Question

a drawing of a town includes the central park, which has vertices at c(-8, -12), d(-7, -10), e(-9, -9), and f(-10, -11). how can the distance formula be used to show whether the park is a rhombus? the side lengths are √709, √617, √761, and √853, so the shape is a rhombus. each side has a length of √5, so the shape is a rhombus. the side lengths are √709, √617, √761, and √853, so the shape is not a rhombus. each side has a length of √5, so the shape is not a rhombus.

Explanation:

Step1: Recall 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: Calculate the length of side $CD$

For points $C(-8,-12)$ and $D(-7,-10)$, $x_1=-8,y_1 = - 12,x_2=-7,y_2=-10$. Then $d_{CD}=\sqrt{(-7+8)^2+(-10 + 12)^2}=\sqrt{1^2+2^2}=\sqrt{1 + 4}=\sqrt{5}$.

Step3: Calculate the length of side $DE$

For points $D(-7,-10)$ and $E(-9,-9)$, $x_1=-7,y_1=-10,x_2=-9,y_2=-9$. Then $d_{DE}=\sqrt{(-9 + 7)^2+(-9+10)^2}=\sqrt{(-2)^2+1^2}=\sqrt{4 + 1}=\sqrt{5}$.

Step4: Calculate the length of side $EF$

For points $E(-9,-9)$ and $F(-10,-11)$, $x_1=-9,y_1=-9,x_2=-10,y_2=-11$. Then $d_{EF}=\sqrt{(-10 + 9)^2+(-11 + 9)^2}=\sqrt{(-1)^2+(-2)^2}=\sqrt{1+4}=\sqrt{5}$.

Step5: Calculate the length of side $FC$

For points $F(-10,-11)$ and $C(-8,-12)$, $x_1=-10,y_1=-11,x_2=-8,y_2=-12$. Then $d_{FC}=\sqrt{(-8 + 10)^2+(-12 + 11)^2}=\sqrt{2^2+(-1)^2}=\sqrt{4+1}=\sqrt{5}$.

Step6: Recall the property of a rhombus

A rhombus has all - sides equal. Since all side - lengths $d_{CD}=d_{DE}=d_{EF}=d_{FC}=\sqrt{5}$, the shape is a rhombus.

Answer:

Each side has a length of $\sqrt{5}$, so the shape is a rhombus.