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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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## QuestionMathsClass 12

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

$$8$$and$$0$$
4.6
4.6

## 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$$0$$.