In $∆\mathrm{ABC}, \mathrm{A}=(1, 2); \mathrm{B}=(5, 5), \angle \mathrm{ACB}={90}^0$. If area of $∆\mathrm{ABC}$ is to be 6.5 sq. units, then the possible number of points for C is

$$\begin{gathered} \mathrm{Given} \mathrm{A}=(1,2), \mathrm{B}=(5,5) \mathrm{and} \\ \angle \mathrm{ACB}={90}^{° } \mathrm{and} \mathrm{AB}=\sqrt{25}=5 \\ \mathrm{Now} \mathrm{area} \mathrm{of} {∆}^{\mathrm{le}}\mathrm{ABC}=\frac{1}{2}(\mathrm{h} \mathrm{cos\theta })(\mathrm{h} \mathrm{sin\theta }) \\ =\frac{{\mathrm{h}}^2}{2}\sin \mathrm{\theta } \cos \mathrm{\theta } \\ =\frac{{\mathrm{h}}^2}{4}\sin 2\mathrm{\theta } \\ =\frac{{\mathrm{h}}^2}{4}(\because -1\le \sin 2\mathrm{\theta }\le 1) \\ \mathrm{maximum} \mathrm{area} =\frac{{\mathrm{h}}^2}{4}=\frac{25}{4}=6.25 \\ \mathrm{But} \mathrm{given} \mathrm{area} \mathrm{of} {∆}^{\mathrm{le}}\mathrm{ABC}=6.5 \\ \mathrm{this} \mathrm{is} \mathrm{impossible}, \mathrm{hence} \mathrm{no} \mathrm{such} \mathrm{points} \mathrm{for} \text{'}\mathrm{c}\text{'} \mathrm{exists}. \\ \Rightarrow \mathrm{Number} \mathrm{of} \mathrm{possible} \mathrm{point}s \mathrm{for} C=0 \end{gathered}$$