to Quadrant II. Reflection in the x-axis behaves the same way, because each shift only changes sign.
6. Midpoint of a Segment
The midpoint M of a segment ST is the point on ST equally far from S and T. NCERT’s exercise asks you to find the connection between the coordinates of M, S and T: each coordinate of M is the average of the matching coordinates of S and T.
For S (x1, y1) and T (x2, y2):
M = ((x1 + x2)/2, (y1 + y2)/2)
Why the average works. Halfway from S to T you have made half of each shift, so the x-coordinate of M is x1 + (x2 − x1)/2 = (x1 + x2)/2. The y-coordinate works the same way.
Check. For S (2, 3) and T (4, 5), M = (3, 4), and SM = MT = √2.
Running it backwards. Knowing M and one end A, apply the shift from A to M once more, starting at M, to reach the other end B.
Points of trisection. P and Q divide AB into three equal parts, P closer to A. P is the midpoint of AQ and Q is the midpoint of PB. To compute them, step one-third of the shift from A to B twice from A.
Vertices from midpoints. If D, E, F are the midpoints of BC, CA, AB, then B + C = 2D, C + A = 2E, A + B = 2F (coordinate by coordinate), so A + B + C = D + E + F. Solved Example 10 uses this.
7. Using Distances: Collinearity, Right Angles, Circles and Squares
With the distance formula, many geometry questions can be settled by arithmetic without plotting.
(a) Three points on one straight line (collinearity). Find all three distances PQ, QR and PR. If the two smaller distances add up exactly to the largest, the points lie on one straight line (the middle point sits on the segment joining the other two). If the sum is larger, the three points form a triangle.
PQ + QR = PR ⇒ P, Q, R are collinear
A second check: from M (−3, −4) to A (0, 0) you move (3, 4), and from A to G (6, 8) you move (6, 8), exactly double. The direction never changes.
(b) Is a triangle right-angled? By the converse of the Baudhāyana-Pythagoras theorem, it is right-angled exactly when the largest side2 equals the sum of the other two squares; the right angle is opposite the longest side. For triangle ADM, 25 + 29 = 54 ≠ 40, so there is no right angle; NCERT calls ADM acute-angled.
(c) Points and a circle centred at the origin. A circle K with centre O (0, 0) and radius r is the set of all points at distance r from O. For a point P (x, y), compare OP2 = x2 + y2 with r2:
| Comparison | Position of P |
|---|---|
| x2 + y2 = r2 | On the circle |
| x2 + y2 < r2 | Inside the circle |
| x2 + y2 > r2 | Outside the circle |
For a circle with centre C, use the distance from C instead. Two circles with centres d apart and radii r1, r2 cross when d is less than r1 + r2 and greater than the difference of the radii.
(d) Identifying quadrilaterals. Four equal sides give a rhombus; add equal diagonals and it is a square, with area side2. Diagonals sharing a midpoint give a parallelogram; a parallelogram with equal diagonals is a rectangle.
Formula and Theorem Sheet
| Result | Statement or Formula | When to use it |
|---|---|---|
| Quadrant signs | I (+, +); II (−, +); III (−, −); IV (+, −) | Naming the quadrant of a point |
| Horizontal distance | (x1, y) to (x2, y) = |x2 − x1| | Same y-coordinate |
| Vertical distance | (x, y1) to (x, y2) = |y2 − y1| | Same x-coordinate |
| Distance formula | √[(x2 − x1)2 + (y2 − y1)2] | Any two points |
| Distance from origin | √(x2 + y2) | Circles centred at O |
| Midpoint | ((x1 + x2)/2, (y1 + y2)/2) | Halfway point; missing end point |
| Points of trisection of AB | A + (1/3)(shift from A to B), A + (2/3)(shift from A to B) | Dividing a segment into three equal parts |
| Reflection in the y-axis | (x, y) → (−x, y) | Lengths preserved |
| Collinearity | PQ + QR = PR | Three points on one line |
| Right-angle test | Largest side2 = sum of squares of the other two | Converse of Baudhāyana-Pythagoras |
| Point and circle (centre O, radius r) | Compare x2 + y2 with r2 | On (=), inside (<), outside (>) |
Read the rest of the chapter →Hide the rest ↑
Important Definitions
| Term | Definition |
|---|---|
| System of coordinates | A structured framework, like the grid lines on a map or graph paper, that lets us use numbers to describe the exact physical locations of points or objects |
| Origin | The point of intersection of the x-axis and the y-axis; its coordinates are (0, 0) |
| Cartesian plane | The plane in which the coordinate axes lie; also called the coordinate plane or the xy-plane |
| Quadrants | The four parts into which the coordinate axes divide the plane, numbered I, II, III and IV |
| x-coordinate | The perpendicular distance of a point from the y-axis, measured along the x-axis, with sign |
| y-coordinate | The perpendicular distance of a point from the x-axis, measured along the y-axis, with sign |
| Distance between two points | The length of the segment joining them; √[(x2 − x1)2 + (y2 − y1)2] |
| Midpoint of a segment | The point on the segment equidistant from its two ends; its coordinates are the averages of the end coordinates |
| Collinear points | Points that lie on one straight line |
| Circle | The set of all points in a plane at a fixed distance (the radius) from a fixed point (the centre) |
Solved Examples (NCERT-Based)
Example 1: The fourth foot of Reiaan’s table (Exercise Set 1.2)
“Place Reiaan’s rectangular study table with three of its feet at the points (8, 9), (11, 9) and (11, 7). (i) Where will the fourth foot of the table be? … (iii) What is the width of the table? The length? Can you make out the height of the table?”
