Q : 5         If the area of triangle is 35 sq units with vertices  \small (2,-6),(5,4)  and \small (k,4). Then k is

                 (A) \small 12      (B) \small -2     (C) \small -12,-2       (D) \small 12,-2
           

Answers (1)

Area of triangle is given by:

\triangle = \frac{1}{2} \begin{vmatrix} 2 &-6 &1 \\ 5& 4 & 1\\ k& 4& 1 \end{vmatrix} = 35\ sq.\ units.

or \begin{vmatrix} 2 &-6 &1 \\ 5& 4 & 1\\ k& 4& 1 \end{vmatrix} = 70\ sq.\ units.

2\begin{vmatrix} 4 &1 \\ 4& 1 \end{vmatrix}-(-6)\begin{vmatrix} 5 &1 \\ k &1 \end{vmatrix}+1\begin{vmatrix} 5 &4 \\ k&4 \end{vmatrix} = 70

2(4-4) +6(5-k)+(20-4k) = \pm70

50-10k = \pm70

 k = 12 or  k = -2 

Hence the possible values of k are 12 and -2.

Therefore option (D) is correct.

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