Question

Name the type of quadriateral 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

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

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

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

$|DC|=\sqrt{(-3+1{)}^2+(0-2{)}^2}=2\sqrt{2}$

$|AD|=\sqrt{(-3+1{)}^2+(0+2{)}^2}=2\sqrt{2}$

Slope of $AB=\frac{0+2}{1+1}=1=m_1$

Slope of $AD=\frac{0+2}{-3+1}=-1=m_2$

$m_1{m}_2=-1$ & $AB=BC=DC=AD$

$\therefore ABCD$ is a square

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

$|PQ|=\sqrt{(3+3{)}^2+(1-5{)}^2}=\sqrt{52}$

$|QP|=\sqrt{(3-0{)}^2+(1-3{)}^2}=\sqrt{13}$

$|RS|=\sqrt{(a+1{)}^2+(3+4{)}^2}=\sqrt{50}$

$|PS|=\sqrt{(5+4{)}^2+(-3+1{)}^2}=\sqrt{85}$

$\therefore$ it is simple quadrilateral

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

$|MN|=\sqrt{(7-4{)}^2+(6-5{)}^2}=\sqrt{10}$

$|NO|=\sqrt{7-4{)}^2+(6-3{)}^2}=\sqrt{18}$

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

$|PM|=\sqrt{(4-1{)}^2+(5-2{)}^2}=\sqrt{18}$

Slope of $MN=\frac{6-5}{7-4}=\frac{1}{3}$

Slope of $MP=\frac{5-2}{4-1}=1$

$MN=OP$ & $NO=PM$

$\therefore$ it is a parallelogram