Question

A sc to . S 12 10 Р 8 6 6 4. 2 - 20-18-16-14-12-10 -8 6 41-2 – 2 2 4 6 8 10 12 14 16 18 20 R. -4 6 S. -8 - 10 - 12 Which transformations of quadrilateral PQRS would result in the image of the quadrilateral being located only in the first quadrant of the coordinate plane?

Answer 1

We are asked to determine which transformation will result in the image being entirely in the I quadrant. This means that we need to determine which transformation will result in each image point of each vertex (x',y') being x' >0 and y'>0.

we can use the following transformation:

A rotation of 90° counterclockwise around the point Q = (4, 12).

The law associated with this transformation is:

$(x^{\prime},y^{\prime})=(-y+a+b,x+b-a)$

Where (a,b) is the point around which the rotation is made, that is (a, b) = (4, 12). Replacing we get:

$(x^{\prime},y^{\prime})=(-y+4+12,x+12-4)$

Solving the operations:

$(x^{\prime},y^{\prime})=(-y+16,x+8)$

Now we transform each vertex. For vertex S = (-3,-7). replacing in the transformation:

$S^{\prime}(x^{\prime},y^{\prime})=(-(-7)+16,-3+8)$

Solving the operation:

$S^{\prime}(x^{\prime},y^{\prime})=(9,5)$

Since both coordinates are positive, this vertex is in the first quadrant.

For vertex P = (-3, 7). Replacing:

$\begin{gathered} P^{\prime}(x^{\prime},y^{\prime})=(-7+16,5+8) \\ P^{\prime}(x^{\prime},y^{\prime})=(9,13) \end{gathered}$

Therefore P' is in the first quadrant.

For vertex Q = (4,12)

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