Question

Find the area of the triangle whose vertices are (3,8), (-4,2), (5,1)​

Answer 1

Answer:Let A(3,8),B(−4,2),C(5,−1) be the vertices of the given △ABC.  

Then,

(x  

1

​  

=3,y  

1

​  

=8),(x  

2

​  

=−4,y  

2

​  

=2),(x  

3

​  

=5,y  

3

​  

=−1)

Area of △ABC =  

2

1

​  

∣[x  

1

​  

(y  

2

​  

−y  

3

​  

)+x  

2

​  

(y  

3

​  

−y  

1

​  

)+x  

3

​  

(y  

1

​  

−y  

2

​  

)]∣

=  

2

1

​  

∣3[2−(−1)]−4(−1−8)+5(8−2)∣

=  

2

1

​  

∣9+36+30∣=  

2

75

​  

=37.5 sq.units

Step-by-step explanation: