Question

How many points (x,y) with integral coordinates are there whose distance from (1,2) is two units?

Answer 1

- Hint: For finding all points in integral coordinates we can apply distance formula.
According to distance formula if we have two points $({{x}_{1}},{{y}_{1}})$ and $({{x}_{2}},{{y}_{2}})$ then distance (d) between these points is
$d=\sqrt{{{({{x}_{1}}-{{x}_{2}})}^{2}}+{{({{y}_{1}}-{{y}_{2}})}^{2}}}$

Complete step-by-step solution -
Given points are (x,y) and (1,2)
Then distance (d) between these points is
$\Rightarrow d=\sqrt{{{(x-1)}^{2}}+{{(y-2)}^{2}}}$
But the given distance should be two units. So we can write
$\Rightarrow \sqrt{{{(x-1)}^{2}}+{{(y-2)}^{2}}}=2$
On squaring both side
$\Rightarrow {{\left( \sqrt{{{(x-1)}^{2}}+{{(y-2)}^{2}}} \right)}^{2}}={{2}^{2}}$
$\Rightarrow {{(x-1)}^{2}}+{{(y-2)}^{2}}=4$
$\Rightarrow {{(x-1)}^{2}}+{{(y-2)}^{2}}=4+0$
Now we can consider case 1
If ${{(x-1)}^{2}}=4$ and ${{(y-2)}^{2}}=0$,
On solving ${{(x-1)}^{2}}=4$
$\Rightarrow (x-1)=\sqrt{4}$
$\Rightarrow (x-1)=2\,or\,(x-1)=-2$
$\Rightarrow x=3\,or\,x=\,-1$
On solving ${{(y-2)}^{2}}=0$
$\Rightarrow {{(y-2)}^{2}}=0$
$\Rightarrow y=2$
Hence possible points are (3,2) and (-1,2)
Now we can consider case 2
If ${{(x-1)}^{2}}=0$ and ${{(y-2)}^{2}}=4$
On solving ${{(x-1)}^{2}}=0$
$\Rightarrow {{(x-1)}^{2}}=0$
$\Rightarrow x=1$
On solving ${{(y-2)}^{2}}=4$
$\Rightarrow {{(y-2)}^{2}}=4$
$\Rightarrow (y-2)=\sqrt{4}=\pm 2$
$\Rightarrow (y-2)=2\,or\,(y-2)=-2$
$\Rightarrow y=4\,or\,y=0$
Hence possible points are (1,4) and (1,0)
So there are a total 4 possible integral coordinates whose distance from (1,2) is two units.

Note: In this we need to consider all possible cases otherwise we will miss possible value of coordinates.
Also if $x=\sqrt{{{a}^{2}}}$ then the possible value of x can be $x=+a\,or\,x=-a$ because the square of negative numbers is also positive.