NTSE Stage 2 SAT 2017 Question Paper

Let the third point be C(x, y)
Then,
⇒ $\mathrm{CA}=\mathrm{CB}=\mathrm{AB}$
⇒ ${\mathrm{CA}}^2={\mathrm{CB}}^2={\mathrm{AB}}^2$
⇒ $2\mathrm{x}-6\mathrm{y}=0$
⇒ $\mathrm{x}=3\mathrm{y}$ ....(1)
∴ ${\left(\mathrm{x}-1\right)}^2+{\left(\mathrm{y}-2\right)}^2=10$
⇒ $2{\mathrm{y}}^2-2\mathrm{y}-1=0$
On solving we get,
⇒ $\mathrm{y}=1\pm \sqrt{3}$
∴ $\mathrm{x}=3\left(1\pm \sqrt{3}\right)$
Hence, C(x, y) lies in either 1st quadrant or 3rd quadrant.
So, C cannot lies in the second quadrant