Area of triangle

Chapter 4 Class 12 Determinants
Concept wise

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 )

## 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

Β

Check the questions below to learn more

Β

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### Transcript

Area of β = 1/2 |β 8(π₯_1&π¦_1&[email protected]π₯_2&π¦_2&[email protected]π₯_3&π¦_3&1)| For points in 3D plane A (x1, y1, z1) B (x2, y2 , z2) C (x3, y3 , z3) Area of βABC = 1/2 |β 8(π₯_1&π¦_1&π§[email protected]π₯_2&π¦_2&π§[email protected]π₯_3&π¦_3&π§_3 )| Example If Area = 1/2 |β 8(9&2&[email protected]&β1&[email protected]&0&β2)| = 1/2 Γ β90 = β45 Therefore, Area = 45 square units 1/2 |β 8(1&3&[email protected]&0&[email protected]π&0&1)| = Β± 3