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Home  >>  CBSE XII  >>  Math  >>  Determinants
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Find the value of $k$ if area of triangle is $4\; sq. units$ and vertices are: $(k, 0), (4, 0), (0, 2)$

$\begin{array}{1 1} 8 \\ 4 \\ 0 \\ -2 \end{array} $

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1 Answer

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Toolbox:
  • The area of a triangle whose vertices are $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ is given by
  • $\bigtriangleup=\begin{vmatrix}x_1 & y_1 & 1\\x_2 & y_2 & 1\\x_3 &y_3 &1\end{vmatrix}$
  • $\mid\bigtriangleup\mid=\frac{1}{2}[x_1(y_2-y_3)-y_1(x_2-x_3)+1(x_2y_3-y_2x_3)]$
Let $(x_1,y_1)=(k,0)$
 
$\;\;\;(x_2,y_2)=(4,0)$
 
$\;\;\;(x_3,y_3)=(0,2)$
 
Given the area of the triangle is
 
$\bigtriangleup=\begin{vmatrix}x_1 & y_1 & 1\\x_2 & y_2 & 1\\x_3 &y_3 &1\end{vmatrix}$
 
But the area given is 4sq.units.
 
Substituting the respective values we get,
 
$4=\frac{1}{2}\begin{vmatrix}k & 0 & 1\\4 & 0 & 1\\0 & 2 & 1\end{vmatrix}$
 
Now expanding along the row $R_1$
 
$4=\frac{1}{2}\begin{bmatrix}k\begin{vmatrix}0 & 1\\2 & 1\end{vmatrix}-0\begin{vmatrix}4 & 1\\0 & 1\end{vmatrix}+1\begin{vmatrix}4 & 0\\0 & 2\end{vmatrix}\end{bmatrix}$
 
$4=\pm[\frac{1}{2}[k(0-2)-0+1(8-0)]]$
 
Since the area can only be positive, the value of the determinant is modulus of | -2k + 8|.
That is it will be either +(-2k+8 ) or - (2k + 8).
 
8=+[-2k+8]
 
$\Rightarrow$ then 2k=0.
 
Hence k=0.
 
If 8=-[-2k+8]
 
8=-[-2k+8]
 
8=2k-8$\Rightarrow 2k=16.$
 
or $k=8.$
 
Hence the value of k is (0,8).

 

answered Feb 22, 2013 by sreemathi.v
edited Feb 24, 2013 by vijayalakshmi_ramakrishnans
 
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