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NCERT Class 9 Mathematics Orienting Yourself: The Use of Coordinates: Exercise Solutions

Step-by-step working for every question of End-of-chapter exercises in NCERT's Class 9 Mathematics book, chapter “Orienting Yourself: The Use of Coordinates”, written by Shishya. Each question is in our own short words; open the NCERT chapter for the exact wording and figures.

1. Exercises Q1

   Coordinates of the point where the two axes meet.

   1. The two axes are the x-axis and the y-axis.
   2. They intersect at the origin of the coordinate plane.
   3. The x-coordinate at the origin is 0.
   4. The y-coordinate at the origin is 0.

   Answer: (0, 0)

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2. Exercises Q2

   W has x-coordinate −5. What can you say about the coordinates of a point H on the line through W parallel to the y-axis, and which quadrants can H be in?

   1. A line through W parallel to the y-axis is a vertical line, and every point on a vertical line has the same x-coordinate.
   2. W has x-coordinate −5, so H = (−5, y) for some number y.
   3. If y > 0, H is to the left of the y-axis and above the x-axis: Quadrant II.
   4. If y < 0, H is to the left and below: Quadrant III.
   5. If y = 0, H = (−5, 0) lies on the x-axis itself, which is in no quadrant.

   Answer: H = (−5, y): it lies in Quadrant II (y > 0) or Quadrant III (y < 0), or on the x-axis at (−5, 0).

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3. Exercises Q3

   R (3, 0), A (0, −2), M (−5, −2), P (−5, 2) joined in order as RAMP: (i) two perpendicular sides, (ii) a side parallel to an axis, (iii) two points that are mirror images in an axis (which axis?). Then plot and verify.

   1. (i) A (0, −2) and M (−5, −2) have the same y-coordinate −2, so AM is a horizontal segment.
   2. (i) M (−5, −2) and P (−5, 2) have the same x-coordinate −5, so MP is a vertical segment.
   3. (i) A horizontal and a vertical segment meet at a right angle, so AM ⊥ MP.
   4. (ii) The side AM has a constant y-coordinate of −2, so it is parallel to the x-axis.
   5. (ii) The side MP has a constant x-coordinate of −5, so it is parallel to the y-axis.
   6. (iii) Compare the coordinates of M(−5, −2) and P(−5, 2); their x-coordinates are equal and y-coordinates are opposites.
   7. (iii) Therefore, M and P are mirror images of each other in the x-axis.
   8. Plot the points R(3, 0), A(0, −2), M(−5, −2), and P(−5, 2) on the coordinate plane and join them in order to form the quadrilateral RAMP.
   9. Verify the geometric properties by observing the plotted figure: AM is a horizontal segment, MP is a vertical segment, and they meet at a right angle at M.

   Answer: (i) AM ⊥ MP; (ii) AM ∥ x-axis, MP ∥ y-axis; (iii) M and P, x-axis

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4. Exercises Q4

   Plot Z (5, −6); make a right-angled triangle IZN and find its three side lengths (answers can differ).

   1. Z (5, −6) is 5 units right of the y-axis and 6 units below the x-axis (Quadrant IV). One easy choice: I = (0, 0) (the origin) and N = (0, −6) on the y-axis, level with Z.
   2. IN is vertical (same x-coordinate 0): IN = 0 − (−6) = 6.
   3. NZ is horizontal (same y-coordinate −6): NZ = 5 − 0 = 5.
   4. IN is vertical and NZ is horizontal, so the angle at N is a right angle.
   5. By the Baudhāyana–Pythagoras theorem, IZ² = IN² + NZ² = 6² + 5² = 36 + 25 = 61, so IZ = √61 ≈ 7.81.
      why?

      In a right-angled triangle, (hypotenuse)² = (one side)² + (other side)² (Baudhāyana–Pythagoras theorem). The hypotenuse is the side opposite the right angle and is the longest side.

