206-NCERT New Syllabus Grade 10 Coordinate Geometry Ex-7.2

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NCERT New Syllabus Mathematics

Class: 10

Exercise 7.2

Topic: Coordinate Geometry 

Understanding Coordinate Geometry: A Key Concept in Class 10 Mathematics

Coordinate Geometry, often referred to as Cartesian Geometry, bridges the gap between algebra and geometry, allowing us to visualize geometric shapes and solve problems using algebraic equations. It’s a fascinating tool for plotting points, lines, and curves on a plane, all through coordinates. In Class 10, you’ll explore concepts like the distance between two points, the section formula, and the area of a triangle, all of which are essential building blocks for higher-level mathematics and real-life applications.

This chapter enhances your spatial reasoning and gives you the foundation to tackle more complex problems in trigonometry, calculus, and beyond. Let's dive into the core concepts and see how algebra and geometry come together to solve some exciting issues!

EXERCISE 7.2

Q 1. Find the coordinates of the point which divides the join of 

(– 1, 7) and (4, – 3) in the ratio 2 : 3.

Explanation:

1) The point P(x, y) divides the segment joining the points A(x₁, y₁) and B(x₂, y₂) in

the ratio m:n, so we have,

x = (mx₂ + nx₁)/(m + n)

y = (my₂ + ny₁)/(m + n)

2) In general P(x, y) = ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)) 

Solution:

1) The point P(x, y) divides the segment joining the points A(– 1, 7) and B(4, – 3) in

the ratio 2 : 3, so,

a) first we will find x-coordinate

x = (mx₂ + nx₁)/(m + n)

x = ((2) (4) + (3) (– 1))/(2 + 3)

x = (8 – 3)/(5)

x = (5)/(5)

x = 1

b) now we will find the y-coordinate 

y = (my₂ + ny₁)/(m + n)

y = ((2) (– 3) + (3) (7))/(2 + 3)

y = (– 6 + 21)/(5)

y = (15)/(5)

y = 3

2) So, the point P(1, 3) divides the segment joining the points 

A(– 1, 7) and B(4, – 3) in the ratio 2 : 3.

Q 2. Find the coordinates of the points of trisection of the line segment

joining (4, –1) and (–2, –3).

Explanation:

1) The point P(x, y) divides the segment joining the points A(x₁, y₁) and B(x₂, y₂) in

the ratio m:n, so we have,

x = (mx₂ + nx₁)/(m + n)

y = (my₂ + ny₁)/(m + n)

2) In general P(x, y) = ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)) 

Solution:

1) The point P(x₁, y₁) divides the segment joining the points 

A(4, – 1) and B(– 2, – 3) in the ratio 1 : 2, so,

a) first we will find x-coordinate of point P(x₁, y₁)

x₁ = (mx₂ + nx₃)/(m + n)

x₁ = ((1) (– 2) + (2) (4))/(1 + 2)

x₁ = (– 2 + 8)/(3)

x₁ = (6)/(3)

x₁ = 2

b) now we will find the y-coordinate 

y₁ = (my₂ + ny₃)/(m + n)

y₁ = ((1) (– 3) + (2) (– 1))/(1 + 2)

y₁ = (– 3 – 2)/(3)

y₁ = (– 5)/(3)

y₁ = – 5/3

2) So, the coordinates of the point P(x₁, y₁) is P(2, – 5/3)

3) The point Q(x₂, y₂) divides the segment joining the points 

A(4, – 1) and B(– 2, – 3) in the ratio 2 : 1, so,

a) first we will find x-coordinate of point Q(x₂, y₂)

x₂ = (mx₁ + nx₃)/(m + n)

x₂ = ((2) (– 2) + (1) (4))/(2 + 1)

x₂ = (– 4 + 4)/(3)

x₂ = (0)/(3)

x₂ = 0

b) now we will find the y-coordinate 

y₂ = (my₁ + ny₃)/(m + n)

y₂ = ((2) (– 3) + (1) (– 1))/(2 + 1)

y₂ = (– 6 – 1)/(3)

y₂ = (– 7)/(3)

y₂ = – 7/3

4) So, the coordinates of the point Q(x₂, y₂) is Q(0, – 7/3).

