Co- ordinate Geometry Notes CBSE NCERT

Class 10th - Coordinate Geometry

Class 10th - Coordinate Geometry


Introduction

➥ Coordinate geometry, also known as Cartesian geometry, is the study of geometry using a coordinate system.

➥ This method allows for the precise description of geometric figures using algebraic equations.

Cartesian Plane

Cartesian Plane: A plane with a rectangular coordinate system that associates each point in the plane with an ordered pair (x, y).

Axes: The two perpendicular lines that define the coordinate system. The horizontal line is the x-axis and the vertical line is the y-axis.

Origin: The point of intersection of the x-axis and y-axis, denoted as O(0, 0).

Quadrants: The coordinate plane is divided into four quadrants:

  1. First Quadrant: (+,+)
  2. Second Quadrant: (−,+)
  3. Third Quadrant: (−,−)
  4. Fourth Quadrant: (+,−)

Distance Formula

➥ The distance between two points (x1, y1) and (x2, y2) is given by:

Section Formula

➥ If a point P(x, y) divides the line segment joining the points (x1, y1) and (x2, y2) in the ratio m:n, then the coordinates of P are given by:

Midpoint Formula

➥ The midpoint of the line segment joining the points (x1, y1) and (x2, y2) is given by:

Area of a Triangle

The area of a triangle with vertices (x1, y1), (x2, y2), and (x3, y3) is given by:

Collinearity of Three Points

➥ Three points A(x1, y1), B(x2, y2), and C(x3, y3) are collinear if the area of the triangle formed by them is zero.





Exercise 7.1 - Coordinate Geometry(NCERT)

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

(i) Points (2, 3) and (4, 1):

The distance formula is given by d = √[(x₂ - x₁)² + (y₂ - y₁)²].

Distance = √[(4 - 2)² + (1 - 3)²] = √[4 + 4] = √8 = 2√2

(ii) Points (–5, 7) and (–1, 3):

Distance = √[(-1 - (-5))² + (3 - 7)²] = √[4 + 16] = √20 = 2√5

(iii) Points (a, b) and (–a, –b):

Distance = √[(–a - a)² + (–b - b)²] = √[(-2a)² + (-2b)²] = √[4a² + 4b²] = 2√(a² + b²)

2. Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in Section 7.2?

Distance = √[(36 - 0)² + (15 - 0)²] = √[1296 + 225] = √1521 = 39

For the distance between towns A and B, refer to Section 7.2 for specific details.

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

To determine if points are collinear, calculate the area of the triangle formed by these points. If the area is zero, the points are collinear.

Area = 1/2 | 1(3 - (-11)) + 2((-11) - 5) + (-2)(5 - 3) | = 1/2 | 1(14) + 2(-16) - 2(2) |

Area = 1/2 | 14 - 32 - 4 | = 1/2 | -22 | = 11

Since the area is not zero, the points are not collinear.

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

Calculate the distances between each pair of points:

Distance between (5, –2) and (6, 4): √[(6 - 5)² + (4 - (-2))²] = √[1 + 36] = √37

Distance between (6, 4) and (7, –2): √[(7 - 6)² + ((-2) - 4)²] = √[1 + 36] = √37

Distance between (7, –2) and (5, –2): √[(7 - 5)² + ((-2) - (-2))²] = √[4] = 2

Two sides are equal, so the triangle is isosceles.

5. In a classroom, 4 friends are seated at the points A, B, C, and D as shown in Fig. 7.8. Champa and Chameli walk into the class, and after observing for a few minutes, Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees. Using the distance formula, find which of them is correct.

To determine if ABCD is a square, calculate the distances between all pairs of adjacent vertices and the diagonals.

For each side to be equal and diagonals to be equal, use the distance formula as described in the previous answers.

Compare all sides and diagonals to verify if all four sides are equal and diagonals are also equal. The correct answer depends on these calculations.

6. Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:

(i) Points (–1, –2), (1, 0), (–1, 2), (–3, 0):

Calculate all side lengths and diagonals to determine the type of quadrilateral.

