Question

Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:
(i) $(-1,-2),(1,0),(-1,2),(-3,0)$
(ii) $(-3,5),(3,1),(0,3),(-1,-4)$
(iii) $(4,5),(7,6),(4,3),(1,2)$

Answer 1

Lets identify the types of Quadrilaterals based on their lengths of sides and their lengths of diagonals.

We know that, the distance between two points $A(x_1,y_1),B(x_2,y_2)$ is $\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$.

(i) $(-1,-2),(1,0),(-1,2),(-3,0)$

Let the given points are $A(-1,-2)$, $B(1,0)$, $C(-1,2)$and $D(-3,0)$ Then,

$AB=\sqrt{(1+1{)}^2+(0+2{)}^2}=\sqrt{2^2+2^2}=\sqrt{4+4}=\sqrt{8}$

$BC=\sqrt{(-1-1{)}^2+(2-0{)}^2}=\sqrt{(2^2+2^2)}=\sqrt{4+4}=\sqrt{8}$

$CD=\sqrt{\left(\right(-3)-(-1){)}^2+(0-2{)}^2}=\sqrt{2^2+(-2{)}^2}=\sqrt{4+4}=\sqrt{8}$

$DA=\sqrt{(-3)-(-1){)}^2+(0-(-2){)}^2}=\sqrt{(-2{)}^2+2^2}=\sqrt{4+4}=\sqrt{8}$

$AC\sqrt{\left(\right(-1)-(-1){)}^2+(2-(-2){)}^2}=\sqrt{0+4^2}=\sqrt{16}=4$

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

Since the four sides $AB,BC,CD$ and $DA$ are equal and the diagonals $AC$ and $BD$ are equal .

$\therefore$ Quadrilateral $ABCD$ is a square.

(ii) $(-3,5),(3,1),(0,3),(-1,-4)$

Let the given points are $A(-3,5)$,$B(3,1)$,$C(0,3)$ and $D(-1,-4)$Then

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

$BC=\sqrt{(3-0{)}^2+(1-3{)}^2}=\sqrt{(3^2+(-2\left)2^2\right)}=\sqrt{9+4}=\sqrt{11}$

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

$DA=\sqrt{(-1)-(-3){)}^2+\left(\right(-4)-5){)}^2}=\sqrt{(2{)}^2+(-9{)}^2}=\sqrt{4+81}=\sqrt{85}$

Here $AB\ne BC\ne CD\ne DA$

$\therefore$ it is a quadrilateral.

(iii) $(4,5),(7,6),(4,3),(1,2)$

Let the given points are $A(4,5)$,$B(7,6)$,$C(4,3)$ and $D(1,2)$Then

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

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

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

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

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

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

Here $AB=CD,BC=DA$ . But $AC\ne BD$

Hence the pairs of opposite sides are equal but diagonal are not equal so it is a parallelogram.