5) The coordinates of the points of trisection of the line segment joining 

(4, –1) and (–2, –3) are P(2, – 5/3) and Q(0, – 7/3).

Q 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, as shown in the following fig., 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 flags? If Rashmi has to post a blue flag halfway between the line segment joining the two flags, where should she post her flag?

Explanation:

1) The point P(x, y) divides the segment joining the points A(x₁, y₁) and B(x₂, y₂) in

the ratio m:n, so we have,

x = (mx₂ + nx₁)/(m + n)

y = (my₂ + ny₁)/(m + n)

2) In general P(x, y) = ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)) 

Solution:

1) According to the problem, the total distance of AD is 100 m.

2) Niharika covers 1/4th of the distance of AD. i.e. (1/4)(100) = 25 on 2nd line. So

Niharika posts a green flag at the coordinates P(2, 25).

3) In the same way, Preet covers 1/5th of the distance of AD. i.e. (1/5)(100) = 20

on 8th line. So Preet posts a red flag at the coordinates Q(8, 20).

4) so using the distance formula, we can find the distance between two flags as

follows:

5) The coordinates of P and Q are P(2, 25) and Q(8, 20), so here

a) x₁ = 2
b) y₁ = 25

c) x₂ = 8
d) y₂ = 20 

6) We know that:

(PQ) = √[(x₁ – x₂)² + (y₁ – y₂)²]

(PQ) = √[(2 – 8)² + (25 – 20)²]

(PQ) = √[(– 6)² + (5)²]

(PQ) = √[36 + 25]

(PQ) = √61

7) The distance between the two flags is √61 m.

8) As Rashmi puts the blue flag in the middle of the green and red flag, i.e.,

R(x, y) = ((x₁ + x₂)/2, (y₁ + y₂)/2)

R(x, y) = ((2 + 8)/2, (25 + 20)/2)

R(x, y) = ((10)/2, (45)/2)
R(x, y) = (5, 22.5)

9) Therefore, 

a) The distance between the two flags is √61 m.

b) Rashmi should post her blue flag at 22.5m on the 5th line. 

Q 4. Find the ratio in which the line segment joining the points (– 3, 10) 

and (6, – 8) is divided by (– 1, 6). 

Explanation:

1) The point P(x, y) divides the segment joining the points A(x₁, y₁) and B(x₂, y₂) in

the ratio m:n, so we have,

x = (mx₂ + nx₁)/(m + n)

y = (my₂ + ny₁)/(m + n)

2) In general P(x, y) = ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)) 

Solution:

1) Let the point P(– 1, 6) divides segment A(– 3, 10) B(6, – 8) in the ratio k : 1, so

using section formula, we have,

x = (mx₂ + nx₁)/(m + n)
(6k – 3)/(k + 1) = – 1

(6k – 3) = – 1(k + 1)

(6k – 3) = – k – 1

(6k + k) = – 1 + 3

7k = 2

k = 2/7

2) The ratio is 2:7.

Q 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.

Explanation:

1) The point P(x, y) divides the segment joining the points A(x₁, y₁) and B(x₂, y₂) in

the ratio m:n, so we have,

x = (mx₂ + nx₁)/(m + n)

y = (my₂ + ny₁)/(m + n)

2) In general P(x, y) = ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)) 

Solution:

1) Let the line segment joining the points A(1, – 5) and B(– 4, 5) get divided by the

x-axis in the ratio k : 1.

2) We know that the y-coordinate of every point on the x-axis is 0, so first we will

find the y-coordinate. 

3) Using the section formula, we have,

y = (my₂ + ny₁)/(m + n)

y = (5k – 5)/(k + 1)

0  = (5k – 5)/(k + 1)

0 (k + 1) = (5k – 5)

(5k – 5) = 0

5k = 5

k = 5/5

k = 1 

4) The x-axis divides the line segment joining the points A(1, – 5) and B(– 4, 5) in

1:1 ratio.

