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Important Questions Class 9 Maths Chapter 1 The Use of Coordinates

This page gives Important Questions Class 9 Maths Chapter 1 The Use of Coordinates. In this chapter, we use two numbers (x, y) to show the place of a point on a flat plane. The questions cover plotting points, finding the distance between two points, finding the midpoint of a line segment, and checking if three points lie on one straight line.

Some questions use real-life ideas, like a computer screen and a city map. Every answer is solved step by step, so students can follow each step easily. Teachers can also use these questions for class practice and revision.

Important Questions Class 9 Maths Chapter 1 The Use of Coordinates

Q1. 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?

Square ABCD with A (2, 1), B (−1, 2), C (−2, −1) and D (1, −2) plotted on a coordinate grid

Answer: Yes, ABCD is a square.

We can check it by finding the lengths of sides and diagonals.

Finding the lengths of all sides:

AB = √[(−1 − 2)² + (2 − 1)²] = √10 units

BC = √[(−2 − (−1))² + (−1 − 2)²] = √10 units

CD = √[(1 − (−2))² + (−2 − (−1))²] = √10 units

DA = √[(2 − 1)² + (1 − (−2))²] = √10 units

All four sides are equal.

Now finding diagonals:

AC = √[(−2 − 2)² + (−1 − 1)²] = √20 units

BD = √[(1 − (−1))² + (−2 − 2)²] = √20 units

Here, diagonals are also equal.

Since, all sides are equal and diagonals are equal, so ABCD is a square.

Area of ABCD = (side)² = (√10)² = 10 square units

Q2. 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.

800 by 600 pixel screen with a circle of radius 80 at A (100, 150) and a circle of radius 100 at B (250, 230)

Answer: (i) For the first circle, the extreme x-coordinates are 100 − 80 = 20 and 100 + 80 = 180; the extreme y-coordinates are 150 − 80 = 70 and 150 + 80 = 230.

For the second circle, the extreme x-coordinates are 250 − 100 = 150 and 250 + 100 = 350; the extreme y-coordinates are 230 − 100 = 130 and 230 + 100 = 330.

All values lie within 0 ≤ x ≤ 800 and 0 ≤ y ≤ 600. Thus neither circle extends outside the screen.

(ii) AB = √[(250 − 100)² + (230 − 150)²] = √(22500 + 6400) = 170 pixels

The radii add to 80 + 100 = 180 pixels and differ by 100 − 80 = 20 pixels. Because 20 < 170 < 180, the two circular boundaries intersect at two points.

Q3. A city has two main roads which cross each other at the centre of the city. These two roads are along the North–South (N–S) direction and East–West (E–W) direction. All the other streets of the city run parallel to these roads and are 200 m apart. There are 10 streets in each direction.

(i) Using 1 cm = 200 m, draw a model of the city in your notebook. Represent the roads/streets by single lines.

(ii) There are street intersections in the model. Each street intersection is formed by two streets — one running in the N–S direction and another in the E–W direction. Each street intersection is referred to in the following manner: If the second street running in the N–S direction and 5th street in the E–W direction meet at some crossing, then we call this street intersection (2, 5). Using this convention, find:

(a) how many street intersections can be referred to as (4, 3).

(b) how many street intersections can be referred to as (3, 4).

Model of the city with two main roads and numbered N–S and E–W streets 1 cm apart, with intersections (4, 3) and (3, 4) marked

Answer: (i) Draw the North–South and East–West main roads perpendicular to one another. Draw 10 other streets parallel to each main road, 1 cm apart from neighbouring roads. For example, place five parallel streets on each side of each main road.

(ii) Number the 10 North–South streets uniquely from 1 to 10 and do the same for the 10 East–West streets. A pair (a, b) identifies the meeting of just the a^th North–South street and the b^th East–West street.

(a) (4, 3): one intersection.

(b) (3, 4): one intersection.

These two crossings are different because their street numbers are in a different order.

Q4. 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.

Triangle ABC with A (1, 7), B (−1, −1), C (11, 3) and midpoints D (5, 1), E (6, 5), F (0, 3)

Answer: Let the vertices of triangle ABC be A (x₁, y₁), B (x₂, y₂), and C (x₃, y₃).

