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
[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
now, we can plug values
[tex]AB=\sqrt{(4-(-2))^2+((-2)-(-2))^2}[/tex]
[tex]AB=6[/tex]
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
[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
now, we can plug values
[tex]CB=\sqrt{(4-(4))^2+((-2)-(6))^2}[/tex]
[tex]CB=8[/tex]
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
[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
now, we can plug values
[tex]AC=\sqrt{(4-(-2))^2+((6)-(-2))^2}[/tex]
[tex]AC=10[/tex]
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
[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
now, we can plug values
[tex]DE=\sqrt{(5-(5))^2+((1)-(7))^2}[/tex]
[tex]DE=6[/tex]
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
[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
now, we can plug values
[tex]EF=\sqrt{(13-(5))^2+((1)-(1))^2}[/tex]
[tex]EF=8[/tex]
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
[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
now, we can plug values
[tex]DF=\sqrt{(13-(5))^2+((1)-(7))^2}[/tex]
[tex]DF=10[/tex]
now, we can compare them
[tex]AB=DE=6[/tex]
[tex]CB=EF=8[/tex]
[tex]AC=DF=10[/tex]
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
