Q

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

Q : 3       Find values of k if area of triangle is 4 sq. units and vertices are

(ii)   $(-2,0), (0,4), (0,k)$

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The area of the triangle is given by the formula:

$\triangle = \frac{1}{2} \begin{vmatrix} -2 &0 &1 \\ 0 & 4 & 1\\ 0& k & 1 \end{vmatrix} = 4\ sq.\ units.$

Now, calculating the area:

$= \frac{1}{2} \left | -2(4-k)-0(0-0)+1(0-0) \right | = \frac{1}{2} \left | -8+2k \right | = 4$

or  $-8+2k =\pm 8$

Therefore we have two possible values of 'k' i.e.,  $k = 8$  or  $k = 0$.

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