If area of a triangle is 35sq,units with vertices (2,-6),(5,4) and (k,4).Then k is\[(A)\;12\quad(B)\;-2\quad(C)\;-12,-2\quad(D)\;12.-2\]
The area of the triangle is given by
$\bigtriangleup=\frac{1}{2}\begin{vmatrix}x_1 & y_1 & 1\\x_2 & y_2 & 1\\x_3 &y_3 &1\end{vmatrix}$
Let (2,-6),(5,4) and (k,4) be ($x_1,y_1),(x_2,y_2),(x_3,y_3)$
Now substituting the respective values,we get
$0=\frac{1}{2}\begin{vmatrix}2 & -6 & 1\\5 & 4 & 1\\k & 4 & 1\end{vmatrix}$
Now expanding along the row $R_1$ we get
$35=\frac{1}{2}\begin{bmatrix}2\begin{vmatrix}4 & 1\\4 & 1\end{vmatrix}-(-6)\begin{vmatrix}5 & 1\\k & 1\end{vmatrix}+1\begin{vmatrix}5 & 4\\k& 4\end{vmatrix}\end{bmatrix}$
70=[2(4-4)+6(5-k)+1(20-4k)]
70=0+30-6k+20-4k
70=-10k+50
20=-10k$\Rightarrow k=-2.$
Hence B is the correct answer.