Note:
Images with coordinates or original and translated triangles are attached.
Answer:
The coordinates for the given triangle are A(6,3), B(6,-3) and C(-2,-3), respectively.
Let us translate this triangle by -6 units along the x-axis and +3 units along the y-axis. This means that 6 will be subtracted from each of the abcissas (x-points) and 3 will be added to each ordinates (y-points). Hence, our translated triangle will have the following coordinates: A'(0,6), B'(0,0) and C'(-8,0), respectively.
Now to prove that both triangles are congruent, we use the SSS postulate, which states that:
"If three sides of one triangle are congruent to three sides of another triangle, then the triangles are congruent."
Distance between two point P1 and P2 will be:
[tex]|P_1P_2|=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
Now using distance formula to find sides of triangle ABC and triangle A'B'C'.
For Triangle ABC:
|AB| = [tex]\sqrt{(6-6)^2+(-3-3)^2} =\;6\;units[/tex]
|AC| = [tex]\sqrt{(-2-6)^2+(-3-3)^2} =\;10\;units[/tex]
|BC| = [tex]\sqrt{(-2-6)^2+[-3-(-3)]^2} =\;8\;units[/tex]
For Triangle A'B'C':
|A'B'| = [tex]\sqrt{(0-0)^2+(0-6)^2} =\;6\;units[/tex]
|A'C'| = [tex]\sqrt{(-8-0)^2+(0-6)^2} =\;10\;units[/tex]
|B'C'| = [tex]\sqrt{(-8-0)^2+(0-0)^2} =\;8\;units[/tex]
Hence, as the distances are equal, the two triangles are congruent by SSS postulate.
