# Q : 3          Find values of k if area of triangle is 4 sq. units and vertices are                  (i) $(k,0), (4,0), (0,2)$

We can easily calculate the area by the formula :

$\triangle = \frac{1}{2} \begin{vmatrix} k &0 &1 \\ 4& 0& 1\\ 0 &2 & 1 \end{vmatrix} = 4\ sq.\ units$

$= \frac{1}{2}\left [ k\begin{vmatrix} 0 &1 \\ 2& 1 \end{vmatrix} -0\begin{vmatrix} 4 &1 \\ 0 & 1 \end{vmatrix}+1\begin{vmatrix} 4 &0 \\ 0& 2 \end{vmatrix} \right ]= 4\ sq.\ units$

$=\frac{1}{2}\left [ k(0-2)-0+1(8-0) \right ] = \frac{1}{2}\left [ -2k+8 \right ] = 4\ sq.\ units$

$\left [ -2k+8 \right ] = 8\ sq.\ units$    or   $-2k +8 = \pm 8\ sq.\ units$

or  $k = 0$    or  $k = 8$

Hence two values are possible for k.

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