If the points A (-2, -1), B(1, 0), C(x, 3) and D(1, y)

Solution:

Given: Vertices of the parallelogram are $A(-2,-1), B(1,0), C(x, 3)$ and $D(1, y)$.

To find: values of $x$ and $y$.

Since, $A B C D$ is a parallelogram, we have $A B=C D$ and $B C=D A$.

$A B=\sqrt{(1+2)^{2}+(0+1)^{2}}=\sqrt{9+1}$

$=\sqrt{10}$ units

$B C=\sqrt{(x-1)^{2}+9}$

$C D=\sqrt{(1-x)^{2}+(y-3)^{2}}$

$D A=\sqrt{9+(1+y)^{2}}$

Since $A B=C D$

$\Rightarrow \sqrt{10}=\sqrt{(1-\mathrm{x})^{2}+(\mathrm{y}-3)^{2}}$

Squaring both sides, we get

$\Rightarrow 10=(1-x)^{2}+(y-3)^{2}$

$\Rightarrow 10=1-2 x+x^{2}+y^{2}-6 y+9$

$\Rightarrow x^{2}+y^{2}-2 x-6 y=0 \ldots \ldots(1)$

Since BC = DA,

$\Rightarrow \sqrt{(x-1)^{2}+9}=\sqrt{9+(1+y)^{2}}$

Squaring both sides,

$\Rightarrow(x-1)^{2}+9=9+(1+y)^{2}$

$\Rightarrow x^{2}-2 x+1=1+2 y+y^{2}$

$\Rightarrow x^{2}-y^{2}-2 x-2 y=0 \ldots \ldots(2)$

Equation 1 - Equation 2 gives us,

$\Rightarrow 2 y^{2}-4 y=0$

$\Rightarrow y^{2}-2 y=0$

$\Rightarrow y(y-2)=0$

$\Rightarrow y=0$ or $y=2$

But $y \neq 0$ because then point $D(1,0)$ is same as $B(1,0)$

Therefore, y = 2

When y = 2, from equation 1,

$\Rightarrow x^{2}+4-2 x-12=0$

$\Rightarrow x^{2}-2 x-8=0$

$\Rightarrow(x-4) \times(x+2)=0$

$\Rightarrow x=4$ or $x=-2$

So, the possible set of values for x and y are:

$x=4, y=2$

$x=-2, y=2$

But when $x=-2$, then $C(-2,3) .$ Then $A B C D$ does not form a parallelogram.

Therefore, the only solution is $x=4$ and $y=2$.