(ii) Points (–3, 5), (3, 1), (0, 3), (–1, –4):

Similarly, calculate side lengths and diagonals to determine the type of quadrilateral.

(iii) Points (4, 5), (7, 6), (4, 3), (1, 2):

Calculate side lengths and diagonals to determine the type of quadrilateral.

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

Let the point on the x-axis be (x, 0). Set up the distance equations:

√[(x - 2)² + (0 - (-5))²] = √[(x + 2)² + (0 - 9)²]

Squaring both sides and solving for x gives the required point.

After calculations, x = -0.8.

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

Use the distance formula:

√[(10 - 2)² + (y - (-3))²] = 10

Square both sides and solve for y:

(10 - 2)² + (y + 3)² = 100

64 + (y + 3)² = 100

(y + 3)² = 36

y + 3 = ±6

y = 3 or y = -9

9. 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.

To find the values of x, we use the distance formula:

The distance between Q(0, 1) and P(5, –3) is:

PR = √[(5 - 0)² + ((-3) - 1)²] = √[25 + 16] = √41

The distance between Q(0, 1) and R(x, 6) should be equal to PR:

QR = √[(x - 0)² + (6 - 1)²] = √[x² + 25]

Setting QR equal to PR:

√[x² + 25] = √41

x² + 25 = 41

x² = 16

x = ±4

For x = 4:

QR = √[(4 - 0)² + (6 - 1)²] = √[16 + 25] = √41

For x = -4:

QR = √[(-4 - 0)² + (6 - 1)²] = √[16 + 25] = √41

Thus, the distances QR and PR are both √41, and the values of x are ±4.

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

Let (x, y) be equidistant from (3, 6) and (–3, 4). We set up the distance formula:

Distance from (x, y) to (3, 6): √[(x - 3)² + (y - 6)²]

Distance from (x, y) to (–3, 4): √[(x + 3)² + (y - 4)²]

Since the distances are equal:

√[(x - 3)² + (y - 6)²] = √[(x + 3)² + (y - 4)²]

Squaring both sides to remove the square roots:

(x - 3)² + (y - 6)² = (x + 3)² + (y - 4)²

Expanding both sides:

x² - 6x + 9 + y² - 12y + 36 = x² + 6x + 9 + y² - 8y + 16

Combining like terms:

-6x - 12y + 45 = 6x - 8y + 25

-12x - 12y = 6x - 8y - 20

-18x - 4y = -20

18x + 4y = 20

9x + 2y = 10

Thus, the relation between x and y is: 9x + 2y = 10


Exercise 7.2 - Coordinate Geometry (NCERT)

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

Let the point dividing the line segment in the ratio 2:3 be P(x, y).

Using the section formula, we have:

x = (m x2 + n x1) / (m + n)

y = (m y2 + n y1) / (m + n)

where (x1, y1) = (–1, 7), (x2, y2) = (4, –3), m = 2, and n = 3.

x = (2 × 4 + 3 × (–1)) / (2 + 3) = (8 – 3) / 5 = 1

y = (2 × (–3) + 3 × 7) / (2 + 3) = (–6 + 21) / 5 = 3

Thus, the coordinates are (1, 3).

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

Let the points of trisection be T1 and T2.

Using the section formula for trisection:

T1 divides the segment in the ratio 1:2:

x = (1 × (–2) + 2 × 4) / (1 + 2) = (–2 + 8) / 3 = 2

y = (1 × (–3) + 2 × (–1)) / (1 + 2) = (–3 – 2) / 3 = –5/3

T2 divides the segment in the ratio 2:1:

x = (2 × (–2) + 1 × 4) / (2 + 1) = (–4 + 4) / 3 = 0

y = (2 × (–3) + 1 × (–1)) / (2 + 1) = (–6 – 1) / 3 = –7/3

Thus, the points of trisection are (2, –5/3) and (0, –7/3).

3. To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1m each. 100 flower pots have been placed at a distance of 1m from each other along AD. Niharika runs 1/4 th the distance AD on the 2nd line and posts a green flag. Preet runs 1/5 th the distance AD on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag?

Let the length of AD be L.

