Question

homework: unit 5 - algebra part 2 question 11, 10.1.77 find the perimeter of the rectangle whose vertices are the points with coordinates (6, - 2), (9, - 2), (6,8), and (9,8). the perimeter of the rectangle is □ units. (simplify your answer.)

Explanation:

Step1: Find the length of one - side

For two points \((x_1,y_1)\) and \((x_2,y_2)\), the distance formula is \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). Consider the points \((6,-2)\) and \((9,-2)\). Here \(x_1 = 6,y_1=-2,x_2 = 9,y_2=-2\). Then \(d_1=\sqrt{(9 - 6)^2+(-2+2)^2}=\sqrt{3^2+0^2}=3\).

Step2: Find the length of the adjacent - side

Consider the points \((6,-2)\) and \((6,8)\). Here \(x_1 = 6,y_1=-2,x_2 = 6,y_2 = 8\). Then \(d_2=\sqrt{(6 - 6)^2+(8 + 2)^2}=\sqrt{0^2+10^2}=10\).

Step3: Calculate the perimeter of the rectangle

The perimeter formula of a rectangle is \(P=2(l + w)\), where \(l\) and \(w\) are the length and width of the rectangle. Substituting \(l = 10\) and \(w = 3\) into the formula, we get \(P=2(10 + 3)=2\times13 = 26\).

Answer:

26