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.…

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
  1. $\{1,3\}$
  2. $\{0,2\}$
  3. $\{-1,3\}$
  4. $\{-3,-2\}$

Solution

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

Asked in: JEE Main 2007

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