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NCERT Class 10 Mathematics Exercise 7.1 Solutions: Coordinate Geometry

Step-by-step working for every question of Exercise 7.1 in NCERT's Class 10 Mathematics book, chapter “Coordinate Geometry”, written by Shishya. Each question is in our own short words; open the NCERT chapter for the exact wording and figures.

1. Ex 7.1 Q1

   Find the distance between (i) (2, 3) and (4, 1) (ii) (−5, 7) and (−1, 3) (iii) (a, b) and (−a, −b).

   1. (i) Use the distance formula d = √((x₂ − x₁)² + (y₂ − y₁)²) for points (2, 3) and (4, 1).
      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) Substitute the coordinates: d = √((4 − 2)² + (1 − 3)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.

   3. (i) Simplify the terms inside the square root: d = √(2² + (−2)²) = √(4 + 4).
      why?

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

   4. (i) Calculate the final value: d = √8 = 2√2.
   5. (ii) Apply the distance formula to points (−5, 7) and (−1, 3).
      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.

   6. (ii) Substitute the coordinates: d = √((−1 − (−5))² + (3 − 7)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. (ii) Simplify the differences: d = √((−1 + 5)² + (−4)²) = √(4² + (−4)²).
      why?

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

   8. (ii) Calculate the final value: d = √(16 + 16) = √32 = 4√2.
   9. (iii) Use the distance formula for points (a, b) and (−a, −b).
      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.

   10. (iii) Substitute the coordinates: d = √((−a − a)² + (−b − b)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.

   11. (iii) Simplify the terms: d = √((−2a)² + (−2b)²) = √(4a² + 4b²).
       why?

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

   12. (iii) Factor out the 4 and simplify: d = √(4(a² + b²)) = 2√(a² + b²).
       why?

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

   Answer: (i) 2√2; (ii) 4√2; (iii) 2√(a² + b²)

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2. Ex 7.1 Q2

   Find the distance between (0, 0) and (36, 15), and so the distance between towns A and B of Section 7.2 (placed at those points, in km).

   1. Identify the coordinates of the two points as (0, 0) and (36, 15).
   2. Use the distance formula d = √((x₂ − x₁)² + (y₂ − y₁)²).
      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. Substitute the values into the formula: d = √((36 − 0)² + (15 − 0)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. Calculate the squares: 36² = 1296 and 15² = 225.
   5. Add the results: 1296 + 225 = 1521.
   6. Take the square root of the sum: √1521 = 39.
   7. The distance between the points is 39 units, which corresponds to 39 km for the towns.

   Answer: 39 km

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3. Ex 7.1 Q3

   Are (1, 5), (2, 3) and (−2, −11) collinear?

   1. Use the section formula to find the ratio in which the line segment joining (1, 5) and (2, 3) is divided by the point (−2, −11).
      why?

      The point dividing the join of (x₁, y₁) and (x₂, y₂) in the ratio m : n is ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)) — a weighted average of the two end points.

   2. Let the ratio be k : 1. The x-coordinate of the dividing point is given by (k × 2 + 1 × 1) / (k + 1).
   3. Set this equal to the x-coordinate of the third point: (2k + 1) / (k + 1) = −2.
   4. Solving for k gives 2k + 1 = −2k − 2, which simplifies to 4k = −3, so k = −3/4.
      why?

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

   5. Now check the y-coordinate using this ratio. The y-coordinate should be (k × 3 + 1 × 5) / (k + 1).
   6. Substitute k = −3/4 into the y-coordinate formula: ((−3/4) × 3 + 5) / (−3/4 + 1).
      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. Calculate the numerator: −9/4 + 20/4 = 11/4.
   8. Calculate the denominator: −3/4 + 4/4 = 1/4.
   9. Divide the numerator by the denominator: (11/4) ÷ (1/4) = 11.
   10. The calculated y-coordinate is 11, but the given y-coordinate of the point is −11.
   11. Since the y-coordinates do not match, the point does not lie on the line segment in the same ratio.
   12. Therefore, the three points are not collinear.

