Find the area of the triangle whose vertices are :
Question.

Find the area of the triangle whose vertices are :

(i) (2,3), (–1, 0), (2, –4)

(ii) (– 5, – 1), (3,–5), (5,2)

Solution:

(i) Let the vertices of the triangles be A(2, 3), B (–1, 0) and C(2, –4)

Here $x_{1}=2, y_{1}=3$

$x_{2}=-1, y_{2}=0$

$x_{3}=2, y_{3}=-4$

$\because \quad$ Area of a $\Delta$

$=\frac{1}{\boldsymbol{2}}\left[\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right]$

$\therefore \quad$ Area of a $\Delta$

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

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

$=\frac{1}{\mathbf{2}}[8+7+6]=\frac{1}{\mathbf{2}}[21]=\frac{\mathbf{2 1}}{\mathbf{2}}$ sq.units

(ii) $\mathrm{A}(-5,-1), \mathrm{B}(3,-5), \mathrm{C}(5,2)$ are the vertices of the given triangle.

$x_{1}=-5, x_{2}=3, x_{3}=5 ; y_{1}=-1, y_{2}=-5, y_{3}=2$

Area of the $\triangle \mathrm{ABC}$

$=\frac{\mathbf{1}}{\mathbf{2}}\left[\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right]$

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

$=\frac{1}{\mathbf{2}}[35+9+20]=\frac{1}{\mathbf{2}}[64]=32$ sq. units
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