Question

One way to prove this quadrilateral is a square is by calculating the slopes of its sides and the slopes of its diagonals.

Answer 1

Answer:

Step-by-step explanation:

To prove quadrilateral is a square:

a)  Slope of CB

   C(-3,-1)  ; B = (0,3)

     $\sf \boxed{Slope = \dfrac{y_2-y_1}{x_2-x_1}}$

                $\sf = \dfrac{3-[-1]}{0-[-3]}\\\\ =\dfrac{3+1}{0+3}\\\\ = \dfrac{4}{3}$

  $\sf \text{\bf slope of CB = \dfrac{4}{3} }$

b) D(1,-4)   ; A(4,0)

    $Slope \ of \ DA = \dfrac{0-[-4]}{4-1}$

                          $= \dfrac{0+4}{3}$

    $\sf \text{\bf Slope of DA = \dfrac{4}{3} }$

Slope of CB = slope of DA

c) C(-3,-1) ; D(1 , -4)

    $\sf Slope \ of \ CD =\dfrac{-4-[-1]}{1-[-3]}$

                       $\sf = \dfrac{-4+1}{1+3}\\\\ = \dfrac{-3}{4}$

$\sf Slope \ of \ CD * Slope of CB = \dfrac{-3}{4}*\dfrac{4}{3}=-1$

So, CD is perpendicular to   CB

d) B(0,3)  ;  D(1,-4)

    $Slope \ of \ BD = \dfrac{-4-3}{1-0}\\\\=\dfrac{-7}{1}\\\\=-7$

e) C(-3,-1) ; A(4,0)

   $\sf Slope \ of \ CA = \dfrac{0-[-1]}{4-[-3]}$

                       $=\dfrac{0+1}{4+3}\\\\=\dfrac{1}{7}$

$\text{Slope of CA *Slope of BD = \dfrac{1}{7} *(-7)}=-1$

So, CA is perpendicular to BD

$\sf \text{\bf Slope of DA = \dfrac{4}{3}}$