Question

the graph shows triangles abc and efg.
is abc congruent to efg? justify your answer.
yes, because a rotation 180° around b followed by a translation down 6 units maps abc onto efg.
yes, because a reflection across the line y = 3 maps abc onto efg.
no, because \\(\overline{ca}\\) and \\(\overline{ge}\\) do not have the same length.
no, because \\(\overline{ab}\\) and \\(\overline{ef}\\) do not have the same length.

Explanation:

Step1: Identify coordinates of triangles

From the graph:

• Triangle ABC: A(-3,9), B(0,6), C(3,7)
• Triangle EFG: E(-3,-3), F(0,0), G(3,-1)

Step2: Calculate side lengths (distance formula)

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

• $AB$: $\sqrt{(0-(-3))^2 + (6-9)^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt{2}$
• $EF$: $\sqrt{(0-(-3))^2 + (0-(-3))^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt{2}$
• $BC$: $\sqrt{(3-0)^2 + (7-6)^2} = \sqrt{9+1} = \sqrt{10}$
• $FG$: $\sqrt{(3-0)^2 + (-1-0)^2} = \sqrt{9+1} = \sqrt{10}$
• $AC$: $\sqrt{(3-(-3))^2 + (7-9)^2} = \sqrt{36+4} = \sqrt{40} = 2\sqrt{10}$
• $EG$: $\sqrt{(3-(-3))^2 + (-1-(-3))^2} = \sqrt{36+4} = \sqrt{40} = 2\sqrt{10}$

Step3: Verify congruence via transformations

Check if rigid transformations map ABC to EFG:

• Reflect ABC over $y=3$: Midpoint of y-coordinates of B(6) and F(0) is 3.
• A(-3,9) → (-3, -3) (since $9 - 2(9-3) = -3$)
• B(0,6) → (0,0) (since $6 - 2(6-3) = 0$)
• C(3,7) → (3,-1) (since $7 - 2(7-3) = -1$)

These are exactly coordinates of E, F, G.

Answer:

Yes, because a reflection across the line y = 3 maps ABC onto EFG.