The area of triangle with vertices (K, 0), (4, 0), (0, 2) is 4 square units, then value of K is

Concept:

The area of a triangle whose vertices are (x₁, y₁), (x₂, y₂) and (x₃, y₃), is given by the expression

\(\rm Area = \frac{1}{2}\begin{vmatrix} x_{1} &x_{2} & 1\\ y_{1}& y_{2} &1 \\ z_{1}&z_{2} & 1 \end{vmatrix}\)  

Calculation:

Given, Area of triangle, A = 4 sq. unit and vertices (K, 0), (4, 0), (0, 2) 

since the area is always Positive But the determinant can be both positive and negative.  

∴ Δ = ± 4 . 

⇒ ± 4 = \(\rm = \frac{1}{2}\begin{vmatrix} k &0 & 1\\ 4& 0 &1 \\ 0&2 & 1 \end{vmatrix}\)  

⇒ ± 4  \(\rm = \frac{1}{2} \left [ k(0-2) - 0 (4-0)+1(8-0)\right ]\) 

⇒ ± 8 = -2k + 8 

So, 8 = -2k + 8 or  -8 = -2k +8 

⇒ k = 0 or  8  .

The correct option is 2.