Chapter 7 : Coordinate Geometry Class 10 Maths NCERT Solutions

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Contents

Class 10 Maths NCERT Solutions - Chapter wise

1

Chapter 1: Real Numbers

2

Chapter 2: Polynomials

3

Chapter 3 : Pair of Linear Equations in Two Variables

4

Chapter 4 : Quadratic Equations

5

Chapter 5 : Arithmetic Progressions

6

Chapter 6 : Triangles

7

Chapter 7 : Coordinate Geometry

8

Chapter 8: Introduction to Trigonometry

9

Chapter 9: Some Applications of Trigonometry

10

Chapter 10 : Circles

11

Chapter 11 : Areas Related to Circles

12

Chapter 12: Surface Areas and Volumes

13

Chapter 13: Statistics

14

Chapter 14 : Probability

Chapter Notes

Chapter 7: Coordinate Geometry — Exercise 7.1

Welcome to the world of mapping! Coordinate Geometry is like a GPS for your notebook—it helps you find the exact distance between any two points on a flat surface using simple logic and a bit of algebra.

Q1. Find the distance between the following pairs of points:

(i) (2, 3), (4, 1) (ii) (-5, 7), (-1, 3) (iii) (a, b), (-a, -b)

  • The Logic: To find the distance between two points, we use the Distance Formula. Think of it as the Pythagorean theorem applied to a graph. We find the difference in ‘x’ values and ‘y’ values, square them (to get rid of negatives), add them, and then take the square root.

  • The Steps (i): Points are (2, 3) and (4, 1).

    • x1 = 2, y1 = 3; x2 = 4, y2 = 1

    • Distance = root[(x2 – x1)^2 + (y2 – y1)^2]

    • Distance = root[(4 – 2)^2 + (1 – 3)^2]

    • Distance = root[(2)^2 + (-2)^2] = root(4 + 4) = root(8)

  • The Result (i): Distance = 2 root 2 units

  • The Steps (ii): Points are (-5, 7) and (-1, 3).

    • Distance = root[(-1 – (-5))^2 + (3 – 7)^2]

    • Distance = root[(-1 + 5)^2 + (-4)^2]

    • Distance = root[(4)^2 + (-4)^2] = root(16 + 16) = root(32)

  • The Result (ii): Distance = 4 root 2 units

  • The Steps (iii): Points are (a, b) and (-a, -b).

    • Distance = root[(-a – a)^2 + (-b – b)^2]

    • Distance = root[(-2a)^2 + (-2b)^2]

    • Distance = root(4a^2 + 4b^2) = root[4(a^2 + b^2)]

  • The Result (iii): Distance = 2 root(a^2 + b^2) units

Q2. Find the distance between the points (0, 0) and (36, 15).

  • The Logic: Whenever one point is the Origin (0,0), the formula becomes even simpler: root(x^2 + y^2). You are essentially finding the length of the diagonal from the center.

  • The Steps:

    • Distance = root[(36 – 0)^2 + (15 – 0)^2]

    • Distance = root(36^2 + 15^2)

    • Distance = root(1296 + 225) = root(1521)

  • The Result: Distance = 39 units

Q3. Determine if the points (1, 5), (2, 3) and (-2, -11) are collinear.

  • The Logic: Collinear means the points lie on the same straight line. If they are on a line, the distance between Point A to B plus B to C must exactly equal the distance from A to C. If the sum is greater, they form a triangle instead!

  • The Steps:

    1. Let A = (1, 5), B = (2, 3), C = (-2, -11).

    2. AB = root[(2-1)^2 + (3-5)^2] = root(1 + 4) = root(5)

    3. BC = root[(-2-2)^2 + (-11-3)^2] = root(16 + 196) = root(212)

    4. AC = root[(-2-1)^2 + (-11-5)^2] = root(9 + 256) = root(265)

    5. Check: Does AB + BC = AC? root(5) + root(212) is not equal to root(265).

  • The Result: The points are not collinear.

Q4. Check whether (5, -2), (6, 4) and (7, -2) are the vertices of an isosceles triangle.

