Area of triangle using determinants

Last updated at April 16, 2024 by

In a 2D plane,

For three points

   A (x 1 , y 1 ),

  B (x 2 , y 2 ),

  C (x 3 , y 3 )

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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
    Area of triangle using determinants - Part 2



    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

Area of triangle using determinants - Part 3

 

Check the questions below to learn more

 

Transcript

Area of ∆ = 1/2 |■8(𝑥_1&𝑦_1&1@𝑥_2&𝑦_2&1@𝑥_3&𝑦_3&1)| For points in 3D plane A (x1, y1, z1) B (x2, y2 , z2) C (x3, y3 , z3) Area of ∆ABC = 1/2 |■8(𝑥_1&𝑦_1&𝑧_1@𝑥_2&𝑦_2&𝑧_2@𝑥_3&𝑦_3&𝑧_3 )| Example If Area = 1/2 |■8(9&2&1@5&−1&6@4&0&−2)| = 1/2 × −90 = −45 Therefore, Area = 45 square units 1/2 |■8(1&3&1@0&0&1@𝑘&0&1)| = ± 3

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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, Social Science, Physics, Chemistry, Computer Science at Teachoo.