   6. Other choices of I and N give other right triangles; the method is the same.

   Answer: With I (0, 0) and N (0, −6): IN = 6, NZ = 5, IZ = √61 (≈ 7.81); other correct triangles are possible.

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5. Exercises Q5

   What would coordinates be like without negative numbers? Could every point of the plane be located?

   1. If we only use positive numbers, we can only describe points in the first quadrant where both x and y are positive.
   2. Points in the second quadrant have negative x-values, so they cannot be located without negative numbers.
   3. Points in the third quadrant have negative x and y-values, making them impossible to locate with only positive numbers.
   4. Points in the fourth quadrant have negative y-values, so they also cannot be located without negative numbers.
   5. Therefore, without negative numbers, we can only locate one-fourth of the plane.
   6. We cannot locate every point of the plane if negative numbers are not allowed.

   Answer: No, every point of the plane cannot be located without negative numbers; only points in the first quadrant can be located.

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6. Exercises Q6

   Are M (−3, −4), A (0, 0) and G (6, 8) on one straight line? Give a method that does not need plotting.

   1. Use the distance formula to find the length of segment MA between (−3, −4) and (0, 0).
      why?

      The distance between (x₁, y₁) and (x₂, y₂) is √[(x₂ − x₁)² + (y₂ − y₁)²]: the two differences are the legs of a right triangle and the distance is its hypotenuse.

   2. MA = √[(0 − (−3))² + (0 − (−4))²] = √(3² + 4²) = √(9 + 16) = √25 = 5.
   3. Find the length of segment AG between (0, 0) and (6, 8).
   4. AG = √[(6 − 0)² + (8 − 0)²] = √(6² + 8²) = √(36 + 64) = √100 = 10.
   5. Find the length of segment MG between (−3, −4) and (6, 8).
   6. MG = √[(6 − (−3))² + (8 − (−4))²] = √(9² + 12²) = √(81 + 144) = √225 = 15.
   7. Check if the sum of the two shorter distances equals the longest distance.
   8. MA + AG = 5 + 10 = 15, which is equal to MG.
   9. Since MA + AG = MG, the points M, A, and G lie on one straight line.

   Answer: MA = 5, AG = 10, MG = 15, yes

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7. Exercises Q7

   Use the same method: are R (−5, −1), B (−2, −5) and C (4, −12) on one straight line?

   1. Calculate the distance between points R(−5, −1) and B(−2, −5) using the distance formula.
      why?

      The distance between (x₁, y₁) and (x₂, y₂) is √[(x₂ − x₁)² + (y₂ − y₁)²]: the two differences are the legs of a right triangle and the distance is its hypotenuse.

   2. RB = √[(−2 − (−5))² + (−5 − (−1))²] = √[3² + (−4)²] = √[9 + 16] = √25 = 5.
   3. Calculate the distance between points B(−2, −5) and C(4, −12).
   4. BC = √[(4 − (−2))² + (−12 − (−5))²] = √[6² + (−7)²] = √[36 + 49] = √85.
   5. Calculate the distance between points R(−5, −1) and C(4, −12).
   6. RC = √[(4 − (−5))² + (−12 − (−1))²] = √[9² + (−11)²] = √[81 + 121] = √202.
   7. Check if the sum of the shorter distances equals the longest distance to verify collinearity.
   8. RB + BC = 5 + √85 ≈ 5 + 9.22 = 14.22.
   9. RC = √202 ≈ 14.21.
   10. To be sure without rounding, square both: (5 + √85)² = 25 + 85 + 10√85 = 110 + 10√85 ≈ 202.2, which is not 202 = (√202)². So RB + BC ≠ RC, and the points do not lie on the same straight line.

   Answer: RB = 5, BC = √85, RC = √202, not on one straight line

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8. Exercises Q8

   With the origin as one vertex, plot (i) a right-angled isosceles triangle, (ii) an isosceles triangle with one vertex in Quadrant III and one in Quadrant IV.