5) Now we will find the x-coordinate point of division with the ratio 1:1

Using the section formula, we have,

x = (mx₂ + nx₁)/(m + n)

x = (x₂ + x₁)/(1 + 1)

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

6) Therefore

a) The x-axis divides the line segment joining the points A(1, – 5) and B(– 4, 5)

in 1:1 ratio.

b) The coordinates of the point of division are (– 3/2, 0).

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

Explanation:

1) The point P(x, y) divides the segment joining the points A(x₁, y₁) and B(x₂, y₂) in

the ratio m:n, so we have,

x = (mx₂ + nx₁)/(m + n)

y = (my₂ + ny₁)/(m + n)

2) In general P(x, y) = ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)) 

Solution:

1) As ABCD is the parallelogram, the diagonals bisect each other.

2) Point O is the midpoint of diagonal AC and BD.

3) Now we will find the mid-point of AC

Mid-point of AC = ((1 + x)/2, (2 + 6)/2)

Mid-point of AC = ((1 + x)/2, (8)/2)
Mid-point of AC = ((1 + x)/2, 4) --------- equation 1

4) Now we will find the mid-point of BD

Mid-point of BD = ((3 + 4)/2, (5 + y)/2)

Mid-point of BD = ((7)/2, (y + 5)/2)
Mid-point of BD = (7/2, (y + 5)/2) --------- equation 2

5) From equations 1 and 2, we will get the x-coordinate,

(1 + x)/2 = 7/2

(1 + x) = 7
x = 7 – 1

x = 6 --------- equation 3

6) From equations 1 and 2,  we will get the y-coordinate,

(y + 5)/2 = 4
(y + 5) = 4 (2)

(y + 5) = 8

y = 8 – 5

y = 3 --------- equation 4

7) From equations 3 and 4, x = 6 and y = 3.

Q 7. Find the coordinates of point A, where AB is the diameter of a circle whose center is (2, – 3) and B is (1, 4).

Explanation:

1) The point P(x, y) divides the segment joining the points A(x₁, y₁) and B(x₂, y₂) in

the ratio m:n, so we have,

x = (mx₂ + nx₁)/(m + n)

y = (my₂ + ny₁)/(m + n)

2) In general P(x, y) = ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)) 

Solution:

1) Let the coordinates of point A be (x, y).

2) AB is the diameter of the circle with center O(2, – 3).

3) So, point O(2, – 3) is the mid-point of the segment joining the points 

A(x, y) and B(1, 4).

4) So, using the mid-point form, we will find the x-coordinate of point A,

(x + 1)/2 = 2

(x + 1) = 2 (2)
(x + 1) = 4
x = 4 – 1

x = 3 --------- equation 1

5) So, using the mid-point form, we will find the y-coordinate of point A,

(y + 4)/2 = – 3

(y + 4) = – 3 (2)
(y + 4) = – 6
y = – 6 – 4

y = – 10 --------- equation 2

6) From equations 1 and 2, x = 3 and y = – 10. So, the coordinates of 

A are A (3, – 10).

Q 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.

Explanation:

1) The point P(x, y) divides the segment joining the points A(x₁, y₁) and B(x₂, y₂) in

the ratio m:n, so we have,

x = (mx₂ + nx₁)/(m + n)

y = (my₂ + ny₁)/(m + n)

2) In general P(x, y) = ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)) 

Solution:

1) Let the coordinates of point P be (x, y).

2) Point P divides AB such that, AP = (3/7) AB, so,

(3/7) AB = AP

(AB)/(AP) = 7/3
(AB – AP)/(AP) = (7 – 3)/3
(PB)/(AP) = (4)/3
(AP)/(PB) = 3/4

3) Here, point P(x, y) divides the segment joining point A(– 2, – 2) and B(2, – 4), So,

we now, will find x coordinate

x = (3 (2) + 4 (– 2))/(3 + 4)

x = (6 – 8)/(7)

x = (– 2)/(7)

x = – 2/7 --------- equation 1

we now, will find y coordinate

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

4) From equations 1 and 2, x = – 2/7 and y = – 20/7. So, the coordinates of 

P are P(– 2/7, – 20/7).

