NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 1 Orienting Yourself: The Use of Coordinates 2026-27

Think and Reflect (NCERT Textbook Page No. 5)

Question 1. What are the standard widths for a room door? Look around your home and in school.

Solution: The usual width of a room door is generally between 36 inches and 48 inches, which is about 3 to 4 feet.

Question 2. Are the doors in your school suitable for people in wheelchairs?

Solution: Students should observe the doors in their school and check whether they are wide enough for a wheelchair to pass through comfortably.

Think and Reflect (NCERT Textbook Page No. 7)

Question 1. What is the x-coordinate of a point on the y-axis?

Solution: The x-coordinate shows the horizontal distance of a point from the y-axis. A point lying on the y-axis has no horizontal distance from it. So, the x-coordinate of every point on the y-axis is 0.

Question 2. Is there a similar generalisation for a point on the x-axis?

Solution: Yes. The y-coordinate shows the vertical distance of a point from the x-axis. A point lying on the x-axis has no vertical distance from it. Hence, the y-coordinate of every point on the x-axis is 0.

Question 3. Does point Q (y, x) ever coincide with point P (x, y)? Justify your answer.

Solution: Yes, points Q (y, x) and P (x, y) will coincide only when their corresponding coordinates are equal. This means y = x and x = y. Therefore, both points coincide only when x = y.

Question 4. If x ≠ y, then (x, y) ≠ (y, x); and (x, y) = (y, x) if and only if x =y. Is this claim true?

Solution: Yes, the claim is true. Two ordered pairs are equal only when their corresponding coordinates are the same. So, (x, y) will be equal to (y, x) only if x = y. If x and y are different, then (x, y) and (y, x) will not be equal.

Think and Reflect (NCERT Textbook Page No. 9)

Question 1. In moving from A (3, 4) to D (7, 1), what distance has been covered along the x-axis? What about the distance along the y-axis?

Solution: From A (3, 4) to D (7, 1), the change along the x-axis is 7 - 3 = 4 units. The change along the y-axis is 4 - 1 = 3 units. So, the distance covered along the x-axis is 4 units and along the y-axis is 3 units.

Question 2. Can these distances help you find the distance AD?

Solution: Yes. The horizontal and vertical distances form a right triangle. So, using the Pythagoras theorem, AD = √(4² + 3²) = √25 = 5 units.

Think and Reflect (NCERT Textbook Page No. 11)

Question 1. What has remained the same and what has changed with this reflection?

Solution: After reflection, the lengths of the corresponding sides remain unchanged. However, the signs of the x-coordinates of the vertices change.

Question 2. Would these observations be the same if AADM is reflected in the x-axis (instead of the y-axis)?

Solution: No. If ΔADM is reflected in the x-axis, the signs of the y-coordinates will change, while the x-coordinates will remain the same.

Class 9 Maths Ganita Manjari Chapter 1 Exercise Set 1.1 Solutions

Figure shows Reiaan's room with points OABC marking its corners. The x- and y-axes are marked in the figure. Point O is the origin.

Image 1

Referring to Figure, answer the following questions:

(i) If D₁R₁ represents the door to Reiaan's room, how far is the door from the left wall (the y-axis) of the room? How far is the door from the x-axis?

Solution: The door begins at point D₁, which lies on the x-axis at a distance of 8 units from the origin O. Therefore, the door is 8 units away from the left wall (the y-axis). As the entire door D₁R₁ lies along the x-axis itself, its distance from the x-axis is 0 units.

(ii) What are the coordinates of D₁?

Solution: Point D₁ lies on the x-axis, so its y-coordinate is 0. It is located 8 units to the right of the origin, giving us the coordinates D₁ = (8, 0).

(iii) If R₁ is the point (11.5, 0), how wide is the door? Do you think this is a comfortable width for the room door? If a person in a wheelchair wants to enter the room, will he/she be able to do so easily?

