NCERT Solutions Class 10 Maths Chapter 7 Coordinate Geometry : Exercise 7.3

NCERT Solutions Class 10 Maths Chapter 7 Exercise 7.3 involves an important topic of Coordinate Geometry, which is the Area of triangles. In this exercise, we will learn a new way of finding the area of a triangle using the coordinates of its vertices. This concept of using coordinates to find the area of a triangle is a fundamental topic in coordinate geometry. The exercise is designed to assist students in grasping the formula for the area of a triangle given its vertex coordinates and how to apply it to various situations. Let's go through this important practice step by step.

1.0Download NCERT Solutions Class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.3 : Free PDF

2.0Introduction to the Area of a Triangle

The area of a triangle is the measure of the space closed within three sides of the triangle. In earlier classes, we have learned many ways of calculating this area, the simplest one being when the base and height of the triangle are given. Mathematically, this can be written as: 

$\text{Area of a triangle}=\frac{1}{2}\times \text{Base}\times \text{Height}$

Another known method to find the area of a triangle is Heron’s Formula used when the lengths of all the sides of a triangle are known. However, in this exercise, we will concentrate on a more basic way of computing the area of a triangle if the coordinates of its vertices are given. Before that, let’s understand some key concepts related to this method. 

3.0Class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.3 : Key Concepts

Area of a Triangle Given its Coordinates: 

The area of a triangle can be calculated using the coordinates of its three vertices, without needing to calculate the side lengths. This is done by means of a formula derived from coordinate geometry, using the vertices' x and y coordinates. The formula for the area of a triangle (ABC) for coordinate points given as A (x₁,y₁), B (x₂,y₂), and C (x₃,y₃) can be written as: 

$\text{Area of}\Delta ABC=\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_2)+x_3(y_1-y_2)|$

This exercise makes it easy and effective for students to calculate the area of any triangle rapidly, even if the lengths of the sides are difficult to find or are in irrational numbers. It makes complicated geometric problems easier, which is why it is a must-learn method in coordinate geometry.

Collinearity of Points: 

The area of a triangle using its coordinates is the easiest way of determining the collinearity of three points. Collinearity of points refers to the condition where three or more points lie on the same straight line. Mathematically, if three points, say A (x₁,y₁), B (x₂,y₂), and C (x₃,y₃), are collinear, then the area of a triangle formed with these points will be equal to zero. The three points follow this rule for being collinear. 

$\text{Area of}\Delta ABC=\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_2)+x_3(y_1-y_2)|=0$

Verification of Median Dividing the Triangle into Two Equal Areas: 

The concepts of this exercise also help in verifying the fact that the median of a triangle divides it into two equal areas of the triangle. To check this, find the area of the two triangles created by the median. The sum of these two areas of the triangles must equal the area of the original triangle. The coordinates of midpoints formed by the median can be calculated using the section formula. 

The formula of the area of a triangle with its coordinates provides a quick and effective way to calculate the area. This makes it an important concept in coordinate geometry, which is why mastering this formula becomes necessary. This can be done by practicing NCERT Solutions Class 10 Maths Chapter 7 Exercise 7.3 with the best resources that you can find here. 

4.0NCERT Class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.3 : Detailed Solutions

1. Find the area of the triangle whose vertices are :

(i) (2, 3), (-1, 0), (2, -4)

(ii) (-5, -1), (3, -5), (5, 2)

Sol. (i) Let the vertices of the triangles be A(2, 3), B(-1, 0) and C(2, -4)

Here x₁ = 2, y₁ = 3,

x₂ = -1, y₂ = 0

x₃ = 2, y₃ = -4

Area of a Δ = (1/2) |x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)|

Area of a Δ = (1/2) |2{0 - (-4)} + (-1){-4 - (3)} + 2{3 - 0}|

= (1/2) |2(0 + 4) + (-1)(-4 - 3) + 2(3)|

= (1/2) |8 + 7 + 6| = (1/2) |21| = 21/2 sq. units

(ii) A(-5, -1), B(3, -5), C(5, 2) are the vertices of the given triangle.

x₁ = -5, x₂ = 3, x₃ = 5; y₁ = -1, y₂ = -5, y₃ = 2.

Area of the ΔABC = (1/2) |x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)|

= (1/2) |-5 * (-5 - 2) + 3 * (2 + 1) + 5 * (-1 + 5)|

= (1/2) |35 + 9 + 20| = (1/2) |64| = 32 sq. units.