   1. (i) Choose vertices O(0, 0), A(4, 0), and B(0, 4) to form a right-angled isosceles triangle with the right angle at the origin.
   2. (i) Plot the points O(0, 0), A(4, 0), and B(0, 4) on the coordinate plane.
   3. (i) Join OA, OB, and AB to complete the triangle.
   4. (ii) Choose vertices O(0, 0), C(−3, −4), and D(3, −4) to form an isosceles triangle with one vertex in QIII and one in QIV.
   5. (ii) Plot the points O(0, 0), C(−3, −4), and D(3, −4) on the coordinate plane.
   6. (ii) Join OC, OD, and CD to complete the triangle.
   7. Verify that OC = OD = 5 units, confirming the triangle is isosceles.

   Answer: (i) Triangle with vertices (0, 0), (4, 0), (0, 4); (ii) Triangle with vertices (0, 0), (−3, −4), (3, −4)

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9. Exercises Q9

   Is M the midpoint of ST? (a) S (−3, 0), M (0, 0), T (3, 0); (b) S (2, 3), M (3, 4), T (4, 5); (c) S (0, 0), M (0, 5), T (0, −10); (d) S (−8, 7), M (0, −2), T (6, −3). What links the coordinates of M, S and T when M is the midpoint?

   1. (a) Calculate the midpoint of ST: ((−3 + 3) ÷ 2, (0 + 0) ÷ 2) = (0, 0).
      why?

      The midpoint of the join of (x₁, y₁) and (x₂, y₂) is ((x₁ + x₂)/2, (y₁ + y₂)/2): each coordinate is the average of the two ends.

   2. (a) The calculated midpoint (0, 0) matches the given point M, so M is the midpoint.
   3. (b) Calculate the midpoint of ST: ((2 + 4) ÷ 2, (3 + 5) ÷ 2) = (3, 4).
      why?

      The midpoint of the join of (x₁, y₁) and (x₂, y₂) is ((x₁ + x₂)/2, (y₁ + y₂)/2): each coordinate is the average of the two ends.

   4. (b) The calculated midpoint (3, 4) matches the given point M, so M is the midpoint.
   5. (c) Calculate the midpoint of ST: ((0 + 0) ÷ 2, (0 + (−10)) ÷ 2) = (0, −5).
      why?

      The midpoint of the join of (x₁, y₁) and (x₂, y₂) is ((x₁ + x₂)/2, (y₁ + y₂)/2): each coordinate is the average of the two ends.

   6. (c) The calculated midpoint (0, −5) does not match the given point M (0, 5), so M is not the midpoint.
   7. (d) Calculate the midpoint of ST: ((−8 + 6) ÷ 2, (7 + (−3)) ÷ 2) = (−1, 2).
      why?

      The midpoint of the join of (x₁, y₁) and (x₂, y₂) is ((x₁ + x₂)/2, (y₁ + y₂)/2): each coordinate is the average of the two ends.

   8. (d) The calculated midpoint (−1, 2) does not match the given point M (0, −2), so M is not the midpoint.
   9. When M is the midpoint, its coordinates are the averages of the corresponding coordinates of S and T.
      why?

      The midpoint of the join of (x₁, y₁) and (x₂, y₂) is ((x₁ + x₂)/2, (y₁ + y₂)/2): each coordinate is the average of the two ends.

   Answer: (a) yes; (b) yes; (c) no, (0, -5); (d) no, (-1, 2)

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10. Exercises Q10

    M (−7, 1) is the midpoint of A (3, −4) and B (x, y). Find B.

    1. The midpoint formula states that the coordinates of the midpoint M are the averages of the coordinates of the endpoints A and B.
       why?

       The midpoint of the join of (x₁, y₁) and (x₂, y₂) is ((x₁ + x₂)/2, (y₁ + y₂)/2): each coordinate is the average of the two ends.

