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 

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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

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.