Solution: Label the feet P (8, 9), Q (11, 9) and R (11, 7).
Step 1: P and Q share y = 9, so PQ is horizontal; Q and R share x = 11, so QR is vertical. The right angle is at Q.
Step 2: The fourth foot S must be directly below P (same x-coordinate, 8) and level with R (same y-coordinate, 7). So S = (8, 7).
Step 3: PQ = |11 − 8| = 3 units and QR = |9 − 7| = 2 units. So the table is 3 units long and 2 units wide. Shalini’s scale is 1 cm : 1 foot and the exercise uses 1 cm = 1 unit, so taking 1 unit as 1 foot, the table is 3 feet by 2 feet.
Step 4: The height cannot be found from a 2-D floor map.
Part (ii) depends on the other furniture in the textbook figure.
Example 2: The quadrilateral RAMP (End-of-Chapter Exercise 3)
“Consider the points R (3, 0), A (0, −2), M (−5, −2) and P (−5, 2). If they are joined in the same order, predict: (i) Two sides of RAMP that are perpendicular to each other. (ii) One side of RAMP that is parallel to one of the axes. (iii) Two points that are mirror images of each other in one axis. Which axis will this be?”
Solution: The sides are RA, AM, MP and PR.
Step 1: A (0, −2) and M (−5, −2) have the same y-coordinate, so AM is horizontal, parallel to the x-axis. Its length is |−5 − 0| = 5.
Step 2: M (−5, −2) and P (−5, 2) have the same x-coordinate, so MP is vertical, parallel to the y-axis. Its length is |2 − (−2)| = 4.
Step 3: A horizontal line and a vertical line meet at a right angle, so AM ⊥ MP. This answers (i). Either AM (parallel to the x-axis) or MP (parallel to the y-axis) answers (ii).
Step 4: M (−5, −2) and P (−5, 2) differ only in the sign of the y-coordinate. So M and P are mirror images of each other in the x-axis. This answers (iii).
Example 3: Sides of triangle ADM (Section 1.4)
Find the lengths of the sides of the triangle with vertices A (3, 4), D (7, 1) and M (9, 6).
Solution:
AD: shifts 7 − 3 = 4 and 1 − 4 = −3. AD = √[42 + (−3)2] = √(16 + 9) = √25 = 5 units.
DM: shifts 9 − 7 = 2 and 6 − 1 = 5. DM = √(4 + 25) = √29 units.
MA: shifts 3 − 9 = −6 and 4 − 6 = −2. MA = √(36 + 4) = √40 = 2√10 units.
Example 4: Three points on a line (End-of-Chapter Exercise 6)
“Are the points M (−3, −4), A (0, 0) and G (6, 8) on the same straight line? Suggest a method to check this without plotting and joining the points.”
Solution: Method: find the three distances and check whether the two smaller ones add up to the largest.
MA = √[(0 + 3)2 + (0 + 4)2] = √(9 + 16) = 5.
AG = √[(6 − 0)2 + (8 − 0)2] = √(36 + 64) = 10.
MG = √[(6 + 3)2 + (8 + 4)2] = √(81 + 144) = √225 = 15.
MA + AG = 5 + 10 = 15 = MG. So M, A and G lie on the same straight line, with A between M and G.
Example 5: Testing R, B and C (End-of-Chapter Exercise 7)
“Use your method (from Problem 6) to check if the points R (−5, −1), B (−2, −5) and C (4, −12) are on the same straight line.”
Solution:
RB: shifts 3 and −4. RB = √(9 + 16) = 5.
BC: shifts 6 and −7. BC = √(36 + 49) = √85.
RC: shifts 9 and −11. RC = √(81 + 121) = √202.
RC is the largest. Is 5 + √85 equal to √202? Square the left side: (5 + √85)2 = 25 + 10√85 + 85 = 110 + 10√85. Since √85 is more than 9.2, this is more than 110 + 92 = 202. So 5 + √85 > √202, and R, B and C are not on the same straight line.
The shifts confirm it: R to B is (3, −4), B to C is (6, −7), and a straight line would need (6, −8). Rounded decimals (14.22 against 14.21) would have misled you.
Example 6: Is M the midpoint of ST? (End-of-Chapter Exercise 9)
“The following table shows the coordinates of points S, M and T. In each case, state whether M is the midpoint of segment ST. Justify your answer.”
Solution: Compute the true midpoint ((xS + xT)/2, (yS + yT)/2) and compare.
| S | M | T | True midpoint | Is M the midpoint? |
|---|---|---|---|---|
| (−3, 0) | (0, 0) | (3, 0) | (0/2, 0/2) = (0, 0) | Yes |
| (2, 3) | (3, 4) | (4, 5) | (6/2, 8/2) = (3, 4) | Yes |
| (0, 0) | (0, 5) | (0, −10) | (0, −10/2) = (0, −5) | No: M is 5 from S and 15 from T, and it is not even between them |
| (−8, 7) | (0, −2) | (6, −3) | (−2/2, 4/2) = (−1, 2) | No |
Example 7: Finding a missing end point (End-of-Chapter Exercise 10)
“Use the connection you found to find the coordinates of B given that M (−7, 1) is the midpoint of A (3, −4) and B (x, y).”