  • The Logic: An Isosceles triangle has at least two equal sides. We calculate all three distances (AB, BC, and AC) and see if any two are the same.

  • The Steps:

    1. AB = root[(6-5)^2 + (4 – (-2))^2] = root(1 + 36) = root(37)

    2. BC = root[(7-6)^2 + (-2-4)^2] = root(1 + 36) = root(37)

    3. AC = root[(7-5)^2 + (-2 – (-2))^2] = root(4 + 0) = 2

  • The Result: Yes, they form an isosceles triangle because AB = BC.

Chapter 7: Coordinate Geometry — Exercise 7.1 (Questions 5 to 10)

Ready for the next set? These questions move away from basic calculations and start looking at how coordinates define shapes like squares and circles.

Q5. In a classroom, 4 friends are seated at the points A, B, C and D… Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees. Using distance formula, find which of them is correct.

  • The Logic: For a shape to be a Square, two things must be true:

    1. All four sides must be equal ($AB = BC = CD = DA$).

    2. Both diagonals must be equal ($AC = BD$).

  • The Steps:

    • Coordinates: A(3, 4), B(6, 7), C(9, 4), D(6, 1).

    • Sides:

      • AB = root[(6-3)^2 + (7-4)^2] = root(9+9) = root(18)

      • BC = root[(9-6)^2 + (4-7)^2] = root(9+9) = root(18)

      • CD = root[(6-9)^2 + (1-4)^2] = root(9+9) = root(18)

      • DA = root[(3-6)^2 + (4-1)^2] = root(9+9) = root(18)

    • Diagonals:

      • AC = root[(9-3)^2 + (4-4)^2] = 6

      • BD = root[(6-6)^2 + (1-7)^2] = 6

  • The Result: Since all sides and both diagonals are equal, ABCD is a square. Champa is correct!

Q6. Name the type of quadrilateral formed, if any, by the following points…

(i) (-1, -2), (1, 0), (-1, 2), (-3, 0)

  • The Logic: We check the lengths of all sides and diagonals.

    • If 4 sides equal + 2 diagonals equal = Square.

    • If 4 sides equal + diagonals NOT equal = Rhombus.

    • If opposite sides equal + 2 diagonals equal = Rectangle.

  • The Steps:

    • Calculating distances, we find all sides = root(8) and both diagonals = 4.

  • The Result: It is a Square.

Q7. Find the point on the x-axis which is equidistant from (2, -5) and (-2, 9).

  • The Logic: A point on the x-axis always has a ‘y’ coordinate of 0. So, let the point be (x, 0). “Equidistant” means the distance from (x, 0) to Point A is the same as the distance to Point B.

  • The Steps:

    1. Distance to (2, -5) = Distance to (-2, 9)

    2. (x – 2)^2 + (0 – (-5))^2 = (x – (-2))^2 + (0 – 9)^2 (Squaring both sides to remove roots)

    3. x^2 – 4x + 4 + 25 = x^2 + 4x + 4 + 81

    4. -4x + 29 = 4x + 85

    5. -8x = 56 => x = -7

  • The Result: The point is (-7, 0).

Q8. Find the values of y for which the distance between the points P(2, -3) and Q(10, y) is 10 units.

  • The Logic: We know the answer (Distance = 10) and need to find the missing coordinate. We set up the distance formula and solve the quadratic equation that follows.

  • The Steps:

    • root[(10 – 2)^2 + (y – (-3))^2] = 10

    • 8^2 + (y + 3)^2 = 100 (Squaring both sides)

    • 64 + (y + 3)^2 = 100

    • (y + 3)^2 = 36

    • y + 3 = 6 or y + 3 = -6

  • The Result: y = 3 or y = -9.

Q9. If Q(0, 1) is equidistant from P(5, -3) and R(x, 6), find the values of x. Also find the distances QR and PR.

  • The Logic: “Equidistant” means QP = QR. We solve for x first, then calculate the specific distances requested.

  • The Steps:

    1. QP^2 = QR^2

    2. (5-0)^2 + (-3-1)^2 = (x-0)^2 + (6-1)^2

    3. 25 + 16 = x^2 + 25

    4. x^2 = 16 => x = 4 or -4.

