Question

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

Answer 1

Area of triangle having vertices as $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ is given as

$=\frac{1}{2}\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|$

(i) Given, area of triangle = 4 sq.units

$\Rightarrow 4=\frac{1}{2}\left|\begin{array}{ccc}k& 0& 1\\ 4& 0& 1\\ 0& 2& 1\end{array}\right|$

$\Rightarrow 4=\frac{1}{2}|k(0-2)-0(4-0)+1(8-0)|$

$\Rightarrow |8-2k|=8$

$\Rightarrow 8-2k=\pm 8$

$\Rightarrow 8-2k=8$ and $8-2k=-8$

$\Rightarrow k=0$ and $-2k=-16$

$\Rightarrow k=0$ and $k=8$

(ii) Given, area of triangle = 4 sq.units

$\Rightarrow 4=\frac{1}{2}\left|\begin{array}{ccc}2& 0& 1\\ 0& 4& 1\\ 0& k& 1\end{array}\right|$

$\Rightarrow 4=\frac{1}{2}|2(4-k)-0(0-0)+1(0-0)|$

$\Rightarrow |8-2k|=8$

$\Rightarrow 8-2k=\pm 8$

$\Rightarrow 8-2k=8$ and $8-2k=-8$

$\Rightarrow k=0$ and $-2k=-16$

$\Rightarrow k=0$ and $k=8$