2. In each of the following, find the value of 'k' for which the points are collinear.

(i) (7, -2), (5, 1), (3, k)

(ii) (8, 1), (k, -4), (2, -5)

Sol. The given three points will be collinear if the area of the triangle formed by them is equal to zero.

(i) Let A(7, -2), B(5, 1), and C(3, k) be the vertices of a triangle.

Therefore, the given points will be collinear if ar(△ABC) = 0.

Or, 7(1 - k) + 5(k + 2) + 3(-2 - 1) = 0

⇒ 7 - 7k + 5k + 10 + (-6) - 3 = 0

⇒ 17 - 9 + 5k - 7k = 0

⇒ 8 - 2k = 0

⇒ 2k = 8

⇒ k = 8/2 = 4

The required value of k = 4.

(ii) A(8, 1), B(k, -4), C(2, -5) are the given points.

x1 = 8, x2 = k, x3 = 2

y1 = 1, y2 = -4, y3 = -5

The condition for the three points to be collinear is:

x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2) = 0

8 × (-4 + 5) + k × (-5 - 1) + 2 × (1 + 4) = 0

⇒ 8 - 6k + 10 = 0

⇒ 6k = 18

⇒ k = 3

3. Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, -1), (2, 1), and (0, 3). Find the ratio of this area to the area of the given triangle.

Sol. Let the vertices of the triangle be A(0, -1), B(2, 1), and C(0, 3).

Let D, E, and F be the mid-points of the sides BC, CA, and AB respectively. Then:

Coordinates of D are ((2 + 0)/2, (1 + 3)/2) i.e., (2/2, 4/2) or (1, 2).

Coordinates of E are ((0 + 0)/2, (3 + (-1))/2) i.e., (0, 1).

Coordinates of F are ((2 + 0)/2, (1 + (-1))/2) i.e., (1, 0).

Now, ar(△ABC) =

= (1/2) |0(1-3) + 2{3-(-1)} + 0(-1-1)|

= (1/2) |0(-2) + 8 + 0(-2)|

= (1/2) |0 + 8 + 0| = (1/2) × 8 = 4 sq. units

Now, ar (△DEF)

= (1/2) |1(1-0) + 0(0-2) + 1(2-1)|

= (1/2) |1(1) + 0 + 1(1)|

= (1/2) |1 + 0 + 1| = (1/2) × 2 = 1 sq. unit

∴ ar(△DEF) / ar(△ABC) = 1/4

∴ ar(△DEF) : ar(△ABC) = 1 : 4.

4. Find the area of the quadrilateral whose vertices taken in order are (-4, -2), (-3, -5), (3, -2) and (2, 3).

Sol. Join A and C. The given points are

A(-4, -2), B(-3, -5), C(3, -2) and D(2, 3)

Area of triangle ABC = (1/2) |(-4)(-5 + 2) - 3(-2 + 2) + 3(-2 + 5)|

= (1/2) |12 + 0 + 9| = 21/2 = 10.5 sq. units

Area of triangle ACD = (1/2) |(-4)(-2 - 3) + 3(3 + 2) + 2(-2 + 2)|

= (1/2) |20 + 15| = 35/2 = 17.5 sq. units.

Area of quadrilateral ABCD = ar(triangle ABC) + ar(triangle ACD)

= (10.5 + 17.5) sq. units = 28 sq. units.

5. A median of a triangle divides it into two triangles of equal areas. Verify this result for triangle ABC whose vertices are A(4, -6), B(3, -2) and C(5, 2).

Sol. Here, the vertices of the triangles are A(4, -6), B(3, -2) and C(5, 2).

Let D be the midpoint of BC.

Therefore, the coordinates of the midpoint D are ((3 + 5)/2, (-2 + 2)/2) or (4, 0).

Since, AD divides the triangle ABC into two parts i.e., triangle ABD and triangle ACD,

Now, ar(△ABD)

= 1/2 | 4{(-2) - 0} + 3(0 + 6) + 4(-6 + 2) |

= 1/2 | (-8) + 18 + (-16) | = 1/2 | -6 |

= 3 sq. units

...(1)

ar(△ADC) = 1/2 | 4(0 - 2) + 4(2 + 6) + 5(-6 - 0) |

= 1/2 | -8 + 32 - 30 | = 1/2 | -6 |

= 3 sq. units

...(2)

From (1) and (2)

ar(△ABD) = ar(△ADC)

Hence, a median divides the triangle into two triangles of equal areas.

5.0Benefits of NCERT Solutions Class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.3

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