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Question
Find values of k, if area of triangle is 4 square units whose vertices are
(k, 0), (4, 0), (0, 2)
Solution
\[\text{ If the area of a triangle with vertices (k, 0), (4, 0) and (0, 2) is 4 square units, then }\]
\[\Delta = \frac{1}{2}\begin{vmatrix} k & 0 & 1\\4 & 0 & 1\\0 & 2 & 1 \end{vmatrix} \]
\[ = \frac{1}{2} \left\{ \left( 2 \right) \times \begin{vmatrix} k & 1\\4 & 1 \end{vmatrix} \right\} \left[\text{ Expanding along }C_2 \right]\]
\[ = \left( k - 4 \right)\]
Since area is always +ve, we take its absolute value, which is given as 4 square units .
\[ \Rightarrow ( k - 4 ) = \pm 4\]
\[ \Rightarrow (k - 4) = 4 or (k - 4 ) = - 4\]
\[ \Rightarrow k - 4 = 4 or k - 4 = - 4\]
\[ \Rightarrow k = 8\text{ or }k = 0\]
\[ \Rightarrow k = 8, 0\]
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