# If area of triangle is $35\; sq\; units$ with vertices $(2, -6), (5, 4)$ and $(k, 4)$. Then $k$ is

$(A) 12\qquad (B) -2\qquad (C)-12,-2\qquad (D) 12,-2$

Toolbox:
• The area of a triangle whose vertices are $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ is given by
• $\bigtriangleup=\frac{1}{2}\begin{vmatrix}x_1 & y_1 & 1\\x_2 & y_2 & 1\\x_3 &y_3 &1\end{vmatrix}$
• On expanding along row $R_1$
• $|\bigtriangleup|=\frac{1}{2}\begin{bmatrix}x_1\begin{vmatrix}y_2 & 1\\y_3 & 1\end{vmatrix}-y_1\begin{vmatrix}x_2 & 1\\x_3 & 1\end{vmatrix}+1\begin{vmatrix}x_2 & y_2\\x_3 & y_3\end{vmatrix}\end{bmatrix}$
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.