Given midpoints: D (5, 1), E (6, 5), F (0, 3)

D is midpoint of BC, so using midpoint formula:

x₂ + x₃ = 10 .....(1)

y₂ + y₃ = 2 .....(2)

E is midpoint of CA, therefore:

x₃ + x₁ = 12 .....(3)

y₃ + y₁ = 10 .....(4)

F is midpoint of AB, therefore:

x₁ + x₂ = 0 .....(5)

y₁ + y₂ = 6 .....(6)

Solving for x-coordinates, from (5):

x₂ = −x₁

Substitute into (1): −x₁ + x₃ = 10

⇒ x₃ = 10 + x₁ .....(7)

Substituting into (3): (10 + x₁) + x₁ = 12

⇒ 2x₁ + 10 = 12

⇒ 2x₁ = 2

⇒ x₁ = 1

Then: x₂ = −1

⇒ x₃ = 10 + 1 = 11

Solving y-coordinates, from (6):

y₂ = 6 − y₁

Substituting into (2): (6 − y₁) + y₃ = 2

⇒ y₃ = y₁ − 4 .....(8)

Substituting into (4): (y₁ − 4) + y₁ = 10

⇒ 2y₁ − 4 = 10

⇒ 2y₁ = 14

⇒ y₁ = 7

Then: y₂ = 6 − 7 = −1

⇒ y₃ = 7 − 4 = 3

Therefore, the coordinates of the points are A (1, 7), B (−1, −1) and C (11, 3).

Q5. (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.

Circle K with centre O (0, 0) through A (1, −8), B (−4, 7) and C (−7, −4), with D (−5, 6) inside and E (0, 9) outside

Answer: (i) Distance OA = √(1² + (−8)²) = √65 units

Distance OB = √((−4)² + 7²) = √65 units

Distance OC = √((−7)² + (−4)²) = √65 units

Since all three points are at the same distance from origin, they lie on a circle centred at (0, 0).

Radius = √65 units

(ii) Distance OD = √((−5)² + 6²) = √61 units

Distance OE = √(0² + 9²) = 9 units

Since, √61 < √65 < 9, D lies inside the circle and E lies outside the circle.

Q6. 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).

Line segment from A (4, 7) to B (16, −2) divided into three equal parts by P (8, 4) and Q (12, 1)

Answer: To trisect AB, P and Q divide the segment into 3 equal parts.

Given: A (4, 7), B (16, −2)

Let P (x₁, y₁) and Q (x₂, y₂)

Here, P is midpoint of A and Q, so

x₁ = (4 + x₂)/2 ......(1)

y₁ = (7 + y₂)/2 ......(2)

So, Q is midpoint of P and B, so

x₂ = (x₁ + 16)/2 ......(3)

y₂ = (y₁ − 2)/2 ......(4)

Solving for x-coordinates, from (1):

x₁ = (4 + x₂)/2 and

From (3): x₂ = (x₁ + 16)/2

Substituting (1) into (3), we get

x₂ = ((4 + x₂)/2 + 16)/2

= (4 + x₂ + 32)/4

= (x₂ + 36)/4

So, 4x₂ = x₂ + 36

⇒ 3x₂ = 36

⇒ x₂ = 12

Then from (1): x₁ = (4 + 12)/2 = 8

Solving for y-coordinates, from (2):

y₁ = (7 + y₂)/2 and

From (4): y₂ = (y₁ − 2)/2

Substituting (2) into (4):

y₂ = ((7 + y₂)/2 − 2)/2

= (7 + y₂ − 4)/4

= (y₂ + 3)/4

So, 4y₂ = y₂ + 3

⇒ 3y₂ = 3

⇒ y₂ = 1

Then from (2): y₁ = (7 + 1)/2 = 4

Therefore, P = (8, 4) and Q = (12, 1).

Alternatively:

The total change from A to B is 16 − 4 = 12 horizontally and −2 − 7 = −9 vertically. Each of the three equal parts therefore changes the coordinates by (4, −3).

P = (4 + 4, 7 − 3) = (8, 4)

Q = (8 + 4, 4 − 3) = (12, 1)

Q7. 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).