Solution:
x-coordinates: (3 + x)/2 = −7, so 3 + x = −14 and x = −17.
y-coordinates: (−4 + y)/2 = 1, so −4 + y = 2 and y = 6.
B = (−17, 6).
Check: the shift from A to M is (−10, 5); the same shift from M gives (−17, 6).
Example 8: Points of trisection (End-of-Chapter Exercise 11)
“Let P, Q be points of trisection of AB, with P closer to A, and Q closer to B. Using your knowledge of how to find the coordinates of the midpoint of a segment, how would you find the coordinates of P and Q? Do this for the case when the points are A (4, 7) and B (16, −2).”
Solution: AP = PQ = QB, so P is the midpoint of AQ and Q is the midpoint of PB. Equivalently, P is one-third of the way from A to B and Q is two-thirds of the way.
Step 1: Shift from A to B = (16 − 4, −2 − 7) = (12, −9).
Step 2: One-third of the shift = (4, −3).
Step 3: P = (4 + 4, 7 − 3) = (8, 4). Q = (8 + 4, 4 − 3) = (12, 1).
Check: midpoint of A and Q is (16/2, 8/2) = (8, 4) = P, and midpoint of P and B is (24/2, 2/2) = (12, 1) = Q.
Example 9: Points on a circle centred at the origin (End-of-Chapter Exercise 12)
“(i) Given the points A (1, −8), B (−4, 7) and C (−7, −4), show that they lie on a circle K whose center is the origin O (0, 0). What is the radius of circle K? (ii) Given the points D (−5, 6) and E (0, 9), check whether D and E lie within the circle, on the circle, or outside the circle K.”
Solution: (i) Use OP2 = x2 + y2.
OA2 = 1 + 64 = 65. OB2 = 16 + 49 = 65. OC2 = 49 + 16 = 65.
All three points are the same distance √65 from O, so they lie on a circle K with centre O and radius √65 units (about 8.06 units).
(ii) OD2 = 25 + 36 = 61 < 65, so D lies inside K. OE2 = 0 + 81 = 81 > 65, so E lies outside K.
Example 10: Vertices from the midpoints of the sides (End-of-Chapter Exercise 13)
“The midpoints of the sides of triangle ABC are the points D, E, and F. Given that the coordinates of D, E, and F are (5, 1), (6, 5), and (0, 3), respectively, find the coordinates of A, B and C.”
Solution: The question does not say which side each point bisects. Take the usual labelling: D is the midpoint of BC (the side opposite A), E of CA and F of AB. Work with x-coordinates and y-coordinates separately.
x-coordinates: xB + xC = 10, xC + xA = 12, xA + xB = 0. Adding: 2(xA + xB + xC) = 22, so xA + xB + xC = 11. Then xA = 11 − 10 = 1, xB = 11 − 12 = −1, xC = 11 − 0 = 11.
y-coordinates: yB + yC = 2, yC + yA = 10, yA + yB = 6. Adding: 2(sum) = 18, so the sum is 9. Then yA = 9 − 2 = 7, yB = 9 − 10 = −1, yC = 9 − 6 = 3.
A = (1, 7), B = (−1, −1), C = (11, 3).
Check: midpoint of BC = (10/2, 2/2) = (5, 1) = D; midpoint of CA = (12/2, 10/2) = (6, 5) = E; midpoint of AB = (0/2, 6/2) = (0, 3) = F.
About the labelling. Any other matching of D, E, F to the sides gives the same three vertices; only the letters A, B, C swap. State the labelling you used.
Example 11: Icons on a computer screen (End-of-Chapter Exercise 15)
“A computer graphics program displays images on a rectangular screen whose coordinate system has the origin at the bottom-left corner. The screen is 800 pixels wide and 600 pixels high. A circular icon of radius 80 pixels is drawn with its centre at the point A (100, 150). Another circular icon of radius 100 pixels is drawn with its centre at the point B (250, 230). Determine: (i) whether any part of either circle lies outside the screen. (ii) whether the two circles intersect each other.”
Solution: The screen runs from x = 0 to x = 800 and from y = 0 to y = 600.
(i) Circle A spans x from 20 to 180 and y from 70 to 230. Circle B spans x from 150 to 350 and y from 130 to 330. All values lie inside 0 to 800 (x) and 0 to 600 (y), so no part of either circle lies outside the screen.
(ii) Distance between the centres: shifts 250 − 100 = 150 and 230 − 150 = 80. AB = √(22500 + 6400) = √28900 = 170 pixels.
Sum of the radii = 80 + 100 = 180 pixels. Difference of the radii = 20 pixels. Since 20 < 170 < 180, the centres are closer than the sum of the radii, and neither circle sits wholly inside the other. So the two circles intersect.
Example 12: Is ABCD a square? (End-of-Chapter Exercise 16)
“Plot the points A (2, 1), B (−1, 2), C (−2, −1), and D (1, −2) in the coordinate plane. Is ABCD a square? Can you explain why? What is the area of this square?”
Solution:
Sides: AB: shifts (−3, 1), AB = √(9 + 1) = √10. BC: shifts (−1, −3), BC = √10. CD: shifts (3, −1), CD = √10. DA: shifts (1, 3), DA = √10.
Diagonals: AC: shifts (−4, −2), AC = √(16 + 4) = √20. BD: shifts (2, −4), BD = √20.