Niharika’s position is at a distance of L/4 on the 2nd line.

Preet’s position is at a distance of L/5 on the 8th line.

The distance between the two flags:

Horizontal distance = (8 - 2) × 1 = 6m

Vertical distance = |L/5 - L/4| = L/20

Distance between the flags = √[(6)² + (L/20)²]

Rashmi’s flag should be positioned halfway between these two points. The midpoint is:

Midpoint = [(L/4 + L/5)/2, (2 + 8)/2] = [9L/20, 5]

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

Let the point (–1, 6) divide the segment in the ratio k:1.

Using the section formula:

x = (-3k + 6) / (k + 1)

y = (10k - 8) / (k + 1)

Given x = -1 and y = 6:

-1 = (-3k + 6) / (k + 1)

Solving gives k = 3.

Thus, the ratio is 3:1.

5. 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.

Let the x-axis be divided by the point (x, 0) in the ratio k:1.

Using the section formula:

y = (k × 5 + 1 × (–5)) / (k + 1) = 0

Solving gives k = 5/2.

The x-coordinate is: x = (k × (–4) + 1 × 1) / (k + 1) = (–10 + 1) / (7/2) = -9 / (7/2) = -18/7

Thus, the ratio is 5:2, and the coordinates are (–18/7, 0).

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

In a parallelogram, diagonals bisect each other. Let the diagonals be (1, 2) to (x, 6) and (4, y) to (3, 5).

The midpoint of both diagonals should be equal:

Midpoint of diagonal (1, 2) to (x, 6): ((1 + x) / 2, (2 + 6) / 2)

Midpoint of diagonal (4, y) to (3, 5): ((4 + 3) / 2, (y + 5) / 2)

Equating the midpoints:

((1 + x) / 2, 4) = (7 / 2, (y + 5) / 2)

x = 6

y = 3

Thus, the values are x = 6 and y = 3.

7. 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).

Since A and B are endpoints of a diameter, the midpoint of AB is the center of the circle.

Using the midpoint formula:

(2, –3) = ((x + 1) / 2, (y + 4) / 2)

Solving:

2 = (x + 1) / 2

4 = x + 1

x = 7

–3 = (y + 4) / 2

–6 = y + 4

y = –10

Thus, the coordinates of A are (7, –10).

8. 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.

Using the section formula:

P divides AB in the ratio 3:4.

x = (-2 × 4 + 2 × 3) / (3 + 4) = (-8 + 6) / 7 = -2/7

y = (-2 × 4 + (-4) × 3) / (3 + 4) = (-8 - 12) / 7 = -20/7

Thus, the coordinates of P are (-2/7, -20/7).

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

The line segment can be divided into four equal parts using the section formula for ratios 1:3, 2:2, and 3:1.

For the points of division:

First point: [(1 × 2 + 3 × (–2)) / 4, (1 × 8 + 3 × 2) / 4] = (-4/4, 14/4) = (-1, 3.5)

Second point: [(2 × 2 + 2 × (–2)) / 4, (2 × 8 + 2 × 2) / 4] = (0, 10)

Third point: [(3 × 2 + 1 × (–2)) / 4, (3 × 8 + 1 × 2) / 4] = (2, 19.5)

Thus, the coordinates are (-1, 3.5), (0, 10), (2, 19.5).

10. Find the area of a rhombus if its vertices are (3, 0), (4, 5), (–1, 4) and (–2, –1) taken in order. [Hint: Area of a rhombus = 1/2 (product of its diagonals)]

To find the area, we first need the lengths of the diagonals. The diagonals are the distances between the pairs of opposite vertices.

Diagonal 1: Between (3, 0) and (–1, 4):

Length = √[(3 - (–1))² + (0 - 4)²] = √[16 + 16] = √32 = 4√2

Diagonal 2: Between (4, 5) and (–2, –1):

Length = √[(4 - (–2))² + (5 - (–1))²] = √[36 + 36] = √72 = 6√2

Area of the rhombus = 1/2 × diagonal 1 × diagonal 2 = 1/2 × 4√2 × 6√2 = 24

Thus, the area of the rhombus is 24 square units.