Question

find the area of a
triangle whose vertices pare
(1,-1) (-4.6) & (-3-5) ​

Answer 1

Solution :-

Let A ( 1 , -1 ), B ( -4 , 6 ) and C ( -3 , -5 ) be the vertices of the triangle.

We know :-

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

Here :-

$\bullet \sf \ x_1 = 1 \ , \ y_1 = -1 \\\\\bullet \ x_2 = -4 \ , \ y_2 = 6 \\\\\bullet \ x_3 = -3 \ , \ y_3 = -5$

Area of triangle ABC :-

$\longrightarrow \sf \dfrac{1}{2} \times [ \ 1(6-(-5)) +-4 (-1-(-5)) +-3 (-1-6) \ ] \\\\\longrightarrow \dfrac{1}{2} \times [ \ 1(6+5)+-4 (-1+5)+-3(-1-6) \ ] \\\\\longrightarrow \dfrac{1}{2} \times [ \ (1 \times 11 ) +(-4 \times -4 ) + ( -3 \times -7 ) \ ] \\\\\longrightarrow \dfrac{1}{2} \times [ \ 11+16+21 \ ] \\\\\longrightarrow \dfrac{1}{2} \times 48 \\\\\longrightarrow \bf 24 \ units$