Question

you do check for understanding
point q with coordinates (4, -7) is rotated to q(7, 4). describe the rotation that maps q onto q and represent it algebraically.

independent practice

1. parallelogram jklm has vertices j(2, 1), k(7, 1), l(6, -3), and m(1, -3). its image has vertices j(1, -2), k(1, -7), l(-3, -6), and m(-3, -1). describe the rotation and represent it algebraically.
2. the segment xy with endpoints x(3, 1) and y(2, -2) is rotated such that x(-3, -1) and y(-2, 2). describe the rotation and represent it algebraically.

Explanation:

Step1: Analyze the coordinate transformation

For a point \((x,y)\) rotated \(90^{\circ}\) counter - clockwise about the origin, the new coordinates \((x',y')\) are given by the rule \((x',y')=(-y,x)\).
For the point \(Q(4, - 7)\), when we apply the rule \((x,y)\to(-y,x)\), we substitute \(x = 4\) and \(y=-7\). Then \(x'=-(-7)=7\) and \(y' = 4\), so \(Q'(7,4)\)

For the parallelogram \(JKLM\) with \(J(2,1)\), applying the rule \((x,y)\to(-y,x)\) gives \(J'(-1,2)\) (there might be a typo in the problem statement as \(J'(1,2)\) is incorrect according to the rule. If we assume it's a \(90^{\circ}\) clock - wise rotation \((x,y)\to(y, - x)\), for \(J(2,1)\) we get \(J'(1,-2)\) which also doesn't match. But if we consider the general rotation rules:
The standard \(90^{\circ}\) counter - clockwise rotation about the origin has the transformation \((x,y)\to(-y,x)\)
For the segment \(XY\) with \(X(3,1)\), applying \((x,y)\to(-y,x)\) gives \(X'(-1,3)\) (again, if we assume the problem has a sign error. But if we consider the relationship between the original and image points in terms of rotation matrices.
The rotation matrix for a \(90^{\circ}\) counter - clockwise rotation about the origin is \(R=

$$\begin{bmatrix}0&-1\\1&0\end{bmatrix}$$

\). If we have a point \(

$$\begin{bmatrix}x\\y\end{bmatrix}$$

\), after rotation \(R

$$\begin{bmatrix}x\\y\end{bmatrix}$$

=

$$\begin{bmatrix}0\times x+(-1)\times y\\1\times x + 0\times y\end{bmatrix}$$

=

$$\begin{bmatrix}-y\\x\end{bmatrix}$$

\)

So, based on the transformation of \(Q(4,-7)\) to \(Q'(7,4)\) (where \(x = 4,y=-7\) and \(x'=-y = 7,y'=x = 4\)), the rotation is a \(90^{\circ}\) counter - clockwise rotation about the origin with the algebraic rule \((x,y)\to(-y,x)\)

Answer:

The rotation is a \(90^{\circ}\) counter - clockwise rotation about the origin. The algebraic rule is \((x,y)\to(-y,x)\)