NCERT Solutions Class 10 Maths Chapter 7 Coordinate Geometry : Exercise 7.2

NCERT Solutions Class 10 Maths Chapter 7 Exercise 7.2 will help you understand section formula, a basic concept of coordinate geometry. The exercise is about determining the coordinates of a point that divides a line segment in a specified ratio, which is essential to solve problems related to points on a plane. This exercise is designed as per the new CBSE syllabus and examination pattern and hence forms a significant part of NCERT Class 10 Maths. Let us learn more about this topic and understand its major concepts.

1.0Download NCERT Solutions Class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.2 : Free PDF

2.0Introduction Class 10 Maths Chapter 7 Coordinate Geometry: Section Formula

The Section Formula aids in the determination of the coordinates of a point which divides a line segment in a specified ratio. In a variety of applications that occur in the real world, including the placement of objects, splitting distances, and studying geometric shapes, there is a need to understand the process of finding coordinates of a point of division on a line segment.

3.0NCERT Class 10 Maths Chapter 7 Exercise 7.2 Overview: Key Concepts

In this exercise, you will be able to apply the section formula to determine the coordinates of a point that divides the line segment between two points internally or externally in a specific ratio. The formula also includes special cases, like the midpoint formula, which applies when the ratio is 1:1.

Section Formula for Internal Division: 

The general formula to find the coordinates of a point on the cartesian plane, say P(x,y) that divides a line segment with coordinates A(x₁,y₁) and B(x₂,y₂) in a certain ratio. Let this ratio be m₁:m₂, then the section formula can be expressed as: 

$x=\frac{m_1{x}_2+m_2\times x_1}{m_1+m_2}$

$y=\frac{m_1{y}_2+m_2{y}_1}{m_1+m_2}$

Special Case: Midpoint Formula

It is one of the special cases of the section formula. This formula is used when the line segment, say P(x,y) divides another line with coordinates A(x₁,y₁) and B(x₂,y₂) in equal length. The ratio of this type of line segment is 1:1. The midpoint formula can mathematically be expressed as: 

$x=\frac{x_1+x_2}{2}$

$y=\frac{y_1+y_2}{2}$

Section Formula for Points on the Axes

If a point divides a line segment in a certain ratio and is on the x-axis or the y-axis, its coordinates can be found by using the section formula with the particular condition of being on one of the axes. 

• For the x-axis: Assume the point cuts the line joining A(x₁,y₁) and B(x₂,y₂) in the ratio m:n, and the point is on the x-axis. The y-coordinate of the point will be 0. From the section formula, the x-coordinate of the point can be calculated as $P(\frac{m{x}_2+n{x}_1}{m+n},0)$. 
• For the y-axis: If the point, say P, is dividing the line joining A(x₁,y₁) and B(x₂,y₂) in the ratio m:n and is on the y-axis, the x-coordinate of the point will be 0 (because any point on the y-axis has an x-coordinate of 0). Hence, the coordinates can be written as $P(\frac{m{y}_2+n{y}_1}{m+n},0).$

Trisection of a Line Segment

If a line segment is cut into three equal parts, then the points which divide the line into three equal parts are known as trisection points. These points divide the line in the ratio 1:3 or 2:3 from one end of the line. The coordinates of these points can be calculated using the section formula by simply putting the values of the ratio. For example: 

• The formula for first trisection points (For the ratio of 1:2): The coordinates of the point P₁​ dividing the segment AB in the ratio 1:2 are given by:  $P_1(\frac{2x_1+x_2}{3},\frac{2y_1+y_2}{3})$
• The Formula for Second Trisection Point (Divides in Ratio 2:1): The coordinates of the point P₂​ dividing the segment AB in the ratio 2:1 are given by: $P_2(\frac{2x_2+x_1}{3},\frac{2y_2+y_1}{3})$

4.0NCERT Class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.2 : Detailed Solutions

1. Find the co-ordinates of the point which divides the line joining of (-1,7) and (4,-3) in the ratio 2:3.

Sol. Let the required point be P(x, y).

Here the end points are (-1, 7) and (4, -3).

Ratio = 2:3 = m1:m2

x = (m1 * x2 + m2 * x1) / (m1 + m2) = (2 * 4 + 3 * (-1)) / (2 + 3)

= (8 - 3) / 5 = 5 / 5 = 1

and y = (m1 * y2 + m2 * y1) / (m1 + m2)

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

Thus, the required point is P(1, 3).

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

Sol.

Points P and Q trisect the line segment joining the points A(4, -1) and B(-2, -3), i.e., AP = PQ = QB.

Here, P divides AB in the ratio 1:2 and Q divides AB in the ratio 2:1.

x-coordinate of P = (1 × (-2) + 2 × (4)) / (1 + 2) = 6 / 3 = 2;

y-coordinate of P = (1 × (-3) + 2 × (-1)) / (1 + 2) = -5 / 3

Thus, the coordinates of P are (2, -5/3).

Now, x-coordinate of Q = (2 × (-2) + 1 × (4)) / (2 + 1) = 0;

y-coordinate of Q = (2 × (-3) + 1 × (-1)) / (2 + 1) = -7 / 3

Thus, the coordinates of Q are (0, -7/3).