All four sides are equal and the two diagonals are equal, so ABCD is a square. A further check on the right angle: AB2 + BC2 = 10 + 10 = 20 = AC2, so angle B is 90°.
Area = side2 = (√10)2 = 10 square units, even though the square is tilted.
Competency-Based Questions (with answers)
1. Case-based: The school campus map
A school draws its campus on a grid with the main gate at the origin. The x-axis points east and the y-axis points north, and 1 unit stands for 10 m. The library is at L (3, 4), the canteen at C (−5, 4), the playground at P (−5, −2) and the science lab at B (3, −2).
(a) What shape is LCPB, and what are its length and width in metres? LC is horizontal (shared y = 4) with length |−5 − 3| = 8 units = 80 m. CP is vertical (shared x = −5) with length |4 − (−2)| = 6 units = 60 m. PB and BL match these, and adjacent sides are horizontal and vertical, so LCPB is a rectangle 80 m by 60 m.
(b) How far is it from the library to the playground in a straight line? LP = √[(−5 − 3)2 + (−2 − 4)2] = √(64 + 36) = √100 = 10 units = 100 m.
(c) A water cooler is placed exactly halfway between the library and the playground. Find its coordinates and its quadrant. Midpoint of LP = ((3 − 5)/2, (4 − 2)/2) = (−1, 1), in Quadrant II, 10 m west and 10 m north of the gate.
2. Case-based: A delivery drone
A drone depot is at the origin of a grid in which 1 unit = 1 km. The drone can fly at most 12 km from the depot. Orders come from A (6, 8) and B (−5, −12).
(a) How far is A from the depot? OA = √(36 + 64) = √100 = 10 km.
(b) In which quadrant is B, and how far is it from the depot? Both coordinates are negative, so B is in Quadrant III. OB = √(25 + 144) = √169 = 13 km.
(c) Which orders can the drone reach? OA2 = 100 < 144, so A is inside and can be served. OB2 = 169 > 144, so B is outside and cannot be served from this depot.
3. Source-based: From city streets to Descartes
Read the passage: “City streets in the Sindhu-Sarasvatī Civilisation ran North-South and East-West about 10 metres apart. Brahmagupta formalised zero and negative numbers. Descartes showed that any point in a plane is defined by two numbers, its distances from two perpendicular axes.”
(a) Which two numbers fix a point in the plane? Its perpendicular distances from the two axes, with signs: the x-coordinate (distance from the y-axis) and the y-coordinate (distance from the x-axis).
(b) Why does the passage connect Brahmagupta’s work to the coordinate plane? The origin is zero, and points to the left of the y-axis or below the x-axis need negative coordinates. Without negative numbers only Quadrant I and the positive parts of the axes could be described.
(c) Taking the street spacing as 10 m, a shop is 3 blocks east and 2 blocks south of the city centre. Give its coordinates in metres, with east and north positive, and its straight-line distance from the centre. Coordinates (30, −20). Distance = √(900 + 400) = √1300 = 10√13 ≈ 36.1 m.
4. Assertion-Reason
Assertion (A): The point (−3, 0) lies in Quadrant II. Reason (R): The x-coordinate of (−3, 0) is negative.
Answer: (d) A is false, R is true. The y-coordinate is 0, so the point lies on the x-axis, and points on an axis belong to no quadrant.
5. Assertion-Reason
Assertion (A): Reflecting a triangle in the y-axis does not change the lengths of its sides. Reason (R): Reflection in the y-axis sends (x, y) to (−x, y), which only changes the sign of each x-shift, and the square of a shift does not depend on its sign.
Answer: (a) Both A and R are true, and R is the correct explanation of A.
6. Assertion-Reason
Assertion (A): The points P (1, 2), Q (2, 3) and R (3, 4) lie on one straight line. Reason (R): PQ = QR.
Answer: (b) Both A and R are true, but R is not the correct explanation of A. PQ = QR = √2 and PR = √8 = 2√2, so PQ + QR = PR, which proves A. Equal distances alone do not prove collinearity.
7. Assertion-Reason
Assertion (A): (4, −7) and (−7, 4) represent the same point. Reason (R): (x, y) = (y, x) only when x = y.
Answer: (d) A is false, R is true. Since 4 ≠ −7, the two ordered pairs are different points: (4, −7) is in Quadrant IV and (−7, 4) is in Quadrant II.
8. Data reasoning: A right-angled garden corner
An architect marks three corners of a triangular garden bed at A (1, 2), B (4, 6) and C (8, 3), in metres. She wants the corner at B to be a right angle.
(a) Is angle B a right angle? AB2 = 32 + 42 = 25. BC2 = 42 + (−3)2 = 25. AC2 = 72 + 12 = 50. Since AB2 + BC2 = 50 = AC2, the triangle is right-angled, and the right angle is at B, opposite the longest side AC.
(b) What else can you say about the triangle? AB = BC = 5 m, so it is a right-angled isosceles triangle.
(c) Find its area. The two legs meeting at the right angle are 5 m each, so area = (1/2) × 5 × 5 = 12.5 m2.
Important Questions for Exams
1-Mark Questions
- What are the coordinates of the point where the two axes intersect? (0, 0), the origin.
- In which quadrant does the point (−4, −9) lie? Quadrant III, since both coordinates are negative.
- What is the x-coordinate of every point on the y-axis? 0. Every point on the y-axis has the form (0, y).
- Find the distance of the point (−6, 8) from the origin. √(36 + 64) = √100 = 10 units.
