Question

Name the quadrilateral formed, if any, by the following points and give reasons for your answers:
$A(-1,-2),B(1,0),C(-1,2),D(-3,0)$

Answer 1

Let $A(-3,5),B(3,1),C(0,3)$

$D(-1,-4)$ be the given points

[distance $={\sqrt{x_2-x_1}}^2+(y_2-y_1{)}^2$]

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

$AB=\sqrt{36+16}=\sqrt{52}=2\sqrt{13}$

$AB=2\sqrt{13}$

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

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

$BC=\sqrt{13}$

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

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

$CD=\sqrt{50}=5\sqrt{2}$

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

$DA=\sqrt{(2{)}^2+(-9{)}^2}=\sqrt{4+81}=\sqrt{85}$

$DA=\sqrt{85}$

Now distances

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

$AC=\sqrt{9+4}=\sqrt{13}$

$AC=\sqrt{13}$

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

$BD=\sqrt{(-4{)}^2+(-5{)}^2}=\sqrt{16+25}$

$BD=\sqrt{41}$

Hence we see that from above sides

$AC+BC=AB$

$\sqrt{13}+\sqrt{13}=2\sqrt{13}$

$A,B,C$ are collinear

Hence, points $A,B,C$ & $D$ does not from a quadrilateral.