Solution: Width of the door = difference of the x-coordinates of R₁ and D₁ = 11.5 − 8 = 3.5 units

Taking 1 unit = 1 foot, the door is 3.5 feet wide. Standard room doors are usually 2 to 3 feet wide, so a 3.5-foot door is more than comfortable. A typical wheelchair measures about 2 to 2.25 feet across, which is well within 3.5 feet — so a person using a wheelchair can enter the room without any difficulty.

(iv) If B₁ (0, 1.5) and B₂ (0, 4) represent the ends of the bathroom door, is the bathroom door narrower or wider than the room door?

Solution: Both B₁ and B₂ lie on the y-axis, so the width of the bathroom door = difference of their y-coordinates = 4 − 1.5 = 2.5 units

The room door is 3.5 units wide, while the bathroom door is only 2.5 units wide. Therefore, the bathroom door is narrower than the room door.

Class 9 Maths Ganita Manjari Chapter 1 Exercise Set 1.2 Solutions

On a graph sheet, mark the x-axis and y-axis and the origin O. Mark points from (−7, 0) to (13, 0) on the x-axis and from (0, −15) to (0, 12) on the y-axis. (Use the scale 1 cm = 1 unit.) Using Figure, answer the given questions.

Image 2

Question 1. 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?

Solution: Label the three given feet as A (8, 9), B (11, 9) and C (11, 7). In a rectangle, opposite sides are parallel to the axes here, so the missing corner D must line up vertically with A and horizontally with C. This means D takes its x-coordinate from A (which is 8) and its y-coordinate from C (which is 7). Hence, the fourth foot of the table is at D = (8, 7).

Image 3

(ii) Is this a good spot for the table?

Solution: Yes, this position works well for the table. It sits fully inside the room without crossing any boundary, stays clear of the door's swing and the walking space, and its position close to the right wall keeps the centre of the room free, making it a convenient study corner.

(iii) What is the width of the table? The length? Can you make out the height of the table?

Solution: Length of the table = difference of x-coordinates of A (8, 9) and B (11, 9) = 11 − 8 = 3 units Width of the table = difference of y-coordinates of B (11, 9) and C (11, 7) = 9 − 7 = 2 units The height cannot be found from the figure. The coordinate plane shows only the top view of the room in two dimensions, and height is a third dimension that a 2D figure does not capture.

Question 2. If the bathroom door has a hinge at B₁ and opens into the bedroom, will it hit the wardrobe? Are there any changes you would suggest if the door is made wider?

Solution: The bathroom door runs from B₁ (0, 1.5) to B₂ (0, 4), so its width = 4 − 1.5 = 2.5 units. When the door swings open into the bedroom with its hinge fixed at B₁, the free end traces an arc of radius 2.5 units around B₁. The wardrobe stands along the line x = 3, between the points (3, 0) and (3, 2). The closest the wardrobe gets to the hinge B₁ (0, 1.5) is 3 units — which is more than the door's reach of 2.5 units. Since the arc never extends as far as the wardrobe, the door will not hit it.

If a wider door is planned, its swing radius would increase and could reach the wardrobe. Two simple adjustments would solve this: make the door open inward into the bathroom instead, or move the wardrobe slightly further to the right.

Question 3. Look at Reiaan's bathroom.

(i) What are the coordinates of the four corners O, F, R, and P of the bathroom?

Solution: Reading from the figure, the bathroom's four corners are: O = (0, 0), F = (0, 9), R = (−6, 9), P = (−6, 0)

(ii) What is the shape of the showering area SHWR in Reiaan's bathroom? Write the coordinates of the four corners.

Solution: In SHWR, exactly one pair of opposite sides is parallel to each other, which makes the showering area a trapezium. Its corners are at: S = (−6, 6), H = (−3, 6), W = (−2, 9), R = (−6, 9)

(iii) Mark off a 3 ft × 2 ft space for the washbasin and a 2 ft × 3 ft space for the toilet. Write the coordinates of the corners of these spaces.