- Find the distance between (−3, 5) and (4, 5). The y-coordinates are equal, so the distance is |4 − (−3)| = 7 units.
2-Mark Questions
- Find the distance between P (−2, −3) and Q (4, 5). Shifts 4 − (−2) = 6 and 5 − (−3) = 8. PQ = √(36 + 64) = √100 = 10 units.
- Find the midpoint of the segment joining A (−4, 7) and B (6, −1). ((−4 + 6)/2, (7 − 1)/2) = (2/2, 6/2) = (1, 3).
- Find the values of k if the point (k, 4) is 5 units from (1, 0). (k − 1)2 + (4 − 0)2 = 25, so (k − 1)2 = 9 and k − 1 = 3 or −3. Hence k = 4 or k = −2.
- “Does point Q (y, x) ever coincide with point P (x, y)? Justify your answer.” Only when x = y. Ordered pairs are equal only when both first and second coordinates match, so x = y is needed. (3, 3) is the same either way; (2, 5) and (5, 2) are different points.
3-Mark Questions
- “Point W has x-coordinate equal to −5. Can you predict the coordinates of point H which is on the line through W parallel to the y-axis? Which quadrants can H lie in?” A line parallel to the y-axis is vertical, so every point on it has x-coordinate −5: H = (−5, y). If y > 0, H is in Quadrant II; if y < 0, in Quadrant III; if y = 0, H lies on the x-axis.
- Find the point on the x-axis that is equidistant from A (2, 5) and B (−2, 9). Let the point be P (x, 0). PA2 = PB2 gives (x − 2)2 + 25 = (x + 2)2 + 81. Expanding: x2 − 4x + 4 + 25 = x2 + 4x + 4 + 81, so −8x = 56 and x = −7. P = (−7, 0). Check: PA2 = 81 + 25 = 106 and PB2 = 25 + 81 = 106.
- “Plot point Z (5, −6) on the Cartesian plane. Construct a right-angled triangle IZN and find the lengths of the three sides.” Answers can differ. One choice: I = (0, 0) and N = (5, 0). IN = 5 units (horizontal) and NZ = |0 − (−6)| = 6 units (vertical), so the right angle is at N. IZ = √(25 + 36) = √61 units.
5-Mark Questions
This unit carries only 4 marks in the annual paper, so a 5-mark question on it is not expected there. These are multi-step practice sets; each part matches 1, 2 or 3-mark working.
- The vertices of quadrilateral ABCD are A (1, 2), B (4, 3), C (6, 6) and D (3, 5). (i) Find the lengths of all four sides. (ii) Find the lengths of both diagonals. (iii) Show that the diagonals have the same midpoint. (iv) Name the quadrilateral and say whether it is a rectangle. (i) AB = √(9 + 1) = √10; BC = √(4 + 9) = √13; CD = √(9 + 1) = √10; DA = √(4 + 9) = √13. (ii) AC = √(25 + 16) = √41; BD = √(1 + 4) = √5. (iii) Midpoint of AC = (7/2, 8/2) = (3.5, 4); midpoint of BD = (7/2, 8/2) = (3.5, 4). They are the same point. (iv) Opposite sides are equal in pairs and the diagonals bisect each other, so ABCD is a parallelogram. The diagonals are unequal (√41 ≠ √5), so it is not a rectangle, and adjacent sides are unequal, so it is not a rhombus.
- A (−3, 5) and B (9, −4) are two points. (i) Find AB. (ii) Find the midpoint M of AB. (iii) Find the points of trisection P and Q of AB, with P closer to A. (iv) Verify that AP is one-third of AB. (i) Shifts 12 and −9, so AB = √(144 + 81) = √225 = 15 units. (ii) M = ((−3 + 9)/2, (5 − 4)/2) = (3, 0.5). (iii) One-third of the shift (12, −9) is (4, −3). P = (−3 + 4, 5 − 3) = (1, 2) and Q = (1 + 4, 2 − 3) = (5, −1). (iv) AP = √(16 + 9) = 5 = 15/3. Also PQ = √(16 + 9) = 5 and QB = √(16 + 9) = 5, so the three parts are equal.
Common Mistakes and Examiner Tips
- Swapping the meaning of x and y. The x-coordinate is the distance from the y-axis, and the y-coordinate is the distance from the x-axis. When plotting (x, y), move along the x-axis first, then parallel to the y-axis.
- Writing coordinates in the wrong order. (2, 5) and (5, 2) are different points. Always write the x-coordinate first. The two orders give the same point only when x = y.
- Placing points on an axis in a quadrant. (0, −4) and (7, 0) lie on the axes and belong to no quadrant. Answer “on the x-axis” or “on the y-axis”.
- Adding the shifts instead of using squares. From (3, 4) to (7, 1) the shifts are 4 and 3, and the distance is √(16 + 9) = 5, never 4 + 3 = 7.
- Forgetting the square root. Writing AD = 25 instead of 5. Write the root sign in your first line of working.
- Dropping a minus sign inside a bracket. For (−9, 6) and (−3, 4), the x-shift is −3 − (−9) = 6, not −3 − 9 = −12. Put negative coordinates in brackets before subtracting.
- Forgetting to divide by 2 in the midpoint. The midpoint of (2, 3) and (4, 5) is (3, 4), not (6, 8). Use half the sum, never half the difference.
