Question

given △def with vertices at d(1,1), e(4,5), and f(7,1), prove that △def is an isosceles triangle. drag and drop the answers to the boxes to complete the proof. statements 1. △def with vertices at d(1,1), e(4,5), and f(7,1). reasons 1. given 2. de = ef = df = 2. distance formula 4. definition of isosceles

Explanation:

Step1: Recall 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 length of $DE$

For points $D(1,1)$ and $E(4,5)$, we have $x_1 = 1,y_1 = 1,x_2=4,y_2 = 5$. Then $DE=\sqrt{(4 - 1)^2+(5 - 1)^2}=\sqrt{3^2+4^2}=\sqrt{9 + 16}=\sqrt{25}=5$.

Step3: Calculate length of $EF$

For points $E(4,5)$ and $F(7,1)$, we have $x_1 = 4,y_1 = 5,x_2=7,y_2 = 1$. Then $EF=\sqrt{(7 - 4)^2+(1 - 5)^2}=\sqrt{3^2+( - 4)^2}=\sqrt{9+16}=\sqrt{25}=5$.

Step4: Calculate length of $DF$

For points $D(1,1)$ and $F(7,1)$, we have $x_1 = 1,y_1 = 1,x_2=7,y_2 = 1$. Then $DF=\sqrt{(7 - 1)^2+(1 - 1)^2}=\sqrt{6^2+0^2}=\sqrt{36}=6$.

Answer:

$DE = 5$; $EF = 5$; $DF = 6$