Solution: Placing the washbasin space (3 ft wide, 2 ft deep) in the lower portion of the bathroom, its corners are: L = (−6, 2), M = (−3, 2), N = (−3, 0), P = (−6, 0)

Placing the toilet space (2 ft wide, 3 ft deep) just above it, its corners are: I = (−6, 5), J = (−4, 5), K = (−4, 2), L = (−6, 2)

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Question 4. Other rooms in the house:

(i) Reiaan's room door leads from the dining room which has the length 18 ft and width 15 ft. The length of the dining room extends from point P to point A. Sketch the dining room and mark the coordinates of its corners.

Solution: P is at (−6, 0) and A is at (12, 0). Distance PA = 12 − (−6) = 18 ft ✓ — this matches the given length of the dining room. The dining room lies below the line PA, extending 15 units downward (its width). So PA forms the top edge, and the bottom edge lies along y = −15. The dining room PAUV therefore has corners: P = (−6, 0), A = (12, 0), U = (12, −15), V = (−6, −15)

(ii) Place a rectangular 5 ft × 3 ft dining table precisely in the centre of the dining room. Write down the coordinates of the feet of the table.

Solution: First, find the centre of the dining room. x-coordinate of centre = (−6 + 12)/2 = 3 y-coordinate of centre = (0 + (−15))/2 = −7.5 So the centre is at (3, −7.5).

For a 5 ft × 3 ft table centred here, measure 2.5 ft (half the length) on either side along x, and 1.5 ft (half the width) on either side along y. The four feet of the table are at: E = (3 − 2.5, −7.5 − 1.5) = (0.5, −9) F = (3 + 2.5, −7.5 − 1.5) = (5.5, −9) G = (3 + 2.5, −7.5 + 1.5) = (5.5, −6) H = (3 − 2.5, −7.5 + 1.5) = (0.5, −6)

Class 9 Maths Chapter 1 End of Chapter Exercises Solutions

Question 1. What are the x-coordinate and y-coordinate of the point of intersection of the two axes?

Solution: The two axes cross each other at the origin, the starting point of the coordinate system. At this point, no distance has been moved along either axis, so both coordinates are zero. Hence, the point of intersection is O = (0, 0), with x-coordinate 0 and y-coordinate 0.

Question 2. 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?

Solution: A line through W parallel to the y-axis is a vertical line, and every point on a vertical line shares the same x-coordinate. So every point H on this line must have x-coordinate −5, i.e., H = (−5, y) for some real number y.

The quadrant depends on the value of y:

• When y is positive, H = (−5, y) has signs (−, +), placing it in Quadrant II.

• When y is negative, the signs are (−, −), placing H in Quadrant III.

• When y = 0, H = (−5, 0) sits on the x-axis itself, belonging to no quadrant.

Question 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.

Solution: Points A (0, −2) and M (−5, −2) share the same y-coordinate (−2), so segment AM runs horizontally. Points M (−5, −2) and P (−5, 2) share the same x-coordinate (−5), so segment MP runs vertically. A horizontal line and a vertical line always meet at right angles, so AM ⊥ MP.

(ii) One side of RAMP that is parallel to one of the axes.

Solution: Segment AM is parallel to the x-axis, since both endpoints have the identical y-coordinate −2. Similarly, segment MP is parallel to the y-axis, since both endpoints have the identical x-coordinate −5.

(iii) Two points that are mirror images of each other in one axis. Which axis will this be? Now plot the points and verify your predictions.

Solution: Look at M (−5, −2) and P (−5, 2): their x-coordinates match, while their y-coordinates are equal in size but opposite in sign. Reflecting a point in the x-axis keeps x the same and reverses the sign of y — which is exactly the relationship here. So M and P are mirror images of each other in the x-axis. Plotting the four points confirms all three predictions.

Image 5

Question 4. Plot point Z (5, −6) on the Cartesian plane. Construct a right-angled triangle IZN and find the lengths of the three sides. (Comment: Answers may differ from person to person.)