    2. For the x-coordinate: x-coordinate of M = (x-coordinate of A + x) / 2.
    3. Substitute the given values into the x-coordinate formula: −7 = (3 + x) / 2.
       why?

       Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

    4. Multiply both sides by 2 to get −14 = 3 + x.
       why?

       Doing the same thing to both sides of an equation (adding, subtracting, multiplying or dividing by the same non-zero number) keeps the two sides equal, so the equation stays true while the unknown is left on its own.

    5. Subtract 3 from both sides to find x = −17.
       why?

       Doing the same thing to both sides of an equation (adding, subtracting, multiplying or dividing by the same non-zero number) keeps the two sides equal, so the equation stays true while the unknown is left on its own.

    6. For the y-coordinate: y-coordinate of M = (y-coordinate of A + y) / 2.
    7. Substitute the given values into the y-coordinate formula: 1 = (−4 + y) / 2.
       why?

       Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

    8. Multiply both sides by 2 to get 2 = −4 + y.
       why?

       Doing the same thing to both sides of an equation (adding, subtracting, multiplying or dividing by the same non-zero number) keeps the two sides equal, so the equation stays true while the unknown is left on its own.

    9. Add 4 to both sides to find y = 6.
       why?

       Doing the same thing to both sides of an equation (adding, subtracting, multiplying or dividing by the same non-zero number) keeps the two sides equal, so the equation stays true while the unknown is left on its own.

    10. The coordinates of point B are (−17, 6).

    Answer: B = (−17, 6)

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11. Exercises Q11

    P and Q trisect AB (P nearer A). Using midpoints, find P and Q for A (4, 7) and B (16, −2).

    1. Since P and Q trisect AB, the segment AP is equal to PQ and PQ is equal to QB.
    2. This means P is the midpoint of the segment AQ, and Q is the midpoint of the segment PB.
    3. Let the coordinates of P be (x₁, y₁) and Q be (x₂, y₂).
    4. Using the midpoint formula for Q (midpoint of PB): x₂ = (x₁ + 16)/2 and y₂ = (y₁ + (-2))/2.
       why?

       The midpoint of the join of (x₁, y₁) and (x₂, y₂) is ((x₁ + x₂)/2, (y₁ + y₂)/2): each coordinate is the average of the two ends.

    5. Using the midpoint formula for P (midpoint of AQ): x₁ = (4 + x₂)/2 and y₁ = (7 + y₂)/2.
       why?

       The midpoint of the join of (x₁, y₁) and (x₂, y₂) is ((x₁ + x₂)/2, (y₁ + y₂)/2): each coordinate is the average of the two ends.

    6. Substitute the expression for x₂ into the equation for x₁: x₁ = (4 + (x₁ + 16)/2)/2.
       why?

       Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

    7. Simplify to find x₁: 2x₁ = 4 + (x₁ + 16)/2, which leads to 4x₁ = 8 + x₁ + 16, so 3x₁ = 24 and x₁ = 8.
       why?

       Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.

    8. Substitute the expression for y₂ into the equation for y₁: y₁ = (7 + (y₁ - 2)/2)/2.
       why?

       Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

    9. Simplify to find y₁: 2y₁ = 7 + (y₁ - 2)/2, which leads to 4y₁ = 14 + y₁ - 2, so 3y₁ = 12 and y₁ = 4.
       why?

       Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.

    10. Thus, the coordinates of P are (8, 4).
    11. Now find x₂ using x₂ = (x₁ + 16)/2: x₂ = (8 + 16)/2 = 24/2 = 12.
    12. Now find y₂ using y₂ = (y₁ - 2)/2: y₂ = (4 - 2)/2 = 2/2 = 1.
    13. Thus, the coordinates of Q are (12, 1).

    Answer: P = (8, 4), Q = (12, 1)

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12. Exercises Q12

    (i) Show A (1, −8), B (−4, 7), C (−7, −4) lie on a circle K with centre O (0, 0); find its radius. (ii) Are D (−5, 6) and E (0, 9) inside, on or outside K?

