Question

find area of triangle ABC whose vertices are A(1,3) B(1,0) C(3,0)​

Answer 1

Solution :-

Given :-

The points A ( 1 , 3 ), B ( 1 , 0 ) and C ( 3 , 0 ) are the vertices of the triangle ABC.

We know :-

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

Here :-

$\bullet \sf \ x_1= 1 \ , \ y_1 = 2 \\\\\bullet \ x_2 = 1 \ , \ y_2 = 0 \\\\\bullet \ x_3 = 3 \ , \ y_3 = 0$

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

Area of triangle ABC = 3 units