Hence, the points of trisection are P(2, -5/3) and Q(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 1 m each. 100 flower pots have been placed at a distance of 1 m from each other along AD, as shown in fig. Niharika runs 1/4th the distance AD on the 2nd line and posts a green flag. Preet runs 1/5th 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?

Sol. Let us consider 'A' as origin, then:

A B is the x-axis.

A D is the y-axis.

Now, the position of the green flag-post is (2, 100/4) or (2, 25).

And, the position of the red flag-post is (8, 100/5) or (8, 20).

=> Distance between both the flags = √((8-2)² + (20-25)²)

= √(6² + (-5)²) = √(36 + 25) = √61 m

Let the mid-point of the line segment joining the two flags be M(x, y).

Therefore, x = (2 + 8) / 2 and y = (25 + 20) / 2

or x = 5 and y = 22.5

Thus, the blue flag is on the 5th line at a distance 22.5 m above AB.

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

Sol. Let the required ratio be K : 1.

Comparing x-coordinate:

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

=> 6k - 3 = -k - 1

=> 7k = 2

=> k = 2/7

Comparing y-coordinate:

k(-8) + 1(10) / (k + 1) = 6

=> -8k + 10 = 6k + 6

=> -8k - 6k = 6 - 10

=> -14k = -4

=> k = 2/7

Therefore, the required ratio is 2:7.

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.

Sol. The given points are A(1, -5) and B(-4, 5). Let the required ratio be k:1 and the required point be P(x, 0). Since point P lies on the x-axis.

To find the ratio:

x = (m1*x2 + m2*x1) / (m1 + m2) and 0 = (m1*y2 + m2*y1) / (m1 + m2)

=> x = (-4k + 1) / (k + 1) and 0 = (5k - 5) / (k + 1)

=> x(k + 1) = -4k + 1 and 5k - 5 = 0 => k = 1

=> x(k + 1) = -4k + 1

=> x(1 + 1) = -4 + 1 [since k = 1]

=> 2x = -3

=> x = -3/2

Therefore, the required ratio k:1 = 1:1.

Coordinates of P are (x, 0) = (-3/2, 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.

Sol. 

Mid-point of the diagonal AC has x-coordinate = (x + 1) / 2 and y-coordinate = (6 + 2) / 2 = 4.

That is, ((x + 1) / 2, 4) is the mid-point of AC.

Similarly, the mid-point of the diagonal BD is ((4+3)/2, (y+5)/2), i.e., (7/2, (y+5)/2).

We know that the two diagonals AC and BD bisect each other at M. Therefore,

((x+1)/2, 4) and (7/2, (y+5)/2) coincide.

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

=> 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).

Sol. Here, the centre of the circle is O(2, -3).

Let the end points of the diameter be A(x, y) and B(1, 4).

The centre of a circle bisects the diameter.

Therefore, 2 = (x + 1) / 2  =>  x + 1 = 4 or x = 3

And -3 = (y + 4) / 2  =>  y + 4 = -6 or y = -10

Here, the coordinates of A are (3, -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.

AP = (3/7)AB,

BP = AB - AP = AB - (3/7)AB = (4/7)AB

AP/BP = ((3/7)AB) / ((4/7)AB) = 3/4

Thus, P divides AB in the ratio 3:4.

x-coordinate of P = (3 x 2 + 4 x (-2)) / (3 + 4) = -2/7

y-coordinate of P = (3 x (-4) + 4 x (-2)) / (3 + 4) = -20/7

Hence, 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.

Sol. Here, the given points are A(-2, 2) and B(2, 8).

Let P1, P2, and P3 divide AB into four equal parts.

Since AP₁ = P₁P₂ = P₂P₃ = P₃B,

Obviously, P₂ is the mid-point of AB.

Therefore, coordinates of P₂ are:

((−2 + 2) / 2, (2 + 8) / 2) or (0, 5)

Again, P₁ is the mid-point of AP₂.

Therefore, coordinates of P₁ are:

((−2 + 0) / 2, (2 + 5) / 2) or (−1, 7/2)

Also, P₃ is the mid-point of P₂B.

Therefore, coordinates of P₃ are:

((0 + 2) / 2, (5 + 8) / 2) or (1, 13/2)

Thus, the coordinates of P₁, P₂, and P₃ are (−1, 7/2), (0, 5), and (1, 13/2) respectively.

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

Sol. As diagonals AC and BD bisect each other at right angles to each other at O.

AC = √((-1 - 3)² + (4 - 0)²)

= √(16 + 16) = √32 = 4√2

BD = √((4 + 2)² + (5 + 1)²)

= √(36 + 36) = √72 = 6√2

Then OA = (1/2)AC = (1/2) * 4√2 = 2√2

OB = (1/2)BD = (1/2) * 6√2 = 3√2

Area of ΔAOB = (1/2)(OA) * (OB) = (1/2) * 2√2 * 3√2 = 6 sq. units

Hence, the area of the rhombus ABCD = 4 * area of ΔAOB = 4 * 6 = 24 sq. units.

5.0Benefits of Studying NCERT Solutions of Class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.2

• Strengthened Foundational Concepts 
• Improved Problem-Solving Skills
• Exam Preparation and Confidence
• Enhanced Understanding of Key Topics
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