Solution: One convenient construction: from Z (5, −6), draw a vertical line up to the x-axis, meeting it at I (5, 0), and a horizontal line across to the y-axis, meeting it at N (0, −6). Joining I and N completes the triangle, with the right angle at Z.

Side lengths:

• IZ is vertical: length = 0 − (−6) = 6 units

• ZN is horizontal: length = 5 − 0 = 5 units

• IN is the hypotenuse. By the distance formula: IN = √[(5 − 0)² + (0 − (−6))²] = √(25 + 36) = √61 units

Image 6

Question 5. 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?

Solution: If only zero and positive numbers existed, every coordinate would have to be 0 or greater. Movement would then be possible only to the right of the origin along the x-axis and only upward along the y-axis.

Such a system could describe just: the origin, the positive x-axis, the positive y-axis, and the points of Quadrant I.

Everything else — Quadrants II, III, and IV, and the negative halves of both axes — would have no address at all. So no, this system cannot locate every point of the 2-D plane; it covers only one quarter of it.

Question 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: compute the three pairwise distances using the distance formula, d = √[(x₂ − x₁)² + (y₂ − y₁)²]. If the two shorter distances add up exactly to the longest one, the three points are collinear.

MA = √[(0 + 3)² + (0 + 4)²] = √(9 + 16) = √25 = 5 units AG = √[(6 − 0)² + (8 − 0)²] = √(36 + 64) = √100 = 10 units MG = √[(6 + 3)² + (8 + 4)²] = √(81 + 144) = √225 = 15 units

Test: MA + AG = 5 + 10 = 15 = MG ✓ The sum of the two smaller distances equals the largest, so M, A and G lie on one straight line.

Question 7. Use your method (from Problem 6) to check if the points R (−5, −1), B (−2, −5) and C (4, −2) are on the same straight line. Now plot both sets of points and check your answers.

Solution: Apply the same distance test to the three points:

RB = √[(−2 + 5)² + (−5 + 1)²] = √(9 + 16) = √25 = 5 units BC = √[(4 + 2)² + (−2 + 5)²] = √(36 + 9) = √45 = 3√5 units RC = √[(4 + 5)² + (−2 + 1)²] = √(81 + 1) = √82 units

Test: RB + BC = 5 + 3√5 ≈ 11.7, while RC = √82 ≈ 9.06. No two distances add up to the third, so R, B and C do not lie on the same straight line.

Plotting confirms both results: the points of Problem 6 fall on a single line, while these three form a triangle.

Question 8. Using the origin as one vertex, plot the vertices of:

(i) A right-angled isosceles triangle.

Solution: One possible choice is O (0, 0), A (−4, 0), B (0, 4). Here OA lies along the x-axis and OB along the y-axis, so the angle at O is 90°. Both arms measure 4 units (OA = OB = 4), making two sides equal. Hence triangle OAB is right-angled as well as isosceles. (Many other answers are possible.)

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

Solution: One possible choice is O (0, 0), P (−3, −4), Q (3, −4). P has signs (−, −), so it lies in Quadrant III; Q has signs (+, −), so it lies in Quadrant IV. OP = √(9 + 16) = 5 units and OQ = √(9 + 16) = 5 units, so two sides are equal and triangle OPQ is isosceles.

Question 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. 

Image 7

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

Solution: For each row, check two things: M must lie on segment ST (SM + MT = ST) and must divide it equally (SM = MT).

Row 1: S (−3, 0), M (0, 0), T (3, 0) SM = √[(0 + 3)² + 0] = 3; MT = √[(3 − 0)² + 0] = 3; ST = √[(3 + 3)² + 0] = 6 SM = MT and SM + MT = 6 = ST → M is the midpoint of ST.

Row 2: S (2, 3), M (3, 4), T (4, 5) SM = √(1 + 1) = √2; MT = √(1 + 1) = √2; ST = √(4 + 4) = 2√2 SM = MT and their sum equals ST → M is the midpoint of ST.

Row 3: S (0, 0), M (0, 5), T (0, −10) SM = 5; MT = √(0 + 225) = 15; ST = 10 SM + MT = 20 ≠ 10, so M does not even lie between S and T → M is not the midpoint.

