Question

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

Answer 1

Explanation:

To prove quadrilateral is a square:

a) Slope of CB

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

`\sf \boxed{Slope = (y_2-y_1)/(x_2-x_1)}`

`\sf = (3-[-1])/(0-[-3])\\\\ =(3+1)/(0+3)\\\\ = (4)/(3)`

`\sf \text{\bf slope of CB = $(4)/(3)$}`

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

`Slope \ of \ DA = (0-[-4])/(4-1)\\`

`= (0+4)/(3)\\\`

`\sf \text{\bf Slope of DA = $(4)/(3)$}`

Slope of CB = slope of DA

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

`\sf Slope \ of \ CD =(-4-[-1])/(1-[-3])`

`\sf = (-4+1)/(1+3)\\\\ = (-3)/(4)\\`

`\sf Slope \ of \ CD * Slope of CB = (-3)/(4)*(4)/(3)=-1`

So, CD is perpendicular to CB

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

`Slope \ of \ BD = (-4-3)/(1-0)\\\\=(-7)/(1)\\\\=-7`

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

`\sf Slope \ of \ CA = (0-[-1])/(4-[-3])\\`

`=(0+1)/(4+3)\\\\=(1)/(7)`

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

So, CA is perpendicular to BD

`\sf \text{\bf Slope of DA = (4)/(3)}`