Answer: Given:

A = (3, −4)

B = (x, y)

M = (−7, 1)

Using the connection from the last problem

Midpoint M = ((x₁ + x₂)/2, (y₁ + y₂)/2)

So, (3 + x)/2 = −7

⇒ 3 + x = −14

⇒ x = −17

and (−4 + y)/2 = 1

⇒ −4 + y = 2

⇒ y = 6

Therefore, coordinates of B:

B = (−17, 6)

Q8. 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.

S	M	T
(−3, 0)	(0, 0)	(3, 0)
(2, 3)	(3, 4)	(4, 5)
(0, 0)	(0, 5)	(0, −10)
(−8, 7)	(0, −2)	(6, −3)

When M is the mid-point of ST, can you find any connection between the coordinates of M, S and T?

Answer:

S	M	T	Is M the midpoint of ST? Yes or No	Reason for your answer
(−3, 0)	(0, 0)	(3, 0)	Yes	SM + MT = ST and SM = MT
(2, 3)	(3, 4)	(4, 5)	Yes	SM + MT = ST and SM = MT
(0, 0)	(0, 5)	(0, −10)	No	SM + MT ≠ ST and SM ≠ MT
(−8, 7)	(0, −2)	(6, −3)	No	SM + MT ≠ ST and SM ≠ MT

A midpoint is halfway between the two endpoints. On each axis, its coordinate is the average of the corresponding endpoint coordinates.

x_m = (x_s + x_t) ÷ 2; y_m = (y_s + y_t) ÷ 2

Q9. Using the origin as one vertex, plot the vertices of:

(i) A right-angled isosceles triangle.

(ii) An isosceles triangle with one vertex in Quadrant III and the other in Quadrant IV

Right-angled isosceles triangle OAB with O (0, 0), A (4, 0) and B (0, 4)

Isosceles triangle OAB with O (0, 0), A (−4, −3) in Quadrant III and B (4, −3) in Quadrant IV

Answer: (i) Take O (0, 0), A (4, 0) and B (0, 4). The axes meet at 90°, OA = OB = 4 units, so △OAB is right-angled and isosceles.

(ii) Take O (0, 0), A (−4, −3) and B (4, −3). A is in Quadrant III; B is in Quadrant IV.

OA = √((−4)² + (−3)²) = 5 units

OB = √(4² + (−3)²) = 5 units

Hence △OAB is isosceles.

Q10. 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. Now plot both sets of points and check your answers.

Points R (−5, −1), B (−2, −5) and C (4, −12) plotted; the path R to B to C bends slightly away from the straight line RC

Answer: To check whether the points R (−5, −1), B (−2, −5) and C (4, −12) lie on the same straight line using the distance formula:

d = √[(x₂ − x₁)² + (y₂ − y₁)²]

RB = √[(−2 + 5)² + (−5 + 1)²] = √(3² + (−4)²) = 5 units

BC = √[(4 + 2)² + (−12 + 5)²] = √(6² + (−7)²) = √85 units

RC = √[(4 + 5)² + (−12 + 1)²] = √(9² + (−11)²) = √202 units

Now: RB + BC = 5 + √85 ≠ √202

Since the sum of two distances is not equal to the third, the points R, B and C do not lie on the same straight line.

Q11. 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.

Answer: To check whether the points M (−3, −4), A (0, 0) and G (6, 8) lie on the same straight line using the distance formula:

d = √[(x₂ − x₁)² + (y₂ − y₁)²]

MA = √[(0 + 3)² + (0 + 4)²] = 5 units

AG = √[(6 − 0)² + (8 − 0)²] = 10 units

MG = √[(6 + 3)² + (8 + 4)²] = √(9² + 12²) = 15 units

Since, MA + AG = 5 + 10 = 15 = MG, the sum of two distances equals the third, the points M, A and G lie on the same straight line.

Q12. What would a system of coordinates be like if we did not have negative numbers? Would this system allow us to locate all the points on a 2-D plane?

Answer: If we did not have negative numbers, then coordinates could only be zero or positive.

So:

On the x-axis, we could mark only the points to the right of the origin.

On the y-axis, we could mark only the points above the origin.

