Check sibling questions

In a 2D plane,

For three points

Β  Β A (x 1 , y 1 ),

Β  B (x 2 , y 2 ),

Β  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

Β 

Check the questions below to learn more

Β 


Transcript

Area of βˆ† = 1/2 |β– 8(π‘₯_1&𝑦_1&1@π‘₯_2&𝑦_2&1@π‘₯_3&𝑦_3&1)| For points in 3D plane A (x1, y1, z1) B (x2, y2 , z2) C (x3, y3 , z3) Area of βˆ†ABC = 1/2 |β– 8(π‘₯_1&𝑦_1&𝑧_1@π‘₯_2&𝑦_2&𝑧_2@π‘₯_3&𝑦_3&𝑧_3 )| Example If Area = 1/2 |β– 8(9&2&1@5&βˆ’1&6@4&0&βˆ’2)| = 1/2 Γ— βˆ’90 = βˆ’45 Therefore, Area = 45 square units 1/2 |β– 8(1&3&1@0&0&1@π‘˜&0&1)| = Β± 3

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.