Question

Coordinate plane with triangles QRS and UTS with Q at negative 6 comma 2, R at negative 2 comma 6, S at negative 2 comma 2, T at negative 2 comma 0, and U at negative 4 comma 2

Which set of transformations would prove ΔQRS ~ ΔUTS?



Reflect ΔUTS over y = 2, and dilate ΔU′T′S′ by a scale factor of 2 from point S.


Reflect ΔUTS over y = 2, and translate ΔU′T′S′ by the rule (x − 2, y + 0).


Translate ΔUTS by the rule (x + 0, y + 6), and reflect ΔU′T′S′ over y = 6.


Translate ΔUTS by the rule (x − 2, y + 0), and reflect ΔU′T′S′ over y = 2.

Answer 1

To prove that ΔQRS is similar to ΔUTS, compare their corresponding angles and sides. The corresponding angles are congruent, and the corresponding sides are proportional.

Step-by-step explanation:

To prove that ΔQRS is similar to ΔUTS, we need to show that their corresponding angles are congruent and their corresponding sides are proportional. Let's compare the angles first:

Angle QRS = Angle UTS (both right angles)

Angle RSQ = Angle TSU (both acute angles)

Angle QSR = Angle TUS (both acute angles)

Therefore, the corresponding angles of the two triangles are congruent.

Now let's compare the sides:

QR = UT (both lengths are 4 units)

RS = TS (both lengths are 4 units)

SQ = SU (both lengths are 2 units)

Therefore, the corresponding sides of the two triangles are proportional.

Based on these comparisons, we can conclude that ΔQRS is similar to ΔUTS.

Answer 2

A set of transformations that would prove ΔQRS ~ ΔUTS is: A. Reflect ΔUTS over y = 2, and dilate ΔU′T′S′ by a scale factor of 2 from point S.

In Mathematics and Euclidean Geometry, a reflection across the line y = k and y = 2 can be modeled by the following transformation rule:

(x, y) → (x, 2k - y)

(x, y) → (x, 4 - y)

By applying a reflection across the line y = 2 to the coordinates of the pre-image (triangle), we have the following image coordinates;

(x, y) → (x, 4 - y)

Q (-6, 2) → (-6, 4 - 2) = Q' (-6, 2)

Next, we would apply a dilation to ΔU′T′S′ by using a scale factor of 2 centered at point S (-2, 2) ≡ (a, b);

(x, y) → (k(x - a) + a, k(y - b) + b)

Coordinate U' = (-4, 2) → (2(-4 - (-2)) + (-2), 2(2 - 2) + 2)

Coordinate U' = (-4, 2) → (2(-2) - 2, 2(0) + 2)

Coordinate U' = (-4, 2) → (-4 - 2, 0 + 2)

Coordinate U' = (-6, 2).

Missing information:

The question is incomplete and the missing figure is shown in the attached picture.