Row 4: S (−8, 7), M (0, −2), T (6, −3) SM = √(64 + 81) = √145; MT = √(36 + 1) = √37; ST = √(196 + 100) = √296 √145 + √37 ≠ √296 → M is not the midpoint.

Connection: In every case where M is the midpoint, each coordinate of M is the average of the corresponding coordinates of S and T. In general, if S = (x₁, y₁) and T = (x₂, y₂), then M = ((x₁ + x₂)/2, (y₁ + y₂)/2)

Question 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: By the midpoint relation, each coordinate of M is the average of the corresponding coordinates of A and B.

For x: (3 + x)/2 = −7 ⟹ 3 + x = −14 ⟹ x = −17 For y: (−4 + y)/2 = 1 ⟹ −4 + y = 2 ⟹ y = 6

Therefore, B = (−17, 6).

Question 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: Key idea: when P and Q trisect AB, the segment splits into three equal parts AP = PQ = QB. This makes P the midpoint of AQ, and Q the midpoint of PB. These two midpoint relations are enough to find both points.

Let P = (x₁, y₁) and Q = (x₂, y₂).

P is the midpoint of A and Q: x₁ = (4 + x₂)/2 …(1) y₁ = (7 + y₂)/2 …(2)

Q is the midpoint of P and B: x₂ = (x₁ + 16)/2 …(3) y₂ = (y₁ − 2)/2 …(4)

Substituting (1) into (3): x₂ = [(4 + x₂)/2 + 16]/2 = (4 + x₂ + 32)/4 = (x₂ + 36)/4 ⟹ 4x₂ = x₂ + 36 ⟹ 3x₂ = 36 ⟹ x₂ = 12 Then from (1): x₁ = (4 + 12)/2 = 8

Substituting (2) into (4): y₂ = [(7 + y₂)/2 − 2]/2 = (7 + y₂ − 4)/4 = (y₂ + 3)/4 ⟹ 4y₂ = y₂ + 3 ⟹ 3y₂ = 3 ⟹ y₂ = 1 Then from (2): y₁ = (7 + 1)/2 = 4

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

Question 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?

Solution: A circle is the set of points at a fixed distance from its centre, so it is enough to show that A, B and C are all the same distance from O.

OA = √[(1)² + (−8)²] = √(1 + 64) = √65 OB = √[(−4)² + (7)²] = √(16 + 49) = √65 OC = √[(−7)² + (−4)²] = √(49 + 16) = √65

All three distances equal √65, so A, B and C lie on the circle K centred at the origin with radius √65 units.

(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: Compare each point's distance from O with the radius √65 (≈ 8.06):

OD = √[(−5)² + (6)²] = √(25 + 36) = √61 ≈ 7.81 Since √61 < √65, point D lies inside circle K.

OE = √[(0)² + (9)²] = √81 = 9 Since 9 > √65, point E lies outside circle K.

Question 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: Let A = (x₁, y₁), B = (x₂, y₂), C = (x₃, y₃), with D (5, 1) the midpoint of BC, E (6, 5) the midpoint of CA, and F (0, 3) the midpoint of AB.

Applying the midpoint formula to each side: x₂ + x₃ = 10 …(1) y₂ + y₃ = 2 …(2) x₃ + x₁ = 12 …(3) y₃ + y₁ = 10 …(4) x₁ + x₂ = 0 …(5) y₁ + y₂ = 6 …(6)

For the x-coordinates, add (1), (3) and (5): 2(x₁ + x₂ + x₃) = 22 ⟹ x₁ + x₂ + x₃ = 11 Subtracting each pair-sum from this total: x₁ = 11 − 10 = 1, x₂ = 11 − 12 = −1, x₃ = 11 − 0 = 11

For the y-coordinates, add (2), (4) and (6): 2(y₁ + y₂ + y₃) = 18 ⟹ y₁ + y₂ + y₃ = 9 Subtracting each pair-sum: y₁ = 9 − 2 = 7, y₂ = 9 − 10 = −1, y₃ = 9 − 6 = 3

Therefore, A = (1, 7), B = (−1, −1) and C = (11, 3). (Verification: midpoint of BC = (5, 1) = D ✓, midpoint of CA = (6, 5) = E ✓, midpoint of AB = (0, 3) = F ✓)

Image 8

Question 14. 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.

