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Find values of \(k\) if area of triangle is\(4\)sq. units and vertices are:\( (-2, 0),(0, 4),(0, k)\).
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Find values of \(k\) if area of triangle is\(4\)sq. units and vertices are:\( (-2, 0),(0, 4),(0, k)\).

Answer

 \(8\)and\(0\)
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Solution

Given: Verticles of triangle are \( (-2, 0),(0, 4),(0, k)\)
To Find: Value of \(k\)
Solution:
We know that
 Area of triangle = \(\frac{1}{2}|\left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1 \\ {{x_2}}&{{y_2}}&1 \\ {{x_3}}&{{y_3}}&1 \end{array}} \right| |\)
\(\Rightarrow \frac{1}{2}|\left\vert \begin{array}{ccc}{x}_{1}& {y}_{1}& 1\\ {x}_{2}& {y}_{2}& 1\\ {x}_{3}& {y}_{3}& 1\end{array}\right\vert| =4\)
\(\Rightarrow \frac{1}{2}\left| {\begin{array}{*{20}{c}} { - 2}&0&1 \\ 0&4&1 \\ 0&k&1 \end{array}} \right| =\pm 4\)
\(\Rightarrow {\frac{1}{2}\left[ { - 2\left( {4 - k} \right) - 0 + 1\left( {0 - 0} \right)} \right]} = \pm4\)
\(\Rightarrow {\frac{1}{2}\left( { - 8 + 2k} \right)} =\pm 4\)
\(\Rightarrow k-4=\pm 4\)
Taking positive sign, \(k-4=4\)
\(\Rightarrow k=8\)
Taking negative sign, \(k-4 = - 4\)
\(\Rightarrow k = 0\)
Hence, the values of k are \(8\)and\(0\).
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