Answer:
[tex]A = (5,3)[/tex]
[tex]B = (-5,3)[/tex]
[tex]C = (-5,-8)[/tex]
[tex]D = (5,-8)[/tex]
Step-by-step explanation:
Required
[tex]Perimeter = 42[/tex]
Construct a rectangle whose perimeter is 42 units and satisfies the given conditions.
First, name the rectangle ABCD.
Such that:
[tex]A = (x_1,y_1)[/tex]
[tex]B = (x_2,y_2)[/tex]
[tex]C = (x_3,y_3)[/tex]
[tex]D = (x_4,y_4)[/tex]
For the rectangle to be either horizontal or vertical, then:
[tex]y_1 = y_2[/tex] and [tex]y_3 = y_4[/tex]
We have that:
[tex]Perimeter = 42[/tex]
Replace perimeter with its formula
[tex]2(AB + BC) = 42[/tex]
Divide both sides by 2
[tex]AB + BC = 21[/tex]
This implies that, the distance between adjacent sides (through the edges) must be equal to 21
Having said that: a set of coordinates that satisfy the given conditions are:
[tex]A = (5,3)[/tex] -- First quadrant
[tex]B = (-5,3)[/tex] -- Second quadrant
[tex]C = (-5,-8)[/tex] -- Third quadrant
[tex]D = (5,-8)[/tex] -- Fourth quadrant
The above quadrants satisfy the condition:
[tex]y_1 = y_2[/tex] and [tex]y_3 = y_4[/tex]
HOW TO KNOW THE PERIMETER IS 42
To do this, we simply calculate the distance between the edges and add them up
Distance is calculated as:
[tex]D = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2[/tex]
For AB
[tex]A = (5,3)[/tex]
[tex]B = (-5,3)[/tex]
[tex]D_1 = \sqrt{(5 - (-5))^2 + (3 - 3)^2}= \sqrt{(10)^2 + (0)^2} = \sqrt{100} = 10[/tex]
For BC
[tex]B = (-5,3)[/tex]
[tex]C = (-5,-8)[/tex]
[tex]D_2 = \sqrt{(-5 - (-5))^2 + (3 - (-8))^2}= \sqrt{(0)^2 + (11)^2} = \sqrt{121} = 11[/tex]
For CD
[tex]C = (-5,-8)[/tex]
[tex]D = (5,-8)[/tex]
[tex]D_3 = \sqrt{(-5 -5)^2 + (-8 - (-8))^2}= \sqrt{(-10)^2 + (0)^2} = \sqrt{100} = 10[/tex]
For DA
[tex]D = (5,-8)[/tex]
[tex]A = (5,3)[/tex]
[tex]D_4 = \sqrt{(5 -5)^2 + (-8 -3)^2}= \sqrt{(0)^2 + (11)^2} = \sqrt{121} = 11[/tex]
So, the perimeter is:
[tex]P = D_1 + D_2 + D_3 + D_4[/tex]
[tex]P = 10 + 11 + 10 +11[/tex]
[tex]P = 42[/tex]
See attachment for rectangle