    5. QR = root[(4-0)^2 + (6-1)^2] = root(41).

  • The Result: x = 4, -4; QR = root(41); PR = root(82) or 9 root 2.

Q10. Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (-3, 4).

  • The Logic: Here, we won’t get a single number for x or y. Instead, we get an Equation (a relation) that represents all points that are exactly between these two locations.

  • The Steps:

    1. (x – 3)^2 + (y – 6)^2 = (x – (-3))^2 + (y – 4)^2

    2. x^2 – 6x + 9 + y^2 – 12y + 36 = x^2 + 6x + 9 + y^2 – 8y + 16

    3. -6x – 12y + 45 = 6x – 8y + 16

    4. 12x + 4y = 29 (or simplified: 3x + y – 5 = 0)

  • The Result: The relation is 3x + y – 5 = 0.

Chapter 7: Coordinate Geometry — Exercise 7.2 (Questions 1 to 5)

Mastering the Section Formula is like learning how to split a piece of thread exactly where you want it. Whether you are finding the center of a bridge or the exact meeting point between two cities, these formulas are your best tools!

Q1. Find the coordinates of the point which divides the join of (-1, 7) and (4, -3) in the ratio 2 : 3.

  • The Logic: We use the Section Formula. Think of it as a balance scale—the ratio (2:3) tells us the point is closer to one side than the other. We multiply the ratio of one side with the coordinates of the opposite side.

  • The Steps:

    • Points: (x1, y1) = (-1, 7) and (x2, y2) = (4, -3). Ratio: m1 = 2, m2 = 3.

    • x = (m1x2 + m2x1) / (m1 + m2)

    • x = (24 + 3-1) / (2 + 3) = (8 – 3) / 5 = 5 / 5 = 1

    • y = (m1y2 + m2y1) / (m1 + m2)

    • y = (2*-3 + 3*7) / (2 + 3) = (-6 + 21) / 5 = 15 / 5 = 3

  • The Result: The coordinates of the point are (1, 3).

Q2. Find the coordinates of the points of trisection of the line segment joining (4, -1) and (-2, -3).

  • The Logic: Trisection means the line is divided into three equal parts. To do this, we need two different points:

    1. The first point (P) divides the line in a 1 : 2 ratio.

    2. The second point (Q) divides the line in a 2 : 1 ratio.

  • The Steps:

    • For Point P (1:2):

      • x = (1*-2 + 2*4) / 3 = 6 / 3 = 2

      • y = (1*-3 + 2*-1) / 3 = -5 / 3

    • For Point Q (2:1):

      • x = (2*-2 + 1*4) / 3 = 0 / 3 = 0

      • y = (2*-3 + 1*-1) / 3 = -7 / 3

  • The Result: The points of trisection are (2, -5/3) and (0, -7/3).

Q3. [The Sports Day Problem] … find the distance between both the flags. If Rashmi has to post a blue flag exactly halfway… where should she post her flag?

  • The Logic: This is a two-part question. First, find the distance between Niharika (2, 25) and Preet (8, 20) using the Distance Formula. Second, find the Midpoint (exactly halfway) where Rashmi should stand.

  • The Steps:

    • Distance: root[(8 – 2)^2 + (20 – 25)^2] = root(6^2 + (-5)^2) = root(36 + 25) = root(61).

    • Midpoint (Halfway): x = (2 + 8) / 2 = 5; y = (25 + 20) / 2 = 22.5.

  • The Result: The distance is root(61) m and Rashmi should post her flag at (5, 22.5), which is on the 5th line at a distance of 22.5 m.

Q4. Find the ratio in which the line segment joining the points (-3, 10) and (6, -8) is divided by (-1, 6).

  • Pro-Tip: When you need to find the Ratio, always assume it is k : 1. It makes the algebra much simpler!

  • The Steps:

    • Let the ratio be k : 1. Using the x-coordinate:

    • -1 = (k6 + 1-3) / (k + 1)

    • -1(k + 1) = 6k – 3

    • -k – 1 = 6k – 3

    • 2 = 7k => k = 2/7

  • The Result: The ratio is 2 : 7.

