Question

∆ABC is translated 2 units down and 1 unit to the left. Then it is rotated 90° clockwise about the origin to form ∆A′B′C′.

The coordinates of vertex A′ of ∆A′B′C′ are (-2,0), OR (2,1), OR (-1,-2), OR (-2,1)
The coordinates of vertex B′ of ∆A′B′C′ are (1,-1), OR (1,0), OR (5,0), OR (2,-3)
The coordinates of vertex C′ of ∆A′B′C′ are (-1,-1), OR (3,0), OR (-1,0), OR (0,-3)

Answer 1

Answer:

The coordinates of image are A'(-2,1), B'(1,0) and C'(-1,0).

Step-by-step explanation:

From the figure it is clear that the coordinates of triangle are A(0,0), B(1,3) and C(1,1).

∆ABC is translated 2 units down and 1 unit to the left.

$(x,y)\rightarrow (x-1,y-2)$

$A(0,0)\rightarrow (0-1,0-2)=(-1,-2)$

$B(1,3)\rightarrow (1-1,3-2)=(0,1)$

$C(1,1)\rightarrow (1-1,1-2)=(0,-1)$

Then it is rotated 90° clockwise about the origin to form ∆A′B′C′.

$(x,y)\rightarrow (y,-x)$

$(-1,-2)\rightarrow A'(-2,1)$

$(0,1)\rightarrow B'(1,0)$

$(0,-1)\rightarrow C'(-1,0)$

Therefore the coordinates of image are A'(-2,1), B'(1,0) and C'(-1,0).

Answer 2

Answer:

A'(-2,1), B'(1,0) and C'(-1,0).

Step-by-step explanation: