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a new school is being designed on a coordinate plane. the plans indicat…

Question

a new school is being designed on a coordinate plane. the plans indicate that the gym has vertices at (g(-3,0)), (h(-6,6)), (j(0,9)), and (k(3,3)). use the distance formula to find the exact lengths of the diagonals to determine whether the gym is a rectangle. enter your answers in the boxes. diagonal 1: diagonal 2:

Explanation:

Step1: Identify the distance formula

The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$.

Step2: Find the length of one diagonal (say $GJ$)

Let $G(-3,0)$ and $J(0,9)$. Then $x_1=-3,y_1 = 0,x_2=0,y_2 = 9$.
\[

$$\begin{align*} d_{GJ}&=\sqrt{(0-(-3))^2+(9 - 0)^2}\\ &=\sqrt{(3)^2+(9)^2}\\ &=\sqrt{9 + 81}\\ &=\sqrt{90}\\ &=3\sqrt{10} \end{align*}$$

\]

Step3: Find the length of the other diagonal (say $HK$)

Let $H(-6,6)$ and $K(3,3)$. Then $x_1=-6,y_1 = 6,x_2=3,y_2 = 3$.
\[

$$\begin{align*} d_{HK}&=\sqrt{(3-(-6))^2+(3 - 6)^2}\\ &=\sqrt{(9)^2+(-3)^2}\\ &=\sqrt{81+9}\\ &=\sqrt{90}\\ &=3\sqrt{10} \end{align*}$$

\]

Answer:

Diagonal 1: $3\sqrt{10}$
Diagonal 2: $3\sqrt{10}$