- Books Name
- Mathmatics Book Based on NCERT
- Publication
- KRISHNA PUBLICATIONS
- Course
- CBSE Class 12
- Subject
- Mathmatics
Area of a Triangle
The area of a triangle formed by the vertices - (x1, y1), (x2, y2) and (x3, y3) is given by
½[x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)] sq.units
It can be expressed in the form of a determinant.
Note: We always take 1 in the last column of the determinant.
Expanding along the first column, we get
= x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)
The area of a triangle formed by the vertices - (x1, y1), (x2, y2) and (x3, y3) –
Note:
In calculating the area of a triangle by using this formula, we need to take the absolute value of the above determinant to avoid negative values, if any.
If area is given, use both positive and negative values of the determinant for the calculation.
The area of a triangle formed by three collinear points is equal to zero.
Example: Find out the area of the triangle whose vertices are given by A(0,0) , B (3,1) and C (2,4).
Solution: Using determinants we can find out the area of the triangle obtained by joining these points using the formula
Example : If (k, 2), (2, 4) and (3, 2) are vertices of the triangle of area 4 square units then determine the value of k.
Solution:
Area of triangle = 4 square units
- (1/2){k [4 – 2] – 2[2 – 3] + 1[4 – 12]} = 4
- k(2) – 2(-1) + 1(-8) = 8
- 2k + 2 – 8 = 8
- 2k – 6 = 8
- 2k = 8 + 6
- 2k = 14 => k = 7
So, the value of k is 7.
- (1/2){k [4 – 2] – 2[2 – 3] + 1[4 – 12]} = -4
- k(2) – 2(-1) + 1(-8) = -8
- 2k + 2 – 8 = -8
- 2k – 6 = -8
- 2k = -8 + 6
- 2k = -2 => k = -1
So, the value of k is -1.