# Find values of k, if area of triangle is 4 square units whose vertices are

Question:

Find values of k, if area of triangle is 4 square units whose vertices are

(i) (k, 0), (4, 0), (0, 2)
(ii) (−2, 0), (0, 4), (0, k)

Solution:

(i)

If the area of a triangle with vertices $(k, 0),(4,0)$ and $(0,2)$ is 4 square units, then

$\Delta=\frac{1}{2} \mid k \quad 0 \quad 1$

$4 \quad 0 \quad 1$

$0 \quad 2 \quad 1 \mid$

$=\frac{1}{2}\{(2) \times \mid \mathrm{k} \quad 1$

$4 \quad 1 \mid\} \quad$ [Expanding along $\mathrm{C}_{2}$ ]

$=(\mathrm{k}-4)$

Since area is always $+$ ve, we take its absolute value, which is given as 4 square units.

$\Rightarrow(\mathrm{k}-4)=\pm 4$

$\Rightarrow(k-4)=4$ or $(k-4)=-4$

$\Rightarrow k-4=4$ or $k-4=-4$

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

$\Rightarrow k=8,0$

(ii)

If the area of a triangle with vertices $(-2,0)(0,4)$ and $(0, k)$ is 4 square units, then

$\Delta_{1}=\frac{1}{2} \mid-2 \quad 0 \quad 1$

$\begin{array}{lll}0 & 4 & 1 \\ 0 & \mathrm{k} & 1\end{array}$

$=\frac{1}{2}\{-2 \times \mid 41$

k $1 \mid\}$

[Expanding along $\mathrm{C}_{1}$ ]

$=-(4-\mathrm{k})$

Since area is always $+$ ve, we take its absolute value, which is given as 4 square units.

$\Rightarrow-(4-\mathrm{k})=\pm 4$

$\Rightarrow-(4-\mathrm{k})=\pm 4$

$\Rightarrow-(4-\mathrm{k})=4$ or $-(4-\mathrm{k})=-4$

$\Rightarrow \mathrm{k}=4+4$ or $\mathrm{k}=-4+4$

$\Rightarrow k=8$ or $\mathrm{k}=0$