AREA OF TRIANGLE USING DETERMINANT FORMULA

About "Area of Triangle Using Determinant Formula"

Area of Triangle Using Determinant Formula :

Here we are going to see, how to find the area of triangle with given vertices using determinant formula.

We know that the area of a triangle whose vertices are  (x1 , y1),(x2 , y2) and (x3 , y3) is equal to the absolute value of

(1/2) [x1 y2 - x2y1 + x2y3 -  x3y2 + x3y1 - x1y3]

This expression can be written in the form of a determinant as the absolute value of

Note :

The area of the triangle formed by three points is zero if and only if the three points are collinear.

Also, we remind the reader that the determinant could be negative whereas area is always non negative.

Area of Triangle Using Determinant Formula - Practice questions

Question 1 :

Find the area of the triangle whose vertices are (0, 0), (1, 2) and (4, 3).

Solution :

Let the point be (x1, y1)  ==>  (0, 0), (x2, y2) ==> (1, 2) and (x3, y3) ==> (4, 3)

  =  (1/2)[3 - 8]

  =  (1/2)[-5]

  =  5/2

=  2.5

Area will not have negative. Hence 2.5 square units is the answer.

Question 2 :

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

Hence the value of k is 7.

Question 3 :

Find the area of the triangle whose vertices are (- 2, - 3), (3, 2), and (- 1, - 8).

  =  (1/2) [ -2(2 + 8) + 3(3+1) + 1(-24 + 2) ]

  =  (1/2) [-2 (10) + 3(4) + 1(-22)]

  =  (1/2) [ -20 + 12 - 22]

  =  (1/2) [ -42 + 12]

  =  (1/2) (-30)

  =  15  square units

After having gone through the stuff given above, we hope that the students would have understood, "Area of Triangle Using Determinant Formula". 

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