(1) the area of a rectangle with vertices at (−6, 3) , (−3, 6) , (1, 2) , and (−2, −1)
To find area of rectangle we need to find the length and width
Length = distance between (−6, 3) and (−2, −1)
Distance = [tex]\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}[/tex]
= [tex]\sqrt{(-2-(-6))^2 + (-1-3)^2}[/tex]
= [tex]\sqrt{(4)^2 + (-4)^2}[/tex]=[tex]\sqrt{(16) + 16}[/tex] =[tex]\sqrt{32}[/tex]
width = Distance between (−6, 3) , (−3, 6)
Distance = [tex]\sqrt{(-3-(-6))^2 + (6-3)^2}[/tex]
=[tex]\sqrt{3^2+3^2}= \sqrt{18}[/tex]
Area = Length * width = [tex]\sqrt{32} *\sqrt{18} = \sqrt{576}= 24[/tex]
(2) the area of a triangle with vertices at (0, −2) , (8, −2) , and (9, 1)
Area of triangle = [tex]\frac{1}{2} * base * height[/tex]
base is the distance between (0,-2) and (8,-2)
Distance = [tex]\sqrt{(8-0)^2 + (-2-(-2))^2}[/tex] = 8
To find out height we take two vertices (8,-2) and (9,1)
Height is the change in y values = 1- (-2) = 3
base = 8 and height = 3
So area of triangle = [tex]\frac{1}{2} * 8 * 3 = 12[/tex]
(3) the perimeter of a polygon with vertices at (−2, 1) , (−2, 4) , (2, 7) , (6, 4) , and (6, 1)
To find perimeter we add the length of all the sides
Distance between (−2, 1) and (−2, 4) = [tex]\sqrt{(-2+2)^2 + (4-1)^2}[/tex]= 3
Distance between(−2, 4) and (2, 7) = [tex]\sqrt{(2+2)^2 + (7-4)^2}[/tex]= 5
Distance between (2, 7) and (6, 4) = [tex]\sqrt{(6 - 2)^2 + (4-7)^2}[/tex]= 5
Distance between (6, 4) and (6, 1) = [tex]\sqrt{(6 - 6)^2 + (1-4)^2}[/tex]= 3
Distance between (6, 1) and (−2, 1) = [tex]\sqrt{(-2-6)^2 + (1-1)^2}[/tex]= 8
Perimeter = 3 + 5 + 5 + 3 + 8 = 24
(4) four coordinates are (-7,-1) (-6,4) (3,-3) and (4,2)
Length = Distance between (3,-3) and (4,2) = [tex]\sqrt{(4-3)^2 + (2+3)^2}[/tex]= [tex]\sqrt{26}[/tex]
Width = Distance between (-6,4) and (4,2) = [tex]\sqrt{(4+6)^2 + (2-4)^2}[/tex]= [tex]\sqrt{104}[/tex]
Perimeter = 2(lenght + width) = 2*( [tex]\sqrt{26}[/tex]+[tex]\sqrt{104}[/tex] )
= 30.6 units
