Question

How to find the perimeter and the area of a rectangle on a coordinate plane using the distance formula? : A(3,8) B(5,4) C(-4,-1) D(-6,3) Round to the nearest tenth if necessary.

Answer 1

I used some site to plot the points. 

First, let's recall the formulas for Perimeter and Area of a Rectangle.
Perimeter = 2(l+w)
Area = l×w
Also, the distance formula is
D = $\sqrt{( x_{2}-x_{1})^2+{(y_{2}-y_{1})^2}$

Now, we need to determine l and w.
So, the length is the distance of AD or BC
and the width is the distance of DC or AB
(I'll just use the sides that I've labelled for ease)

So first to determine the length, we need to calculate the distance of AD
Points are A(3,8) and D(-6,3) 
$x_{1} =3, x_{2} =8, y_{1} =6, y_{2} =-3$
$D_{AD}= \sqrt{( 6-3)^2+{(-3-8)^2}$
$D_{AD}= \sqrt{(3)^2+(-11)^2}$
$D_{AD}= \sqrt{9+121}$
$D_{AD}= \sqrt{130} ≈ 11.40$
So the length of the rectangle is 11.40 units.

Now, the width!
So first to determine the width, we need to calculate the distance of DC
Points are D(-6,3) and C(-4,-1)
$x_{1} =6, x_{2} =-4, y_{1} =-3, y_{2} =-1$
$D_{DC}= \sqrt{(-4-6)^2+{(-1- -3)^2}$
$D_{DC}= \sqrt{(-4-6)^2+{(-1+3)^2}$
$D_{DC}= \sqrt{(-10)^2+(2)^2}$
$D_{AD}= \sqrt{100+4}$
$D_{AD}= \sqrt{104} ≈ 10.20$
So the width of the rectangle is ≈10.20 units.

Let's now solve for Perimeter and Area using
l = 11.40
w = 10.20

Perimeter = 2(l+w)
Perimter = 2(11.40+10.20)
Perimter = 2(21.60)
Perimter = 43.2 units

Area = l×w
Area = (11.40)(10.20)
Area = 116.28 
Area = 116.3 units² (rounding)

In conclusion, given points A(3,8) B(5,4) C(-4,-1) D(-6,3), the Perimeter is 43.2 units and the Area is  ≈116.3 units² using the distance formula.