In a 2D plane,

For three points

   A (x 1 , y 1 ),

  B (x 2 , y 2 ),

  C (x 3 , y 3 )

9.jpg


Find area of triangle with vertices (3, 8), (-4, 2), (5, 1)

Check solution - Example 17


For points in 3D plane

A (x 1 , y 1 , z 1 )

B (x 2 , y 2 , z 2 )

C (x 3 , y 3 , z 3 )

10.jpg

There are some points to note:-

  1. If Area of triangle = 0,
    then the three points are collinear

  2. If value of determinant comes negative, we will take the
    positive value as area

    Example
    11.jpg


    Therefore,
    Area = 45 square units

  3. .If area is given,
    We take both positive and negative value of determinant
    Example
    If Area = 3 square units

12.jpg

 

Check the questions below to learn more

 

  1. Chapter 4 Class 12 Determinants
  2. Concept wise

Transcript

Area of โˆ† = 1/2 |โ– 8(๐‘ฅ_1&๐‘ฆ_1&1@๐‘ฅ_2&๐‘ฆ_2&1@๐‘ฅ_3&๐‘ฆ_3&1)| For points in 3D plane A (x1, y1, z1) B (x2, y2 , z2) C (x3, y3 , z3) Area of โˆ†ABC = 1/2 |โ– 8(๐‘ฅ_1&๐‘ฆ_1&๐‘ง_1@๐‘ฅ_2&๐‘ฆ_2&๐‘ง_2@๐‘ฅ_3&๐‘ฆ_3&๐‘ง_3 )| Example If Area = 1/2 |โ– 8(9&2&1@5&โˆ’1&6@4&0&โˆ’2)| = 1/2 ร— โˆ’90 = โˆ’45 Therefore, Area = 45 square units 1/2 |โ– 8(1&3&1@0&0&1@๐‘˜&0&1)| = ยฑ 3

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.