Question

On a coordinate plane, 2 triangles are shown. Triangle F D E is a right triangle and has points (1, 0), (1, 3), and (5, 0). Triangle Q R P is a right triangle and has points (negative 1, negative 2), (negative 1, negative 5), and (negative 5, negative 2). Are ΔDEF and ΔRPQ congruent? Yes. ΔDEF can be mapped to ΔRPQ by a reflection across the x-axis followed by a translation 2 units left. Yes. ΔDEF can be mapped to ΔRPQ by a 180° rotation about the origin followed by a translation 2 units down. No. ΔDEF cannot be mapped to ΔRPQ by a rigid transformation or combination of rigid transformations.

Answer 1

Yes

ΔDEF can be mapped to ΔRPQ by a 180° rotation about the origin followed by a translation 2 units down

Explanation:

Answer 2

Yes

ΔDEF can be mapped to ΔRPQ by a 180° rotation about the origin followed by a translation 2 units down

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.