Question

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

• A. 0 or 8
• B. 0 or −8
• C. $0$
• D. $8$

Answer 1

Answer: (A)

Let $A(K,0),B(4,0),C(0,2)$
Now, we have
ar $(\Delta ABC)=4$
$\therefore \frac{1}{2}|K(0-2)+4(2-0)+0(0-0)|=4$
$\Rightarrow \frac{1}{2}|-2K+8|=4$
$\Rightarrow |-2K+8|=8$
$\Rightarrow -2K+8=\pm 8$
$\Rightarrow -2K=\pm 8-8$
$\Rightarrow -2K=0$ or 16
$\Rightarrow K=0$ or 8