Solution: With the scale 1 cm = 200 m, the 200 m gap between neighbouring streets becomes 1 cm on paper. Draw 10 vertical lines for the N-S streets and 10 horizontal lines for the E-W streets, keeping consecutive lines 1 cm apart. The result is a square grid in which the two main roads cross at the city centre, and every other street runs parallel to one of them.

Image 9

(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).

Solution: In this naming system, the first number always identifies the N-S street and the second number the E-W street. A particular N-S street crosses a particular E-W street at exactly one point.

(a) The name (4, 3) belongs to the single crossing of the 4th N-S street with the 3rd E-W street — so only one intersection has this name. 

(b) Likewise, (3, 4) names the single crossing of the 3rd N-S street with the 4th E-W street — again only one intersection. Note that (4, 3) and (3, 4) are two different crossings, which is why the order of the numbers matters.

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Question 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: (i) A circle stays fully on screen if its centre is at least one radius away from every edge. Circle A (centre (100, 150), radius 80): distance to left edge = 100, to bottom = 150, to right = 800 − 100 = 700, to top = 600 − 150 = 450. All exceed 80. Circle B (centre (250, 230), radius 100): distances are 250, 230, 550 and 370 — all exceed 100. So no part of either circle lies outside the screen.

(ii) Distance between the centres: AB = √[(250 − 100)² + (230 − 150)²] = √(22500 + 6400) = √28900 = 170 pixels Sum of the radii = 80 + 100 = 180, and difference = 20. Since 20 < 170 < 180, the distance between centres is less than the sum of radii but more than their difference — so the two circles intersect each other at two points.

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Question 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: Test for a square: all four sides must be equal AND both diagonals must be equal.

Sides: AB = √[(−1 − 2)² + (2 − 1)²] = √(9 + 1) = √10 units 

BC = √[(−2 + 1)² + (−1 − 2)²] = √(1 + 9) = √10 units 

CD = √[(1 + 2)² + (−2 + 1)²] = √(9 + 1) = √10 units 

DA = √[(2 − 1)² + (1 + 2)²] = √(1 + 9) = √10 units All four sides measure √10 units.

Diagonals: AC = √[(−2 − 2)² + (−1 − 1)²] = √(16 + 4) = √20 units 

BD = √[(1 + 1)² + (−2 − 2)²] = √(4 + 16) = √20 units Both diagonals measure √20 units.

Equal sides rule out an ordinary parallelogram, and equal diagonals rule out a rhombus that isn't a square - so ABCD is a square. Area = (side)² = (√10)² = 10 square units.

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Learn Coordinates Easily with Vedantu’s Class 9 Maths Chapter 1 NCERT Solutions

• Class 9 Maths Chapter 1 Orienting Yourself: The Use of Coordinates helps students understand how points are placed and located on a Cartesian plane. 

• This chapter builds a strong base for coordinate geometry by explaining the x-axis, y-axis, origin, ordered pairs, quadrants, plotting of points, reflection, distance, and midpoint concepts.

• Vedantu’s NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 1 are useful for students who want clear, step-by-step answers for textbook exercises. 

• These solutions also connect coordinates with real-life examples like room layouts, doors, furniture placement, maps, and screen positions, making the chapter easier to understand.

Access Exercise Wise NCERT Solutions for Chapter 1 Maths Class 9

CBSE Class 9 Maths Chapter 1 Orienting Yourself: The Use of Coordinates Other Study Materials

Chapter-Specific NCERT Solutions for Class 9 Maths

Given below are the chapter-wise NCERT Solutions for Class 9 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

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