Q5. Find the ratio in which the line segment joining A(1, -5) and B(-4, 5) is divided by the x-axis. Also find the coordinates of the point of division.

  • The Logic: Any point on the x-axis has a y-coordinate of 0. We use the y-part of the section formula to find the ratio ‘k’ first.

  • The Steps:

    1. Let ratio be k : 1. y = 0.

    2. 0 = (k5 + 1-5) / (k + 1)

    3. 0 = 5k – 5 => 5k = 5 => k = 1.

    4. Since k = 1, it is the Midpoint.

    5. x = (1 + (-4)) / 2 = -3 / 2.

  • The Result: The ratio is 1 : 1 (midpoint) and the coordinates are (-3/2, 0).

Q6. If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.

  • The Logic: In a parallelogram, the diagonals bisect each other. This means the midpoint of the diagonal joining (1, 2) and (x, 6) is the same as the midpoint of the diagonal joining (4, y) and (3, 5).

  • The Steps:

    1. Midpoint of AC = Midpoint of BD

    2. For x: (1 + x) / 2 = (4 + 3) / 2

      • 1 + x = 7 => x = 6

    3. For y: (2 + 6) / 2 = (y + 5) / 2

      • 8 = y + 5 => y = 3

  • The Result: x = 6 and y = 3.

Q7. Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, -3) and B is (1, 4).

  • The Logic: The Centre of a circle is always the Midpoint of its diameter. We know the midpoint and one end, so we just need to find the “starting point” A.

  • The Steps:

    • Let A be (x, y). Centre is (2, -3) and B is (1, 4).

    • Midpoint x: 2 = (x + 1) / 2 => 4 = x + 1 => x = 3

    • Midpoint y: -3 = (y + 4) / 2 => -6 = y + 4 => y = -10

  • The Result: The coordinates of point A are (3, -10).

Q8. If A and B are (-2, -2) and (2, -4), respectively, find the coordinates of P such that AP = 3/7 AB and P lies on the line segment AB.

  • Human Tip: If AP is 3/7 of the total length AB, then the remaining part PB must be 4/7. This means the ratio AP : PB is 3 : 4.

  • The Steps:

    • m1 = 3, m2 = 4. Points: (-2, -2) and (2, -4).

    • x = (32 + 4-2) / 7 = (6 – 8) / 7 = -2/7

    • y = (3*-4 + 4*-2) / 7 = (-12 – 8) / 7 = -20/7

  • The Result: The coordinates of P are (-2/7, -20/7).

Q9. Find the coordinates of the points which divide the line segment joining A(-2, 2) and B(2, 8) into four equal parts.

  • The Logic: To divide into 4 parts, we find 3 points.

    1. Point 2 is the Midpoint of AB.

    2. Point 1 is the Midpoint of A and Point 2.

    3. Point 3 is the Midpoint of Point 2 and B.

  • The Steps:

    • Point 2 (Midpoint AB): ((-2+2)/2, (2+8)/2) = (0, 5)

    • Point 1 (Midpoint of A & P2): ((-2+0)/2, (2+5)/2) = (-1, 3.5)

    • Point 3 (Midpoint of P2 & B): ((0+2)/2, (5+8)/2) = (1, 6.5)

  • The Result: The points are (-1, 3.5), (0, 5), and (1, 6.5).

Q10. Find the area of a rhombus if its vertices are (3, 0), (4, 5), (-1, 4) and (-2, -1) taken in order.

  • Pro-Tip: The area of a rhombus is 1/2 x (Product of its diagonals). Use the distance formula to find the lengths of the diagonals AC and BD first.

  • The Steps:

    1. Diagonal 1 (AC): root[(-1-3)^2 + (4-0)^2] = root(16+16) = 4 root 2

    2. Diagonal 2 (BD): root[(-2-4)^2 + (-1-5)^2] = root(36+36) = 6 root 2

    3. Area = 1/2 x 4 root 2 x 6 root 2 = 1/2 x 24 x 2 = 24.

  • The Result: The area of the rhombus is 24 square units.

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