Question

i Find the area of the triangle =hose vertices are 3.8) (42) & (5.1) by using​

Answer 1

Answer:

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