- Deciding collinearity from rounded decimals. 5 + √85 ≈ 14.22 and √202 ≈ 14.21 look equal, but R (−5, −1), B (−2, −5), C (4, −12) are not collinear. Square both sides or compare the shifts to settle it exactly.
- Treating two equal distances as proof of collinearity. PQ = QR only says Q is equally far from P and R. You need PQ + QR = PR.
- Using the diagonal for the area of a square. Area = side2. For the square with side √10 and diagonal √20, the area is 10 square units, not 20.
Quick Revision Points
- The 2-D Cartesian system uses two number lines at right angles: the horizontal x-axis and the vertical y-axis
- The axes meet at the origin O = (0, 0); distances are marked in equal units on both axes
- Right of O and above O are positive; left of O and below O are negative
- Points on the x-axis are (x, 0); points on the y-axis are (0, y)
- x-coordinate = distance from the y-axis; y-coordinate = distance from the x-axis
- Quadrant signs: I (+, +), II (−, +), III (−, −), IV (+, −); points on the axes lie in no quadrant
- (x, y) = (y, x) if and only if x = y
- Same y: distance = |x2 − x1|; same x: distance = |y2 − y1|
- Distance formula: √[(x2 − x1)2 + (y2 − y1)2], from the Baudhāyana-Pythagoras theorem
- Distance of (x, y) from the origin = √(x2 + y2)
- NCERT triangle A (3, 4), D (7, 1), M (9, 6): AD = 5, DM = √29, MA = √40
- Reflection in the y-axis: (x, y) → (−x, y); in the x-axis: (x, y) → (x, −y); lengths are preserved
- Midpoint = ((x1 + x2)/2, (y1 + y2)/2)
- Missing end B from midpoint M and end A: solve (xA + x)/2 = xM and (yA + y)/2 = yM
- Points of trisection: step one-third of the shift from A to B, twice
- Collinear: the two smaller distances add up exactly to the largest
- Right-angled: the largest side2 equals the sum of the other two squares; right angle opposite the longest side
- Circle with centre O and radius r: compare x2 + y2 with r2 for on, inside or outside
- Square: four equal sides and equal diagonals; area = side2
- A floor map is 2-D, so it cannot show heights such as windows or the height of a table
Weightage and Exam Pattern
Class 9 has no board exam. This chapter is assessed in school exams: periodic tests, the half-yearly exam and the annual exam. In the CBSE Class IX Mathematics course structure for 2026-27, Unit III, Coordinate Geometry, carries 4 marks out of the 80-mark theory paper and is allotted 6 teaching periods. The remaining 20 marks of the year are internal assessment: pen-paper tests and multiple assessment (10), portfolio (5) and lab practical (5).
| Unit | Marks (out of 80) |
|---|---|
| I. Number System | 07 |
| II. Algebra | 20 |
| III. Coordinate Geometry | 04 |
| IV. Geometry | 25 |
| V. Mensuration | 14 |
| VI. Statistics and Probability | 10 |
| Total | 80 |
The course structure lists the competencies for this unit: specify locations and the position of one point relative to another using coordinates, represent a floor plan on a grid, compute the distance between two points, decide whether three points lie on a straight line, find the midpoint of a segment, check whether a triangle is right-angled, and model situations on the coordinate plane to verify geometric properties. The 1, 2, 3 and 5-mark grouping above is a practice ladder for practice, from recall to multi-step working.
The CBSE Class IX question paper design weights questions by thinking skill: about 54% remembering and understanding, 24% applying, and 22% analysing, evaluating and creating.
This chapter is the foundation for Class 10, where the distance formula and the midpoint return.
Class 9 Maths · Chapter 1 – swipe through all 9 cards to understand the whole chapter.
Why Coordinates
A system of coordinates uses numbers to describe the exact location of points or objects.
A floor map is 2-D: it cannot show heights such as windows.
- Sindhu-Sarasvatī streets ran N-S and E-W about 10 m apart.
- Brahmagupta’s zero and negatives make four quadrants possible.
- Descartes (1637): two numbers fix any point in a plane.
Axes and Origin
Two number lines at right angles: the horizontal x-axis and the vertical y-axis.
Right and up are positive; left and down are negative.
- Points on the x-axis are (x, 0).
- Points on the y-axis are (0, y).
- B (4.5, 0) is 4.5 units right of O; G (0, −4.5) is 4.5 units below.
The Four Quadrants
The axes divide the Cartesian plane into four quadrants, numbered anticlockwise from the top right.
Points on an axis lie in no quadrant.
- S (3, −5) is in Quadrant IV.
- Q (−5, 3) is in Quadrant II.
- (x, y) = (y, x) only when x = y.
Horizontal and Vertical Distance
When one coordinate is shared, the distance is the absolute difference of the other.
Absolute value keeps a distance positive.
- (0, 1.5) to (0, 4) is 2.5 units.
- Table feet (8, 9), (11, 9), (11, 7): fourth foot at (8, 7).
- That table is 3 units long and 2 units wide.
Distance Formula
The shifts along the two axes are the legs of a right triangle; the segment is its hypotenuse.
From the Baudhāyana-Pythagoras theorem.
- A (3, 4) to D (7, 1): √(16 + 9) = 5.
- Distance from the origin: √(x2 + y2).
- Signs of the shifts do not matter once squared.
Reflection Keeps Lengths
Reflecting in an axis flips one sign, and every side length stays the same.
The position changes; the shape and size do not.
