Answer:
Yes
ΔDEF can be mapped to ΔRPQ by a 180° rotation about the origin followed by a translation 2 units down
Step-by-step explanation:
The coordinates of the vertices of the triangles are;
F(1, 0), D(1, 3), and E(5, 0), and Q(-1, -2), R(-1, -5), and P(-5, -2),
Length of FD = √((3 - 0)² + (1 - 1)²) = 3
Length of DE = √((5 - 1)² + (0 - 3)²) = 5
Length of FE = √((5 - 1)² + (0 - 0)²) = 4
For triangle QRP, we have;
Length of QR = √((-5 - (-2))² + (-1 - (-1))²) = 3
Length of RP = √((-5 - (-1))² + (-2 - (-4))²) = 5
Length of PQ = √((-5 - (-1))² + (-2 - (-2))²) = 4
For a reflection across the x-axis, we have;
For a rotation of 180° about the origin, we have;
For a coordinate of the preimage before reflection of (x, y), the coordinates of the image after reflection is (-x, -y)
When followed by a translation 2 units down, we have the coordinates of the image will be (-x, -y-2)
Therefore;
For the points F(1, 0), D(1, 3), and E(5, 0) on triangle FDE, we have;
Rotation 180° about the origin gives;
F(1, 0) → Rotation by 180° → F'(-1, 0)
D(1, 3) → Rotation by 180° → D'(-1, -3)
E(5, 0) → Rotation by 180° → E'(-5, 0)
For a translation 2 unites of the coordinates of the vertices of the above triangle ΔF'D'E',
F'(-1, 0) → Translation 2 units down → F''(-1, -2)
D'(-1, -3) → Translation 2 units down → D''(-1, -5)
E'(-5, 0) → Translation 2 units down → E''(-5, -2)
The coordinates of the vertices of triangle ΔF''D''E'' are equivalent to the coordinates of the ΔQRP
Therefore, ΔDEF, and ΔRPQ are congruent, because ΔDEF can be mapped to ΔRPQ by a 180° rotation about the origin followed by a translation 2 units down.
