Question

Basic Mathematics (22103)
Question 18 of 40
:7 point
The points (-1, - 2), (1, 0), (-1, 2), (-3,0)
forms a quadrilateral of type
Square
Rectangle
O Parallelogram
Rhombus​

Answer 1

$\underline{\underline{\sf{\maltese\:\:Question}}}$

The points (-1, - 2), (1, 0), (-1, 2), (-3,0) forms a  

• Square

• Rectangle

• Parallelogram

• Rhombus​

$\underline{\underline{\sf{\maltese\:\:Given\:Points}}}$

• A (-1, - 2)

• B (1, 0)

• C (-1, 2)

• D (-3, 0)

$\underline{\underline{\sf{\maltese\:\:To\:\:Find}}}$

• The given points are vertices of

$\underline{\underline{\sf{\maltese\:\:Answer}}}$

• The given points are vertices of a Square

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\multiput(0,0)(4,0){2}{\line(0,1){4}}\multiput(0,0)(0,4){2}{\line(1,0){4}}\put(-0.5,-0.5){\bf D (-3,0)}\put(-0.5,4.2){\bf A (-1,-2)}\put(4.2,-0.5){\bf C (-1,2)}\put(4.2,4.2){\bf B (1,0)}\put(1.5,-0.6){\bf\large }\put(4.4,2){\bf\large }\end{picture}$

$\underline{\underline{\sf{\maltese\:\:Calculations}}}$

Apply Distance Formula : $\sf{\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}}$

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

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

AB =  $\sf{\sqrt{(2)^{2}+(2)^{2}}}$

AB =  $\sf{\sqrt{4+4}}$

AB = $\sf{\sqrt{8}}$ units

________________________________________

Apply Distance Formula : $\sf{\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}}$

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

BC = $\sf{\sqrt{(-2)^{2}+(2)^{2}}}$

BC = $\sf{\sqrt{4+4}}$

BC = $\sf{\sqrt{8}}$ units

________________________________________

Apply Distance Formula : $\sf{\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}}$

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

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

CD = $\sf{\sqrt{(-2)^{2}+(-2)^{2}}}$

CD = $\sf{\sqrt{4+4}}$

CD = $\sf{\sqrt{8}}$ units

________________________________________

Apply Distance Formula : $\sf{\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}}$

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

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

AD = $\sf{\sqrt{(-2)^{2}+(2)^{2}}}$

AD = $\sf{\sqrt{4+4}}$

AD = $\sf{\sqrt{8}}$ units

________________________________________

As all sides are equal, The given points are vertices of a Square

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\multiput(0,0)(4,0){2}{\line(0,1){4}}\multiput(0,0)(0,4){2}{\line(1,0){4}}\put(-0.5,-0.5){\bf D (-3,0)}\put(-0.5,4.2){\bf A (-1,-2)}\put(4.2,-0.5){\bf C (-1,2)}\put(4.2,4.2){\bf B (1,0)}\put(1.5,-0.6){\bf\large }\put(4.4,2){\bf\large }\end{picture}$

________________________________________

Note :

Please check web version for diagrams

Are you an app User ?

If yes, just swipe left to see whole answer