- A (3, 4), D (7, 1), M (9, 6) reflect to A′ (−3, 4), D′ (−7, 1), M′ (−9, 6).
- Sides stay 5, √29 and √40.
- The triangle moves from Quadrant I to Quadrant II.
Midpoint of a Segment
Each coordinate of the midpoint is the average of the matching coordinates of the ends.
Average means half the sum, never half the difference.
- M (−7, 1), A (3, −4) give the other end B (−17, 6).
- Trisection of A (4, 7), B (16, −2): P (8, 4), Q (12, 1).
- Midpoints (5, 1), (6, 5), (0, 3) give A (1, 7), B (−1, −1), C (11, 3).
Collinearity Test
Three points lie on one straight line when the two smaller distances add up to the largest.
Compare exactly: square both sides, never trust rounded decimals.
- M (−3, −4), A (0, 0), G (6, 8): 5 + 10 = 15, collinear.
- R (−5, −1), B (−2, −5), C (4, −12): 5 + √85 > √202, not collinear.
- Equal distances alone do not prove collinearity.
Right Angles, Circles, Squares
Squared distances settle right angles, circle positions and square checks.
Circle at O: compare x2 + y2 with r2.
- A (1, −8), B (−4, 7), C (−7, −4) lie on a circle of radius √65.
- A (2, 1), B (−1, 2), C (−2, −1), D (1, −2): sides √10, diagonals √20, a square.
- That square has area 10 square units.
📝 Practice Coordinate Geometry - 10 questions
Concept-check questions · with answers & explanations
Start →Close ✕
Why D: The x-coordinate of (0, -4.5) is 0, so the point is at zero distance from the y-axis: it lies on the y-axis, 4.5 units below O. So A is false. R is true, because (x, 0) lies on the x-axis and (0, y) lies on the y-axis.
Why not A: A is false, so this option cannot be right, and R certainly cannot explain a false statement
Why not B: This option needs both statements to be true, but A is false
Why not C: This option says A is true and R is false, which is the exact opposite of the truth
Remember: x-coordinate 0: on the y-axis. y-coordinate 0: on the x-axis.
Why D: Every point on a line parallel to the y-axis is the same distance from the y-axis, so H = (-5, y). With x negative, H is in Quadrant II if y > 0, in Quadrant III if y < 0, and on the x-axis if y = 0.
Why not A: A line parallel to the y-axis fixes the x-coordinate, not the y-coordinate
Why not B: Since x = -5 is negative, H can never be to the right of the y-axis, so Quadrants I and IV are impossible
Why not C: H can also be below the x-axis, where y is negative, putting it in Quadrant III
Remember: Parallel to the y-axis: x stays fixed, y can be anything.
Why C: Step 1 (a line parallel to the x-axis keeps y fixed): P = (x, 2). Step 2 (distance along the line = |x2 – x1|): |x – (-5)| = 7, so x + 5 = 7 or x + 5 = -7, giving x = 2 or x = -12. So P is (2, 2) or (-12, 2).
Why not A: These keep x fixed and move up or down, which is along a line parallel to the y-axis
Why not B: These are 7 units from the y-axis, but the 7 units must be measured from (-5, 2), not from O
Why not D: P can also be 7 units to the left of (-5, 2), at (-12, 2)
Remember: Moving along a horizontal line changes x only, in either direction.
Why B: Step 1 (shifts along the axes): 250 – 100 = 150 and 230 – 150 = 80. Step 2 (distance formula): AB = square root of (22500 + 6400) = square root of 28900 = 170 pixels. Step 3 (compare with the sum of the radii): 80 + 100 = 180, and 170 is less than 180 but more than 100 – 80 = 20, so the circles overlap and their edges cross.
Why not A: 180 is the sum of the radii, not the distance between the centres
Why not C: The gap must be compared with the sum of both radii, 180, since each circle reaches out towards the other
Why not D: 230 is 150 + 80, the path along the grid, not the straight distance between the centres
Remember: Circles overlap when the centre gap is less than the sum of the radii.
Why B: Step 1 (midpoint formula for each side): B + C = 2D = (10, 2), C + A = 2E = (12, 10), A + B = 2F = (0, 6). Step 2 (add all three): 2(A + B + C) = (22, 18), so A + B + C = (11, 9). Step 3 (subtract B + C): A = (11 – 10, 9 – 2) = (1, 7). Check: B = (-1, -1), C = (11, 3), and the midpoint of AB is (0, 3) = F.
Why not A: (11, 3) is vertex C, found by subtracting A + B instead of B + C
Why not C: (-1, -1) is vertex B, found by subtracting C + A instead of B + C
Why not D: This averages the three midpoints; the average of D, E, F is not a vertex of the triangle
Remember: A = E + F – D: the two midpoints on A’s sides, minus the one opposite A.
Why A: Step 1 (distance formula): RB = square root of (9 + 16) = 5, BC = square root of (36 + 49) = √(85), RC = square root of (81 + 121) = √(202). Step 2 (compare exactly, by squaring): (5 + √(85)) squared = 110 + 10√(85), which is about 202.2, more than 202. So RB + BC > RC and the points form a triangle. Step 3 (check with shifts): R to B is 3 right, 4 down; B to C is 6 right, 7 down, but a straight line would need 8 down.
