Question

parallelogram ( j k l m ) has vertices ( j(2,1), k(7,1), l(6,-3) ), and ( m(1,-3) ). its image has vertices ( j^{prime}(1,-2), k^{prime}(1,-7), l^{prime}(-3,-6) ), and ( m^{prime}(-3,-1) ). describe the rotation ( (x, y) \to (quad,quad) )

Explanation:

Step1: Analyze the rotation rule

For a point \((x,y)\) rotated \(270^{\circ}\) clockwise about the origin, the transformation rule is \((x,y)\to(y, -x)\).
Let's check with point \(J(2,1)\):
If we apply \((x,y)\to(y, -x)\), we get \((1,- 2)\) which is \(J'\).
For point \(K(7,1)\):
Applying \((x,y)\to(y, -x)\), we get \((1,-7)\) which is \(K'\).
For point \(L(6,-3)\):
Applying \((x,y)\to(y, -x)\), we get \((-3,-6)\) which is \(L'\).
For point \(M(1,-3)\):
Applying \((x,y)\to(y, -x)\), we get \((-3,-1)\) which is \(M'\).

Answer:

\((x,y)\to(y, -x)\)