Let $A(h, k), B(1,1)$ and $C(2,1)$ be the vertices of a right angled triangle with $A C$ as its hypotenuse. If the area of the triangle is $1$, then the set of values which ' $\mathrm{k}$ ' can take is given by

• A

  $\{1,3\}$
• B

  $\{0,2\}$
• Correct answer
  C

  $\{-1,3\}$
• D

  $\{-3,-2\}$

Solution

$\frac{1}{2} \times 1(k-1)=\pm 1$ $k-1=\pm 2$ $k=3$ $k=-1$

Hence, the correct answer is option: 'C'.