This means we could locate points only in:

Quadrant I

the positive part of the x-axis

the positive part of the y-axis

and the origin

We would not be able to locate:

points in Quadrant II

points in Quadrant III

points in Quadrant IV

points on the negative parts of the axes

Therefore, such a system would not allow us to locate all the points on a 2-D plane.

Related Study Material on Extramarks

Q.1 Rationalise the denominator of

$\frac{2}{\sqrt{7} + \sqrt{5}}$

Marks:2
Ans

$\begin{array}{l} \frac{2}{\sqrt{7} + \sqrt{5}} — \frac{\sqrt{7} \sqrt{5}}{\sqrt{7} \sqrt{5}} \\ = \frac{2 (\sqrt{7} \sqrt{5})}{7 5} µ (\sqrt{a} + \sqrt{b}) (\sqrt{a} \sqrt{b}) = a b \\ = \frac{2 (\sqrt{7} \sqrt{5})}{2} \\ = \sqrt{7} \sqrt{5} \end{array}$

Q.2 Represent

$\sqrt{5}$

on the number line.
Marks:5
Ans

Step 1: Draw a number line ‘l ‘.

Step 2: Mark O as origin to represent

OD = 2 units
Step 3: Draw DE perpendicular to OD, Such that

DE = 1 unit.

Step 4: Join OE
By Pythagoras theorem in right angle triangle ODE

OE² = OD² + DE²
= 2² + 1²
= 4 + 1 = 5
OE =

\sqrt{5}

Step 5 Using a compass, with O as centre and radius OE, draw an arc
which intersect l at some point on the number line.
Thus, the distance from O to the intersecting point will be of 5 units.

Q.3 Find, whether

$\frac{1}{625}$

is a terminating or non terminating decimal number.
Marks:1
Ans

$\begin{array}{l} \frac{1}{625} = \frac{1}{5 — 5 — 5 — 5} Therefore, \frac{1}{625} is a terminating decimal \\ number as denominator can be expressed in the form 5^{4} . \end{array}$

Q.4

$\begin{array}{l} Simplify \\ (i) (3 + \sqrt{3}) (2 + \sqrt{2}) \\ (ii) (5 + \sqrt{7}) (2 + \sqrt{5}) \end{array}$

Marks:4
Ans

\begin{array}{l} (i) (3 + \sqrt{3}) (2 + \sqrt{2}) \\ = 3 — 2 — 3 — \sqrt{2} + \sqrt{3} — 2 + \sqrt{3} — \sqrt{2} \\ = 6 + 3 \sqrt{2} + 2 \sqrt{3} + \sqrt{6} \\ (ii) (5 + \sqrt{7}) (2 + \sqrt{5}) \\ = 5 — 2 + 5 — \sqrt{5} + \sqrt{7} — 2 + \sqrt{7} — \sqrt{5} \\ = 10 + 5 \sqrt{5} + 2 \sqrt{7} + 35 \end{array}

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FAQs (Frequently Asked Questions)

Class 9 Maths Chapter 1 in the 2026 syllabus is The Use of Coordinates from Ganita Manjari. It covers Cartesian plane, x-axis, y-axis, origin, quadrants, coordinates of points, points on axes, distance between two points, and real-life coordinate applications.

No. In the updated 2026 Class 9 Maths book, Chapter 1 is The Use of Coordinates. Many older resources still show Number Systems as Chapter 1, but the current Ganita Manjari mapping starts with coordinate geometry.

The distance between two points (x₁, y₁) and (x₂, y₂) is √[(x₂ – x₁)² + (y₂ – y₁)²]. This formula comes from the Baudhāyana-Pythagoras theorem and works for positive, negative, and zero coordinates.

Check the signs of x and y. Quadrant I is (+, +), Quadrant II is (-, +), Quadrant III is (-, -), and Quadrant IV is (+, -). Points on the x-axis or y-axis are not in any quadrant.

Common questions include finding coordinates of points, identifying quadrants, plotting points, finding distance between two points, finding midpoints, checking collinearity, locating mirror images, and solving room layout or city-grid coordinate problems.

Write coordinates in the correct order, x first and y second. Check quadrant signs before plotting. Remember that points on axes have no quadrant. In distance questions, subtract coordinates first, then square the differences before adding.

Important Questions Class 9 Maths Chapter 1 The Use of Coordinates