   Answer: not collinear

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4. Ex 7.1 Q4

   Are (5, −2), (6, 4) and (7, −2) the vertices of an isosceles triangle?

   1. Let the vertices be A(5, −2), B(6, 4), and C(7, −2).
   2. Calculate the length of side AB using the distance formula: √[(6 − 5)² + (4 − (−2))²] = √[1² + 6²] = √(1 + 36) = √37.
      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 using the distance formula: √[(7 − 6)² + (−2 − 4)²] = √[1² + (−6)²] = √(1 + 36) = √37.
      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.

   4. Calculate the length of side AC using the distance formula: √[(7 − 5)² + (−2 − (−2))²] = √[2² + 0²] = √4 = 2.
      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.

   5. Compare the side lengths: AB = √37, BC = √37, and AC = 2.
   6. Since AB = BC, the triangle has two equal sides.
   7. Therefore, the triangle is isosceles.

   Answer: yes, √37

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5. Ex 7.1 Q5

   Four friends sit at A(3, 4), B(6, 7), C(9, 4) and D(6, 1) (Fig. 7.8). Champa says ABCD is a square; Chameli disagrees. Who is right?

   Uses a figure printed in the NCERT book — open the chapter above to see it.

   1. Calculate the length of side AB using the distance formula: √((6−3)² + (7−4)²) = √(9 + 9) = √18 = 3√2.
      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. Calculate the length of side BC: √((9−6)² + (4−7)²) = √(9 + 9) = √18 = 3√2.
   3. Calculate the length of side CD: √((6−9)² + (1−4)²) = √(9 + 9) = √18 = 3√2.
   4. Calculate the length of side DA: √((3−6)² + (4−1)²) = √(9 + 9) = √18 = 3√2.
   5. Since all four sides are equal in length, the quadrilateral ABCD is a rhombus.
   6. Calculate the length of diagonal AC: √((9−3)² + (4−4)²) = √(36 + 0) = √36 = 6.
   7. Calculate the length of diagonal BD: √((6−6)² + (1−7)²) = √(0 + 36) = √36 = 6.
   8. Since the diagonals AC and BD are equal in length, the rhombus ABCD is a square.
   9. Therefore, Champa is right and Chameli is wrong.

   Answer: Champa is right; side 3√2; diagonals 6

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6. Ex 7.1 Q6

   Name the quadrilateral, if any, formed by: (i) (−1, −2), (1, 0), (−1, 2), (−3, 0) (ii) (−3, 5), (3, 1), (0, 3), (−1, −4) (iii) (4, 5), (7, 6), (4, 3), (1, 2).

   1. (i) A(−1, −2), B(1, 0), C(−1, 2), D(−3, 0): AB² = 2² + 2² = 8, BC² = (−2)² + 2² = 8, CD² = (−2)² + (−2)² = 8, DA² = 2² + (−2)² = 8, so all sides are √8 = 2√2.
   2. (i) Diagonals: AC² = 0² + 4² = 16 and BD² = (−4)² + 0² = 16, so AC = BD = 4. Equal sides and equal diagonals: a square.
   3. (ii) A(−3, 5), B(3, 1), C(0, 3), D(−1, −4): AC² = 3² + (−2)² = 13, CB² = 3² + (−2)² = 13, AB² = 6² + (−4)² = 52.
   4. (ii) So AC = CB = √13 and AB = 2√13 = AC + CB: the points A, C, B lie on one line. Three of the four points are collinear, so no quadrilateral is formed.
   5. (iii) A(4, 5), B(7, 6), C(4, 3), D(1, 2): AB² = 3² + 1² = 10, BC² = (−3)² + (−3)² = 18, CD² = (−3)² + (−1)² = 10, DA² = 3² + 3² = 18.
   6. (iii) Opposite sides are equal (AB = CD = √10, BC = DA = 3√2), and the diagonals AC = 2 and BD = √52 are unequal, so it is a parallelogram (not a rectangle).

