Find values of k if area of triangle is 4 sq. units and vertices are
(i) $(k,0),(4,0),(0,2)$
(ii) $( - 2,0),(0,4),(0,k)$

(i) Area of triangle $=$ 4 sq.

units
Also,
area $= \cfrac{1}{2}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}k&0&1\\4&0&1\\0&2&1\end{array}} \right|$

$= \cfrac{1}{2}[k(0 - 2) - 0 + 1(8 - 0)]$

[Expanding along${R_1}$ ]
$= \cfrac{1}{2}[ - 2k + 8] = - k + 4$

Now, $- k + 4 = \pm 4 \Rightarrow - k + 4 = 4$ or $- k + 4 = - 4$

$\Rightarrow$ $k = 0$ or $k = 8$ $\Rightarrow$ $k = 0,8$

(ii) Area of triangle $=$ 4 sq. units

Also, area $= \cfrac{1}{2}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 2}&0&1\\0&4&1\\0&k&1\end{array}} \right|$

[Expanding along${R_1}$]
$= \cfrac{1}{2}[ - 2(4 - k) - 0 + 1(0)] = - 4 + k$

Now,
$- 4 + k = \pm 4 \Rightarrow - 4 + k = + 4$ or $- 4 + k = - 4$

$\Rightarrow$ $k = 0$ or $k = 8$ $\Rightarrow$ $k = 0,8$