Question

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. ( (x,y)\to )

Explanation:

Step1: Recall rotation rules

The general rule for a \(90^{\circ}\) clockwise rotation about the origin is \((x,y)\to(y, -x)\), for a \(90^{\circ}\) counter - clockwise rotation about the origin is \((x,y)\to(-y,x)\), for a \(180^{\circ}\) rotation about the origin is \((x,y)\to(-x,-y)\).
Let's take a vertex, say \(J(2,1)\).

Step2: Check the transformation

If we consider a \(90^{\circ}\) clockwise rotation about the origin. For a point \((x,y)\), the new point is \((y,-x)\). But for \(J(2,1)\), if we use \((x,y)\to(x - 1,y-3)\) (translation), it's not a rotation.
Let's use the rotation formula. Let the center of rotation be the origin \((0,0)\).
The formula for a rotation of \(\theta\) about the origin is \(

$$\begin{pmatrix}x'\\y'\end{pmatrix}$$

=

$$\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}$$

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

\)
For \(J(2,1)\) and \(J'(1, - 2)\)
If \(\theta = 270^{\circ}\) (or \(- 90^{\circ}\)), \(\cos\theta = 0\), \(\sin\theta=- 1\)
\(x'=x\cos\theta-y\sin\theta=0\times x-(-1)y=y\)
\(y'=x\sin\theta + y\cos\theta=-1\times x+0\times y=-x\)
For \(J(2,1)\): \(x = 2,y = 1\), \(x'=1,y'=-2\) (matches \(J'(1,-2)\))
For \(K(7,1)\): \(x = 7,y = 1\), \(x'=1,y'=-7\) (matches \(K'(1,-7)\))
For \(L(6,-3)\): \(x = 6,y=-3\), \(x'=-3,y'=-6\) (matches \(L'(-3,-6)\))
For \(M(1,-3)\): \(x = 1,y=-3\), \(x'=-3,y'=-1\) (matches \(M'(-3,-1)\))

Answer:

\((x,y)\to(y,-x)\) (a \(270^{\circ}\) counter - clockwise rotation about the origin or a \(90^{\circ}\) clockwise rotation about the origin)