Why not B: R to B is 3 right and 4 down, so twice that is 6 right and 8 down, but B to C is 6 right and 7 down
Why not C: Sharing a quadrant does not put points on one line, and C is in Quadrant IV anyway
Why not D: The rounded values 5 + 9.22 = 14.22 and 14.21 look equal, but the exact comparison shows RB + BC is larger
Remember: Never trust rounded decimals in a collinearity check.
Why C: Every point on the y-axis is at zero distance from the y-axis, so its x-coordinate is 0 and it has the form (0, y). Only (0, -7) has x-coordinate 0; it is 7 units below O.
Why not A: Both coordinates of (7, 7) are non-zero, so it lies in Quadrant I, not on an axis
Why not B: (3, 0) has y-coordinate 0, so it lies on the x-axis, 3 units to the right of O
Why not D: Both coordinates of (-2, 5) are non-zero, so it is not on either axis; it lies in Quadrant II
Remember: x-coordinate 0 means the point is ON the y-axis.
Why C: In Quadrant III both coordinates are negative, so a < 0 and b < 0. Then -a > 0 while b is still negative. The pattern (+, -) belongs to Quadrant IV. This is the mirror image of (a, b) in the y-axis.
Why not A: This assumes -a is still negative, but the negative of a negative number is positive
Why not B: Quadrant II has the pattern (-, +), which needs b positive, but b is negative
Why not D: Quadrant I would need both -a and b positive; only the sign of the first coordinate has changed
Remember: Changing the sign of x flips a point across the y-axis.
Why B: A and M both have y = -2, so AM is parallel to the x-axis. M and P both have x = -5, so MP is parallel to the y-axis. The axes are perpendicular, so AM and MP are perpendicular.
Why not A: Both PR and RA are slanted sides meeting at R; meeting at a vertex does not make them perpendicular, and here they are not
Why not C: RA joins (3, 0) to (0, -2); both coordinates change, so RA is slanted, not vertical, and cannot be perpendicular to the horizontal side AM
Why not D: RA is slanted, not horizontal, so it is not perpendicular to the vertical side MP
Remember: Same y: horizontal. Same x: vertical. Horizontal meets vertical at 90°.
Why A: Step 1 (shifts along the axes, keeping signs): x-shift = -3 – (-9) = 6, y-shift = 4 – 6 = -2. Step 2 (distance formula): M’A’ = square root of (36 + 4) = square root of 40 = 2√(10) units.
Why not B: 148 = 144 + 4 comes from an x-shift of 12, which treats -9 as +9
Why not C: 8 is 6 + 2, the path along the grid, not the straight distance
Why not D: 4√(2) is the square root of 32 = 36 – 4, subtracting the squares instead of adding them
Remember: Subtract coordinates with their signs; the squares take care of direction.
Chapter Navigation
Next: Introduction to Polynomials Class 9 Notes
Related Chapters in Class 9 Maths
- Introduction to Polynomials Class 9 Notes
- Number System Class 9 Notes
- Exploring Algebraic Identities Class 9 Notes
- Circles Class 9 Notes
- Area and Perimeter Class 9 Notes
- Introduction to Probability Class 9 Notes
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Frequently Asked Questions
It is the plane in which the two coordinate axes lie: the horizontal x-axis and the vertical y-axis, meeting at right angles at the origin O (0, 0). It is also called the coordinate plane or the xy-plane. The name honours René Descartes, who formalised the idea that any point in a plane can be described by two numbers, its distances from two perpendicular axes.
Look at the signs of the two coordinates. Quadrant I is (+, +), Quadrant II is (−, +), Quadrant III is (−, −) and Quadrant IV is (+, −). So (3, −5) is in Quadrant IV and (−5, 3) is in Quadrant II. If either coordinate is zero, the point lies on an axis and belongs to no quadrant.
The x-coordinate is the perpendicular distance of the point from the y-axis, measured along the x-axis. The y-coordinate is the perpendicular distance of the point from the x-axis, measured along the y-axis. Both carry signs: right and up are positive, left and down are negative. The x-coordinate is always written first.
Join the two points and draw a horizontal line through one and a vertical line through the other to form a right triangle. The legs are the shifts x₂ − x₁ and y₂ − y₁. By the Baudhāyana-Pythagoras theorem, the segment is the hypotenuse, so its length is √[(x₂ − x₁)² + (y₂ − y₁)²]. The signs of the shifts do not matter because they are squared.
Find the three distances between them with the distance formula. If the two smaller distances add up exactly to the largest, the points are collinear. For M (−3, −4), A (0, 0), G (6, 8): MA = 5, AG = 10 and MG = 15, and 5 + 10 = 15. Compare exactly, never with rounded decimals, because nearly equal sums can mislead.
The midpoint of the segment joining (x₁, y₁) and (x₂, y₂) is ((x₁ + x₂)/2, (y₁ + y₂)/2): each coordinate is the average of the matching coordinates of the ends. For (2, 3) and (4, 5) the midpoint is (3, 4). Run it backwards to find a missing end, and step one-third of the shift twice to find points of trisection.
Compare x² + y² with r², where r is the radius. Equal means the point is on the circle, smaller means inside, larger means outside. For the circle through A (1, −8), r² = 65. D (−5, 6) gives 61, so it is inside; E (0, 9) gives 81, so it is outside.
In the CBSE Class IX Mathematics course structure for 2026-27, Unit III, Coordinate Geometry, carries 4 marks out of the 80-mark theory paper and is allotted 6 teaching periods. Class 9 has no board exam, so it is tested in school exams. The distance formula and the midpoint come back in Class 10.
