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Home  >>  CBSE XII  >>  Math  >>  Determinants
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Find the area of the triangle where vertices are $A(11,7),B(5,5)\; and\; C(-1,3)$

$\begin{array}{1 1} 10 \\ 8 \\ 0 \\ 4 \end{array} $

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  • Area of the triangle where veritices are $A(x_1,y_1), B(x_2,y_2),C(x_3,y_3)$ is given by
  • $\Delta=\large\frac{1}{2}\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} $
Let the vertices of the triangle be $A(x_1y_1)$ be $(11,7)$ $B(x_2y_2)$ be $(5,5)$ and $C(x_3y_3)$ be $(-1,3)$
we know the area of the triangle is
$\Delta=\large\frac{1}{2}\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} $
Now substituting for these values we get
$\Delta=\large\frac{1}{2}\begin{vmatrix} 11 & 7 & 1 \\ 5 & 5 & 1 \\ -1 & 3 & 1 \end{vmatrix} $
Apply $R_2 \to R_2 -R_3 \;and \; R_3 \to R_3 -R_1$
$\Delta=\large\frac{1}{2}\begin{vmatrix} 11 & 7 & 1 \\ 6 & 2 & 0 \\ -12 & -4 & 0 \end{vmatrix} $
Now expanding along $R_1$ we get
$\Delta=\bigg[11(0)-7(0)+1(-24+24)\bigg]$
$=0$
Hence the points are collinear
answered Apr 8, 2013 by meena.p
 

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