1.11 quiz use algebra to describe geometry part 2

Answer : rotation of 90 counterclockwise° about the origin and then translation 2 unit up.

Explanation : In Δ RST, it is an right triangle with ∠R being the right angle. When we draw this triangle we see that long side of the Δ is close to the origin and the short one is below ∠R.

Now, Δ R'S'T', is also a right triangle. But he long side is parallel to the y axis and ∠S' is now above ∠R' instead of ∠S being to right of ∠R in the original Δ.

which clearly shows that the Δ is rotated at 90 degrees counter clockwise.

On drawing a new Δ; The point R" is at (1,-4), S" at (1,-1), and T" at (2,-4).

All three of these points are 2 units below the points R' S' T'.

We need to translate the triangle 2 units higher.

Therefore, the answer is to perform rotation of 90 counterclockwise° about the origin and then translation 2 unit up.

a negative exponent is translated as follows:

so, we have

step-by-step explanation:

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