    1. (i) Calculate the distance from the centre O(0, 0) to point A(1, −8) using the distance formula: √((1−0)² + (−8−0)²) = √(1 + 64) = √65.
       why?

       The distance between (x₁, y₁) and (x₂, y₂) is √[(x₂ − x₁)² + (y₂ − y₁)²]: the two differences are the legs of a right triangle and the distance is its hypotenuse.

    2. (i) Calculate the distance from the centre O(0, 0) to point B(−4, 7): √((−4−0)² + (7−0)²) = √(16 + 49) = √65.
    3. (i) Calculate the distance from the centre O(0, 0) to point C(−7, −4): √((−7−0)² + (−4−0)²) = √(49 + 16) = √65.
    4. (i) Since the distances OA, OB, and OC are all equal to √65, points A, B, and C lie on a circle K with centre O and radius √65.
    5. (ii) Calculate the distance from the centre O(0, 0) to point D(−5, 6): √((−5−0)² + (6−0)²) = √(25 + 36) = √61.
    6. (ii) Compare the distance OD with the radius: since √61 < √65, point D is inside the circle K.
    7. (ii) Calculate the distance from the centre O(0, 0) to point E(0, 9): √((0−0)² + (9−0)²) = √(0 + 81) = √81 = 9.
    8. (ii) Compare the distance OE with the radius: since 9 > √65 (because 81 > 65), point E is outside the circle K.

    Answer: (i) A, B, C lie on circle K with radius √65; (ii) D inside, E outside

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13. Exercises Q13

    D (5, 1), E (6, 5), F (0, 3) are the midpoints of the sides of triangle ABC. Find A, B and C.

    1. Take D, E, F as the midpoints of BC, CA and AB. A midpoint's coordinates are the averages of the end points, so B + C = 2D, C + A = 2E and A + B = 2F (adding x-coordinates and y-coordinates separately).
       why?

       The midpoint of the join of (x₁, y₁) and (x₂, y₂) is ((x₁ + x₂)/2, (y₁ + y₂)/2): each coordinate is the average of the two ends.

    2. Adding all three: 2(A + B + C) = 2(D + E + F), so A + B + C = D + E + F = (5 + 6 + 0, 1 + 5 + 3) = (11, 9).
    3. A = (A + B + C) − (B + C) = (11, 9) − (10, 2) = (1, 7), since 2D = (10, 2).
    4. B = (11, 9) − (C + A) = (11, 9) − (12, 10) = (−1, −1), since 2E = (12, 10).
    5. C = (11, 9) − (A + B) = (11, 9) − (0, 6) = (11, 3), since 2F = (0, 6).
    6. Check: midpoint of BC = ((−1 + 11)/2, (−1 + 3)/2) = (5, 1) = D; of CA = ((11 + 1)/2, (3 + 7)/2) = (6, 5) = E; of AB = ((1 − 1)/2, (7 − 1)/2) = (0, 3) = F.
       why?

       The midpoint of the join of (x₁, y₁) and (x₂, y₂) is ((x₁ + x₂)/2, (y₁ + y₂)/2): each coordinate is the average of the two ends.

    Answer: A = (1, 7), B = (−1, −1), C = (11, 3) (taking D, E, F as the midpoints of BC, CA, AB).

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14. Exercises Q14

    A city: main N–S and E–W roads cross at the centre; other streets parallel to them, 200 m apart, 10 in each direction. (i) Draw a model with 1 cm = 200 m. (ii) An intersection (a, b) = a-th N–S street with b-th E–W street: how many intersections can be called (4, 3), and how many (3, 4)?