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

Explanation:

1) The point P(x, y) divides the segment joining the points A(x₁, y₁) and B(x₂, y₂) in

the ratio m:n, so we have,

x = (mx₂ + nx₁)/(m + n)

y = (my₂ + ny₁)/(m + n)

2) In general P(x, y) = ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)) 

Solution:

1) The point Q(x₂, y₂) the mid-point of the segment joining the points 

A(– 2, 2), B(2, 8), so the coordinates of point Q will be:

Q(x₂, y₂) = ((– 2 + 2)/2, (2 + 8)/2)

Q(x₂, y₂) = ((0)/2, (10)/2)
Q(x₂, y₂) = (0, 5) -------- equation 1

2) The point P(x₁, y₁) the mid-point of the segment joining the points 

A(– 2, 2), Q(0, 5), so the coordinates of point Q will be:

P(x₁, y₁) = ((– 2 + 0)/2, (2 + 5)/2)

P(x₁, y₁) = ((– 2)/2, (7)/2)

P(x₁, y₁) = (– 1, 7/2) -------- equation 2

3) The point R(x₃, y₃) the mid-point of the segment joining the points 

Q(0, 5), B(2, 8), so the coordinates of point Q will be:

R(x₃, y₃) = ((0 + 2)/2, (5 + 8)/2)

R(x₃, y₃) = ((2)/2, (13)/2)

R(x₃, y₃) = (1, 13/2) -------- equation 3

 4) From equations 1, 2, and 3, the coordinates of points P, Q, and R are as follows.

P(x₁, y₁) = P(– 1, 7/2)

Q(x₂, y₂) = Q(0, 5)

R(x₃, y₃) = R(1, 13/2).

Q 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)].

Solution:

1) We know that the area of the rhombus = (1/2) (product of diagonals).

2) First we will find AC and BD using the distance formula.

3) First we will find AC with A(3, 0), C(– 1, 4)

a) x₁ = 3

b) y₁ = 0

c) x₂ = – 1

d) y₂ = 4 

4) We know that:

(AC) = √[(x₁ – x₂)² + (y₁ – y₂)²]

(AC) = √[(3 – (– 1))² + (0 – 4)²]

(AC) = √[(4)² + (– 4)²] 

(AC) = √[16 + 16]
(AC) = 4√2 ------------- equation 1

5) First we will find BD with B(4, 5), C(– 2, – 1)

a) x₁ = 4

b) y₁ = 5

c) x₂ = – 2

d) y₂ = – 1 

6) We know that:

(BD) = √[(x₁ – x₂)² + (y₁ – y₂)²]

(BD) = √[(4 – (– 2))² + (5 – (– 1))²]

(BD) = √[(6)² + (6)²] 

(BD) = √[36 + 36]
(BD) = 6√2 ------------- equation 2

7) From equations 1 and 2, we can find the area of the rhombus as follows.

Area of the rhombus = (1/2) (product of diagonals)

Area of the rhombus = (1/2) (4√2) (6√2)

Area of the rhombus = (2√2) (6√2)
Area of the rhombus = (2) (6) (√2) (√2)
Area of the rhombus = (12) (2)
Area of the rhombus = 24 square units.

Conclusion

Coordinate Geometry offers a powerful way to link algebra with geometric concepts, providing a deeper understanding of spatial relationships. The techniques covered in this chapter, such as finding distances, midpoints, and areas, lay a solid foundation for more advanced topics in mathematics. By mastering these concepts, students enhance their problem-solving skills and develop a critical toolset that will be invaluable in fields like engineering, physics, and computer science.

Continue practicing these methods, and you’ll find that coordinate geometry is useful and an exciting area of math that transforms abstract numbers into meaningful visual solutions.

Keep learning, keep exploring!

#CoordinateGeometry #MathsWithCoordinates #Class10Maths #NCERTSolutions #MathMadeEasy #GeometryAndAlgebra #SpatialReasoning #MathematicsForAll #MathSkills #STEMLearning #EducationForFuture

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Anil Satpute