Question

What is the area, in square units, of a triangle with vertices at (-1,-1.)( 3, -1)and (2, 2)

Answer 1

Solution :-

Let,

Points of the vertices of the triangle be A ( -1 , -1 ), B ( 3 , -1 ) and C ( 2 , 2 ).

$\bf Area \ of \ triangle = \dfrac{1}{2} \times [ \ x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2) \ ]$

Here,

$\bullet\sf \ x_1= -1 \ , \ y_1 = -1 \\\\\bullet \ x_2 = 3 \ , \ y_2 = -1 \\\\\bullet \ x_3 = 2 \ , \ y_3 = 2$

Area of triangle ABC :-

$\longrightarrow \sf \dfrac{1}{2} \times [ \ -1(-1-2)+3(2-(-1))+2(-1-(-1) \ ] \\\\\longrightarrow \dfrac{1}{2} \times [ \ -1(-1-2)+3(2+1)+2(-1+1) \ ] \\\\\longrightarrow \dfrac{1}{2} \times [ \ (-1 \times - 3)+( 3 \times 3 ) +( 2 \times 0 ) \ ] \\\\\longrightarrow \dfrac{1}{2} \times [ \ 3+9+0) \ ] \\\\\longrightarrow \dfrac{1}{2} \times 12 \\\\\longrightarrow 6 \ sq.units$

Area of triangle ABC = 6 sq.units

Answer 2

Answer:

Solution :-

Let,

Points of the vertices of the triangle be A ( -1 , -1 ), B ( 3 , -1 ) and C ( 2 , 2 ).

\bf Area \ of \ triangle = \dfrac{1}{2} \times [ \ x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2) \ ]Area of triangle=

2

1

×[ x

1

(y

2

−y

3

)+x

2

(y

3

−y

1

)+x

3

(y

1

−y

2

) ]

Here,

\begin{gathered}\bullet\sf \ x_1= -1 \ , \ y_1 = -1 \\\\\bullet \ x_2 = 3 \ , \ y_2 = -1 \\\\\bullet \ x_3 = 2 \ , \ y_3 = 2\end{gathered}

∙ x

1

=−1 , y

1

=−1

∙ x

2

=3 , y

2

=−1

∙ x

3

=2 , y

3

=2

Area of triangle ABC :-

\begin{gathered}\longrightarrow \sf \dfrac{1}{2} \times [ \ -1(-1-2)+3(2-(-1))+2(-1-(-1) \ ] \\\\\longrightarrow \dfrac{1}{2} \times [ \ -1(-1-2)+3(2+1)+2(-1+1) \ ] \\\\\longrightarrow \dfrac{1}{2} \times [ \ (-1 \times - 3)+( 3 \times 3 ) +( 2 \times 0 ) \ ] \\\\\longrightarrow \dfrac{1}{2} \times [ \ 3+9+0) \ ] \\\\\longrightarrow \dfrac{1}{2} \times 12 \\\\\longrightarrow 6 \ sq.units\end{gathered}

⟶

2

1

×[ −1(−1−2)+3(2−(−1))+2(−1−(−1) ]

⟶

2

1

×[ −1(−1−2)+3(2+1)+2(−1+1) ]

⟶

2

1

×[ (−1×−3)+(3×3)+(2×0) ]

⟶

2

1

×[ 3+9+0) ]

⟶

2

1

×12

⟶6 sq.units

hope it helps you