Question

Find the area of the triangle whose vertices are :
(i) $(2,3),(-1,0),(2,-4)$
(ii) $(-5,-1),(3,-5),(5,2)$

Answer 1

(i) Let $A(2,3)$, $B(-1,0)$ and $C(2,-4)$ are the vertices of $△ABC$

Area of triangle=$\frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]$

Here $(x_1,y_1)=(2,3)$

$(x_2,y_2)=(-1,0)$

$(x_3,y_3)=(2,-4)$

$=\frac{1}{2}\left[2\right(0+4)-1(-4-3)+2(3-0\left)\right]$

$=\frac{1}{2}[8+7+6]$

$=\frac{21}{2}$ sq.unit

(i) Let $A(-5,1)$, $B(3,-5)$ and $C(5,2)$ are the vertices of $△ABC$

Area of triangle=$\frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]$

Here $(x_1,y_1)=(-5,-1)$

$=(x_2,y_2)=(3,-5)$

$=(x_3,y_3)=(5,2)$

$=\frac{1}{2}[-5(-5-2)+3(2+1)+5(-1+5\left)\right]$

$=\frac{1}{2}[35+9+20]$

$=\frac{64}{2}=32$ sq.unit