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

We can prove that it has 2 pairs of parallel sides , one pair of consecutive sides perpendicular and perpendicular diagonals , then the quadrilateral shown above is a square.
By examining it we found that

a . Slope of CB is

b . Slope of DA is_ so DA is paralleled to CB.

c . Slope of CD is_ so CD is perpendicular to CB

d . Slope of BD is

e . Slope of CA is ___ so CA is perpendicular to BD

Question

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Respuesta :

devishri1977 devishri1977 · Answer 1 · 2022-06-22T16:54:28+02:00

Answer:

Step-by-step explanation:

a) Slope of CB

C(-3,-1) ; B = (0,3)

[tex]\sf \boxed{Slope = \dfrac{y_2-y_1}{x_2-x_1}}[/tex]

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

[tex]\sf \text{\bf slope of CB = $\dfrac{4}{3}$}[/tex]

b) D(1,-4) ; A(4,0)

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

[tex]= \dfrac{0+4}{3}\\\[/tex]

[tex]\sf \text{\bf Slope of DA = $\dfrac{4}{3}$}[/tex]

Slope of CB = slope of DA

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

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

[tex]\sf = \dfrac{-4+1}{1+3}\\\\ = \dfrac{-3}{4}\\[/tex]

[tex]\sf Slope \ of \ CD * Slope of CB = \dfrac{-3}{4}*\dfrac{4}{3}=-1[/tex]

So, CD is perpendicular to CB

d) B(0,3) ; D(1,-4)

[tex]Slope \ of \ BD = \dfrac{-4-3}{1-0}\\\\=\dfrac{-7}{1}\\\\=-7[/tex]

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

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

[tex]=\dfrac{0+1}{4+3}\\\\=\dfrac{1}{7}[/tex]

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

So, CA is perpendicular to BD

[tex]\sf \text{\bf Slope of DA = \dfrac{4}{3}}[/tex]