# Find the value of $k$ if area of triangle is $4\; sq. units$ and vertices are: $(k, 0), (4, 0), (0, 2)$

$\begin{array}{1 1} 8 \\ 4 \\ 0 \\ -2 \end{array}$

Toolbox:
• The area of a triangle whose vertices are $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ is given by
• $\bigtriangleup=\begin{vmatrix}x_1 & y_1 & 1\\x_2 & y_2 & 1\\x_3 &y_3 &1\end{vmatrix}$
• $\mid\bigtriangleup\mid=\frac{1}{2}[x_1(y_2-y_3)-y_1(x_2-x_3)+1(x_2y_3-y_2x_3)]$
Let $(x_1,y_1)=(k,0)$

$\;\;\;(x_2,y_2)=(4,0)$

$\;\;\;(x_3,y_3)=(0,2)$

Given the area of the triangle is

$\bigtriangleup=\begin{vmatrix}x_1 & y_1 & 1\\x_2 & y_2 & 1\\x_3 &y_3 &1\end{vmatrix}$

But the area given is 4sq.units.

Substituting the respective values we get,

$4=\frac{1}{2}\begin{vmatrix}k & 0 & 1\\4 & 0 & 1\\0 & 2 & 1\end{vmatrix}$

Now expanding along the row $R_1$

$4=\frac{1}{2}\begin{bmatrix}k\begin{vmatrix}0 & 1\\2 & 1\end{vmatrix}-0\begin{vmatrix}4 & 1\\0 & 1\end{vmatrix}+1\begin{vmatrix}4 & 0\\0 & 2\end{vmatrix}\end{bmatrix}$

$4=\pm[\frac{1}{2}[k(0-2)-0+1(8-0)]]$

Since the area can only be positive, the value of the determinant is modulus of | -2k + 8|.
That is it will be either +(-2k+8 ) or - (2k + 8).

8=+[-2k+8]

$\Rightarrow$ then 2k=0.

Hence k=0.

If 8=-[-2k+8]

8=-[-2k+8]

8=2k-8$\Rightarrow 2k=16.$

or $k=8.$

Hence the value of k is (0,8).

edited Feb 24, 2013