   Answer: (i) square; (ii) no quadrilateral (three points are collinear); (iii) parallelogram

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7. Ex 7.1 Q7

   Find the point on the x-axis equidistant from (2, −5) and (−2, 9).

   1. Let the required point on the x-axis be P(x, 0).
   2. The distance from P to (2, −5) is √((x − 2)² + (0 + 5)²).
   3. The distance from P to (−2, 9) is √((x + 2)² + (0 − 9)²).
   4. Since P is equidistant from both points, we equate the squares of the distances: (x − 2)² + 25 = (x + 2)² + 81.
   5. Expanding both sides gives x² − 4x + 4 + 25 = x² + 4x + 4 + 81.
      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. Simplifying the equation results in x² − 4x + 29 = x² + 4x + 85.
      why?

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

   7. Subtracting x² from both sides yields −4x + 29 = 4x + 85.
      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.

   8. Rearranging terms gives 8x = 29 − 85.
   9. Solving for x gives 8x = −56, so x = −7.
   10. The coordinates of the point are (−7, 0).

   Answer: (-7, 0)

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8. Ex 7.1 Q8

   Find y if the distance between P(2, −3) and Q(10, y) is 10 units.

   1. Use the distance formula: d = √[(x₂ − x₁)² + (y₂ − y₁)²].
      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. Substitute the given points P(2, −3) and Q(10, y) and distance 10: 10 = √[(10 − 2)² + (y − (−3))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.

   3. Simplify the expression inside the square root: 10 = √[8² + (y + 3)²].
      why?

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

   4. Square both sides to remove the square root: 100 = 64 + (y + 3)².
      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 64 from both sides: (y + 3)² = 36.
      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. Take the square root of both sides: y + 3 = ±6.
      why?

      If two non-negative quantities are equal, their square roots are equal. A square has two roots, + and −; a length or a count cannot be negative, so the negative root is dropped there.

   7. Solve for y when y + 3 = 6: y = 3.
   8. Solve for y when y + 3 = −6: y = −9.

   Answer: y = 3 or y = -9

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9. Ex 7.1 Q9

   Q(0, 1) is equidistant from P(5, −3) and R(x, 6). Find x, and the distances QR and PR.

   1. QP² = (5 − 0)² + (−3 − 1)² = 25 + 16 = 41.
   2. Q is equidistant from P and R, so QR² = 41: (x − 0)² + (6 − 1)² = 41, x² + 25 = 41, x² = 16, x = 4 or x = −4.
   3. QR = QP = √41.
   4. For x = 4, R = (4, 6): PR² = (4 − 5)² + (6 + 3)² = 1 + 81 = 82, so PR = √82.
   5. For x = −4, R = (−4, 6): PR² = (−4 − 5)² + (6 + 3)² = 81 + 81 = 162, so PR = √162 = 9√2.

   Answer: x = ±4; QR = √41; PR = √82 or 9√2

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10. Ex 7.1 Q10

    Find a relation between x and y so that (x, y) is equidistant from (3, 6) and (−3, 4).

    1. Use the distance formula to set the distance from (x, y) to (3, 6) equal to the distance from (x, y) to (−3, 4).
       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. Write the equation: (x − 3)² + (y − 6)² = (x + 3)² + (y − 4)².
    3. Expand both sides: x² − 6x + 9 + y² − 12y + 36 = x² + 6x + 9 + y² − 8y + 16.
       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.

    4. Simplify by cancelling x² and y² from both sides: −6x − 12y + 45 = 6x − 8y + 25.
       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. Move all terms to one side: −12x − 4y + 20 = 0.
    6. Divide the entire equation by −4 to get the simplest form: 3x + y − 5 = 0.

    Answer: 3x + y - 5 = 0

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