Question

Use the given coordinates to determine if triangle ABC is congruent to triangle DEF.

Answer 1

We can find distances

Length of AB:

A=(-2,-2)

so, x1=-2 , y1=-2

B=(4,-2)

so, x2=4 , y2=-2

now, we can use distance formula

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

now, we can plug values

$AB=\sqrt{(4-(-2))^2+((-2)-(-2))^2}$

$AB=6$

Length of CB:

C=(4,6)

so, x1=4 , y1=6

B=(4,-2)

so, x2=4 , y2=-2

now, we can use distance formula

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

now, we can plug values

$CB=\sqrt{(4-(4))^2+((-2)-(6))^2}$

$CB=8$

Length of AC:

A=(-2,-2)

so, x1=-2 , y1=-2

C=(4,6)

so, x2=4 , y2=6

now, we can use distance formula

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

now, we can plug values

$AC=\sqrt{(4-(-2))^2+((6)-(-2))^2}$

$AC=10$

Length of DE:

D=(5,7)

so, x1=5 , y1=7

E=(5,1)

so, x2=5 , y2=1

now, we can use distance formula

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

now, we can plug values

$DE=\sqrt{(5-(5))^2+((1)-(7))^2}$

$DE=6$

Length of EF:

E=(5,1)

so, x1=5 , y1=1

F=(13,1)

so, x2=13 , y2=1

now, we can use distance formula

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

now, we can plug values

$EF=\sqrt{(13-(5))^2+((1)-(1))^2}$

$EF=8$

Length of DF:

D=(5,7)

so, x1=5 , y1=7

F=(13,1)

so, x2=13 , y2=1

now, we can use distance formula

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

now, we can plug values

$DF=\sqrt{(13-(5))^2+((1)-(7))^2}$

$DF=10$

now, we can compare them

$AB=DE=6$

$CB=EF=8$

$AC=DF=10$

now, we can draw both triangles

Since, sides of both triangles are equal

so, by using SSS rule

triangle ABC and triangle DEF are congruent

so, this is TRUE........Answer