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Question
Find values of k, if area of triangle is 4 square units whose vertices are
(−2, 0), (0, 4), (0, k)
Solution
\[\text{If the area of a triangle with vertices ( - 2, 0) (0, 4) and (0, k) is 4 square units, then }\]
\[ ∆_1 = \frac{1}{2}\begin{vmatrix} - 2 & 0 & 1\\ 0 & 4 & 1\\ 0 & k & 1 \end{vmatrix}\]
\[ = \frac{1}{2} \left\{ - 2 \times \begin{vmatrix} 4 & 1\\k & 1\end{vmatrix} \right\} \left[\text{ Expanding along }C_1 \right]\]
\[ = - \left( 4 - k \right)\]
Since area is always +ve, we take its absolute value, which is given as 4 square units .
\[ \Rightarrow - \left( 4 - k \right) = \pm 4\]
\[ \Rightarrow - \left( 4 - k \right) = \pm 4\]
\[ \Rightarrow - \left( 4 - k \right) = 4 or - \left( 4 - k \right) = - 4\]
\[ \Rightarrow k = 4 + 4 or k = - 4 + 4\]
\[ \Rightarrow k = 8 \text{ or }k = 0\]
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