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Home  >>  CBSE XII  >>  Math  >>  Determinants
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Find area of the triangle with vertices at the point given in each of the following (-2, -3), (3, 2), (-1, -8)

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  • The area of a triangle whose vertices are $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ is given by
  • $\bigtriangleup=\frac{1}{2}\begin{vmatrix}x_1 & y_1 & 1\\x_2 & y_2 & 1\\x_3 &y_3 &1\end{vmatrix}$
  • $\bigtriangleup=\frac{1}{2}[x_1(y_2-y_3)-y_1(x_2-x_3)+1(x_2y_3-y_2x_3)]$
Let $(x_1,y_1)$ be=(-2,-3)
 
Let $(x_2,y_2)$ be=(3,2)
 
Let $(x_3,y_3)$ be=(-1,-8)
 
Now area of a triangle is given by
 
$\bigtriangleup=\frac{1}{2}\begin{vmatrix}x_1 & y_1 & 1\\x_2 & y_2 & 1\\x_3 & y_3 & 1\end{vmatrix}$
 
Substituting the respective values we get,
 
$\bigtriangleup=\frac{1}{2}\begin{vmatrix}-2 & -3 & 1\\3 & 2 & 1\\-1 & -8 & 1\end{vmatrix}$
 
Now expanding along the row $R_1$ we get
 
$\mid\bigtriangleup\mid=\frac{1}{2}\begin{bmatrix}-2\begin{vmatrix}2 & 1\\-8 & 1\end{vmatrix}-(-3)\begin{vmatrix}3 & 1\\-1 & 1\end{vmatrix}+1\begin{vmatrix}3 & 2\\-1 & -8\end{vmatrix}\end{bmatrix}$
 
$=\frac{1}{2}[-2(2+8)+3(3+1)+1(-24+2)]$
 
$=\frac{1}{2}[-20+12-22]=-30/2=-15.$
 
Hence the area is 15sq.units.

 

answered Feb 23, 2013 by pady_1
edited Feb 24, 2013 by vijayalakshmi_ramakrishnans
 

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