Answer:
Step-by-step explanation:
Point F is plotted at (5, 4)
Translate ΔABC so that point B maps to point G.
B(1, -1) → G(0, 3)
Rule for the translation,
(x, y) → [x - 1, y + 4]
By this rule,
Point C will map the point,
C(-1, -2) → C'[(-1 - 1), (-2 + 4)]
→ C'(-2, 2)
A(-4, 0) → A'[(-4 - 1), (0 + 4)]
→ A'(-5, 4)
Now reflect these points across y-axis.
Rule for the reflection across y-axis,
(x, y) → (-x, y)
Points after reflection will be,
A'(-5, 4) → F(5, 4)
B'(0, 3) → G(0, 3)
C'(-2, 2) → H(2, 2)
Distance formula,
d = [tex]\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}[/tex]
AB = [tex]\sqrt{(1+4)^2+(-1-0)^2}[/tex] = [tex]\sqrt{26}[/tex]
BC = [tex]\sqrt{(-2+1)^2+(-1-1)^2}[/tex] = [tex]\sqrt{5}[/tex]
CA = [tex]\sqrt{(-4+1)^2+(-2-0)^2}[/tex] = [tex]\sqrt{13}[/tex]
FG = [tex]\sqrt{(5-0)^2+(4-3)^2} =\sqrt{26}[/tex]
GH = [tex]\sqrt{(2-0)^2+(2-3)^2}=\sqrt{5}[/tex]
HF = [tex]\sqrt{(4-2)^2+(5-2)^2}=\sqrt{13}[/tex]
AB ≅ FG, BC ≅ GH, CA ≅ HF
By the SSS property Triangle congruence theorem, ΔABC ≅ ΔFGH
