Check sibling questions

In a 2D plane,

For three points

   A (x 1 , y 1 ),

  B (x 2 , y 2 ),

  C (x 3 , y 3 )


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

Check solution - Example 17


Important Points

There are some points to note:-

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

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

    Example



    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

 

Check the questions below to learn more

 


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

  1. Chapter 4 Class 12 Determinants
  2. Concept wise

About the Author

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo