Name the quadrilateral formed, if any,

Solution:

(i) A (−1,−2) , B(1,0), C(−1,2), D(−3,0)

Let A, B, C and D be the four vertices of the quadrilateral ABCD.

We know the distance between two points Pand Qis given by distance formula:

$\mathrm{PQ}=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

Hence

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

$\Rightarrow A B=\sqrt{(2)^{2}+(2)^{2}}$

$\Rightarrow A B=\sqrt{4+4}$

$\Rightarrow A B=\sqrt{4}$

 

$\Rightarrow A B=2 \sqrt{2}$

Similarly,

$\Rightarrow B C=\sqrt{((-1)-1)^{2}+(2-0)^{2}}$

$\Rightarrow B C=\sqrt{(-2)^{2}+(2)^{2}}$

$\Rightarrow B C=\sqrt{4+4}$

$\Rightarrow B C=\sqrt{8}$

 

$\Rightarrow B C=2 \sqrt{2}$

Similarly,

$\Rightarrow C D=\sqrt{((-3)-(-1))^{2}+(0-(2))^{2}}$

$\Rightarrow C D=\sqrt{(-2)^{2}+(-2)^{2}}$

$\Rightarrow C D=\sqrt{4+4}$

$\Rightarrow C D=\sqrt{8}$

 

$\Rightarrow C D=2 \sqrt{2}$

Also,

$\Rightarrow D A=\sqrt{((-1)-(-3))^{2}+(0-(-2))^{2}}$

$\Rightarrow D A=\sqrt{(2)^{2}+(2)^{2}}$

$\Rightarrow D A=\sqrt{4+4}$

$\Rightarrow D A=\sqrt{8}$

 

$\Rightarrow D A=2 \sqrt{2}$

Hence from above we see that all the sides of the quadrilateral are equal. Hence it is a square.

(ii) A (−3,5) , B(3,1), C(0,3), D(−1,−4)

Let A, B, C and D be the four vertices of the quadrilateral ABCD.

We know the distance between two points Pand Qis given by distance formula:

$\mathrm{PQ}=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

Hence

$\Rightarrow A B=\sqrt{(3-(-3))^{2}+(1-(5))^{2}}$

$\Rightarrow A B=\sqrt{(6)^{2}+(4)^{2}}$

$\Rightarrow A B=\sqrt{36+16}$

$\Rightarrow A B=\sqrt{52}$

 

$\Rightarrow A B=2 \sqrt{13}$

Similarly,

$\Rightarrow B C=\sqrt{(0-3)^{2}+(3-1)^{2}}$

$\Rightarrow B C=\sqrt{(-3)^{2}+(2)^{2}}$

$\Rightarrow B C=\sqrt{9+4}$

 

$\Rightarrow B C=\sqrt{13}$

Similarly,

$\Rightarrow C D=\sqrt{((-1)-0)^{2}+((-4)-(3))^{2}}$

$\Rightarrow C D=\sqrt{(-1)^{2}+(-7)^{2}}$

$\Rightarrow C D=\sqrt{1+49}$

$\Rightarrow C D=\sqrt{50}$

 

$\Rightarrow C D=5 \sqrt{2}$

Also,

$\Rightarrow D A=\sqrt{((-1)-(-3))^{2}+((-4)-5)^{2}}$

$\Rightarrow D A=\sqrt{(2)^{2}+(-9)^{2}}$

$\Rightarrow D A=\sqrt{4+81}$

 

$\Rightarrow D A=\sqrt{85}$

(iii) A (4, 5), B (7,6), C(4,3), D(1,2)

Let A, B, C and D be the four vertices of the quadrilateral ABCD.

We know the distance between two points Pand Qis given by distance formula:

$\mathrm{PQ}=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

Hence

$\Rightarrow A B=\sqrt{(7-4)^{2}+(6-5)^{2}}$

$\Rightarrow A B=\sqrt{(3)^{2}+(1)^{2}}$

$\Rightarrow A B=\sqrt{9+1}$

 

$\Rightarrow A B=\sqrt{10}$

Similarly,

$\Rightarrow B C=\sqrt{(4-7)^{2}+(3-6)^{2}}$

$\Rightarrow B C=\sqrt{(-3)^{2}+(-3)^{2}}$

$\Rightarrow B C=\sqrt{9+9}$

 

$\Rightarrow B C=\sqrt{18}$

Similarly,

$\Rightarrow C D=\sqrt{(1-4)^{2}+(2-3)^{2}}$

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

$\Rightarrow C D=\sqrt{9+1}$

 

$\Rightarrow C D=\sqrt{10}$

Also,

$\Rightarrow D A=\sqrt{(1-4)^{2}+(2-5)^{2}}$

$\Rightarrow D A=\sqrt{(-3)^{2}+(-3)^{2}}$

$\Rightarrow D A=\sqrt{9+9}$

 

$\Rightarrow D A=\sqrt{18}$

Hence from above we see that

AB = CD and BC = DA

Here opposite sides of the quadrilateral is equal. Hence it is a parallelogram.