31. Find the area of a triangle ABC whose vertices are A(2, 2) B(3,4) and C (–1,3).
Answers
Step-by-step explanation:
Area of triangle with vertices A (2,2) B(3,4) C(-1,3)
= 1/2[ 2(4-3) + 3(3-2) + (-1)( 2-4) ]
= 1/2 [ 2(1) + 3(1) - 1(-2) ]
= 1/2 [ 2+3+2]
= 7/2
Given,
For a ∆ABC,
Coordinates of point A = (2,2)
Coordinates of point B = (3,4)
Coordinates of point C = (-1,3)
To find,
The area of the ∆ABC.
Solution,
We can simply solve this mathematical problem using the following process:
Mathematically, according to the coordinate geometry formula;
The area of a triangle whose x and y coordinates of the three vertices are known is calculated as follows:
Area = 1/2{Ax(By - Cy) + Bx(Cy - Ay) + Cx(Ay - By)}
{Statement-1}
Now, according to the question;
For the given ∆ABC,
Ax = 2
Ay = 2
Bx = 3
By = 4
Cx = -1
Cy = 3
Now, according to the statement-1;
The area of the ∆ABC
= 1/2{Ax(By - Cy) + Bx(Cy - Ay) + Cx(Ay - By)}
= 1/2{(2)(4 - 3) + (3)(3 - 2) + (-1)(2 - 4)}
= 1/2{(2)(1) + (3)(1) + (-1)(-2)}
= 1/2 × (2+3+2) = 7/2 square units
= 3.5 square units
Hence, the area of the ∆ABC is equal to 3.5 square units.