NTSE Stage 2 SAT 2018 Exam Question Paper

Since, the distance between the points (x, y) and (1, 2) is 2 units, then
${\left(\mathit{x}-1\right)}^2+{\left(\mathit{y}-2\right)}^2=4$
This is possible only when,
Case 1
${\left(\mathit{x}-1\right)}^2=0$ and ${\left(\mathit{y}-2\right)}^2=4$
$\Rightarrow \mathit{x}=1$ and $\mathit{y}=0,4$
So the points are (1, 0), (1, 4)
Case II
${\left(\mathit{x}-1\right)}^2=4$ and ${\left(\mathit{y}-2\right)}^2=0$
$\Rightarrow \mathit{x}=3,-1$ and $\mathit{y}=2$
So the points are $\left(3,2\right),\left(3,\mathit{–}1\right)$
Hence, there are total 4 points with integral co-ordinates.