    1. (i) Draw one vertical line (the main N–S road) and one horizontal line (the main E–W road) crossing at the centre.
    2. (i) Draw the other streets as lines parallel to these, 1 cm apart (1 cm = 200 m), 10 in each direction.
    3. (ii) Number the N–S streets 1st, 2nd, 3rd, … from one fixed side (say from the west) and the E–W streets 1st, 2nd, … from one fixed side (say from the south).
    4. (ii) The 4th N–S street is one single line and the 3rd E–W street is one single line; two crossing lines meet in exactly one point, so exactly one intersection is called (4, 3).
    5. (ii) In the same way exactly one intersection is called (3, 4), and it is a different crossing from (4, 3): the order of the pair matters, just as (x, y) and (y, x) are different points when x ≠ y.

    Answer: (i) a grid of lines 1 cm apart; (ii) exactly one intersection is (4, 3) and exactly one, a different one, is (3, 4).

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15. Exercises Q15

    Screen 800 × 600 pixels, origin at the bottom-left. Circle 1: centre A (100, 150), radius 80; circle 2: centre B (250, 230), radius 100. (i) Does any part of either circle go off the screen? (ii) Do the circles intersect?

    1. (i) For circle 1, the leftmost x is 100 − 80 = 20 and the rightmost x is 100 + 80 = 180, both within 0 to 800.
    2. (i) The bottom y is 150 − 80 = 70 and the top y is 150 + 80 = 230, both within 0 to 600.
    3. (i) For circle 2, the leftmost x is 250 − 100 = 150 and the rightmost x is 250 + 100 = 350, both within 0 to 800.
    4. (i) The bottom y is 230 − 100 = 130 and the top y is 230 + 100 = 330, both within 0 to 600.
    5. (i) Since all extreme points of both circles lie within the screen boundaries, no part is off the screen.
    6. (ii) The distance between centres A and B is √((250 − 100)² + (230 − 150)²) = √(150² + 80²) = √(22500 + 6400) = √28900 = 170.
    7. (ii) The sum of the radii is 80 + 100 = 180.
    8. (ii) The difference of the radii is 100 − 80 = 20.
    9. (ii) Since 20 < 170 < 180, the distance between centres is between the difference and the sum of the radii.
    10. (ii) Therefore, the circles intersect.

    Answer: (i) no part is off the screen; (ii) AB = 170, 180, they intersect

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16. Exercises Q16

    Plot A (2, 1), B (−1, 2), C (−2, −1), D (1, −2). Is ABCD a square? Why? Find its area.

    1. Plot the points A(2, 1), B(−1, 2), C(−2, −1), and D(1, −2) on the coordinate plane and join them in order to form quadrilateral ABCD.
    2. Calculate the length of side AB using the distance formula: √[(2 − (−1))² + (1 − 2)²] = √[3² + (−1)²] = √[9 + 1] = √10.
       why?

       The distance between (x₁, y₁) and (x₂, y₂) is √[(x₂ − x₁)² + (y₂ − y₁)²]: the two differences are the legs of a right triangle and the distance is its hypotenuse.

    3. Calculate the length of side BC: √[(−1 − (−2))² + (2 − (−1))²] = √[1² + 3²] = √[1 + 9] = √10.
    4. Calculate the length of side CD: √[(−2 − 1)² + (−1 − (−2))²] = √[(−3)² + 1²] = √[9 + 1] = √10.
    5. Calculate the length of side DA: √[(1 − 2)² + (−2 − 1)²] = √[(−1)² + (−3)²] = √[1 + 9] = √10.
    6. Since all four sides are equal (AB = BC = CD = DA = √10), ABCD is a rhombus.
    7. Calculate the length of diagonal AC: √[(2 − (−2))² + (1 − (−1))²] = √[4² + 2²] = √[16 + 4] = √20.
    8. Calculate the length of diagonal BD: √[(−1 − 1)² + (2 − (−2))²] = √[(−2)² + 4²] = √[4 + 16] = √20.
    9. Since the diagonals are equal (AC = BD = √20) and the sides are equal, ABCD is a square.
    10. The area of a square is the square of its side length: (√10)² = 10.

    Answer: ABCD is a square; side √10; diagonal √20; area 10

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