Question

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

(−2, 0), (0, 4), (0, k)

Answer 1

\[\text{If the area of a triangle with vertices ( - 2, 0) (0, 4) and (0, k) is 4 square units, then }\] 

\[ ∆_1 = \frac{1}{2}\begin{vmatrix} - 2 & 0 & 1\\ 0 & 4 & 1\\ 0 & k & 1 \end{vmatrix}\] 

\[ = \frac{1}{2} \left\{ - 2 \times \begin{vmatrix} 4 & 1\\k & 1\end{vmatrix} \right\} \left[\text{ Expanding along }C_1 \right]\] 
\[ = - \left( 4 - k \right)\] 
Since area is always +ve, we take its absolute value, which is given as 4 square units . 
\[ \Rightarrow - \left( 4 - k \right) = \pm 4\] 
\[ \Rightarrow - \left( 4 - k \right) = \pm 4\] 
\[ \Rightarrow - \left( 4 - k \right) = 4 or - \left( 4 - k \right) = - 4\] 
\[ \Rightarrow k = 4 + 4 or k = - 4 + 4\] 
\[ \Rightarrow k = 8 \text{ or }k = 0\]