Question

Name the quadrilateral formed, if any, by the following points, and give reasons for your answer.
(i) $A(-1,-2),B(1,0),C(-1,2),D(-3,0)$
(ii) $A(-3,5),B(3,1),C(0,3),D(-1,-4)$
(iii) $A(4,5),B(7,6),C(4,3),D(1,2)$.

Answer 1

$\left(i\right)$

Let $A=(-1,-1),B=(1,0),C=(-1,2),D=(-3,0)$

$AB=\sqrt{[1-(-1){]}^2+[0-(-2){]}^2}=\sqrt{(2{)}^2+(2{)}^2}=\sqrt{8}=2\sqrt{2}unit$

$BC=\sqrt{(-1-1{)}^2+(2-0{)}^2}=\sqrt{(-2{)}^2+(2{)}^2}=\sqrt{8}=2\sqrt{2}unit$

$CD=\sqrt{[-3-(-1){]}^2+[0-2{]}^2}=\sqrt{(-2{)}^2+(-2{)}^2}=\sqrt{8}=2\sqrt{2}unit$

$AD=\sqrt{[-3-(-1){]}^2+[0-(-2){]}^2}=\sqrt{(-2{)}^2+(2{)}^2}=\sqrt{8}=2\sqrt{2}unit$

Now, let us calculate the diagonals.

$AC=\sqrt{[-1(-1){]}^2+[2-(-2){]}^2}=\sqrt{0+(4{)}^2}=\sqrt{16}=4unit$

$BD=\sqrt{(-3-1{)}^2+(0-0{)}^2}=\sqrt{(-4{)}^2}=\sqrt{16}=4unit$

Here, $AB=BC=CD=AD$ and $AC=BD$

Since, all sides are equal and diagonals are equal, so given quadrilateral is a square.

$\left(ii\right)$

Let $A=(-3,5),B=(3,1),C=(0,3)$ and $D=(-1,-4)$

$AB=\sqrt{[3-(-3){]}^2+[1-5{]}^2}=\sqrt{(6{)}^2+(-4{)}^2}=\sqrt{36+16}=\sqrt{52}unit$

$BC=\sqrt{(0-3{)}^2+(3-1{)}^2}=\sqrt{(-3{)}^2+(2{)}^2}=\sqrt{9+4}=\sqrt{13}unit$

$CD=\sqrt{(-1-0{)}^2+(-4-3{)}^2}=\sqrt{(-1{)}^2+(-7{)}^2}=\sqrt{1+49}=\sqrt{50}=5\sqrt{2}unit$

$AD=\sqrt{(-1+3{)}^2+(-4-5{)}^2}=\sqrt{(2{)}^2+(-9{)}^2}=\sqrt{4+81}=\sqrt{85}unit$

Here, $AC\ne BC\ne CD\ne DA$

Since, sides are not equal, thus only a general and not a specific quadrilateral

such as square or parallelogram is formed.

$\left(iii\right)$

Let $A=(4,5),B=(7,6),C=(4,3),D=(1,2)$

$AB=\sqrt{(7-4{)}^2+(6-5{)}^2}=\sqrt{(3{)}^2+(1{)}^2}=\sqrt{9+1}=\sqrt{10}unit$

$BC=\sqrt{(4-7{)}^2+(3-6{)}^2}=\sqrt{(-3{)}^2+(-3{)}^2}=\sqrt{9+9}=\sqrt{18}unit$

$CD=\sqrt{(1-4{)}^2+(2-3{)}^2}=\sqrt{(-3{)}^2+(-1{)}^2}=\sqrt{9+1}=\sqrt{10}unit$

$AD=\sqrt{(1-4{)}^2+(2-5{)}^2}=\sqrt{(-3{)}^2+(-3{)}^2}=\sqrt{9+9}=\sqrt{18}unit$

Now let us calculate the diagonals

$AC=\sqrt{(4-4{)}^2+(3-5{)}^2}=\sqrt{(0{)}^2+(-2{)}^2}=\sqrt{0+4}=\sqrt{4}=2unit$

$BD=\sqrt{(1-7{)}^2+(2-6{)}^2}=\sqrt{(-6{)}^2+(-4{)}^2}=\sqrt{36+16}=\sqrt{52}=2\sqrt{13}unit$

Here, $AB=CD$ and $BC=AD$

$AC\ne BD$

Here, opposite sides are equal but diagonals are not equal, thus with the given points a parallelogram is formed.