Question

graph the polygon with the vertices ( q(-6,-3), r(-5,0), s(-3,0), t(-1,-3) ) and its image after a clockwise rotation of ( 270^{circ} ) about the origin.

Explanation:

Step1: Recall the rotation rule

When a point \((x,y)\) is rotated \(270^{\circ}\) clockwise about the origin, the transformation rule is \((x,y)\to(y, -x)\).

Step2: Apply the rule to each vertex

• For \(Q(-6,-3)\):

Using the rule \((x,y)\to(y, -x)\), we substitute \(x = - 6\) and \(y=-3\). So the image \(Q'\) is \((-3,6)\).

• For \(R(-5,0)\):

Substitute \(x=-5\) and \(y = 0\) into the rule \((x,y)\to(y, -x)\). The image \(R'\) is \((0,5)\).

• For \(S(-3,0)\):

Substitute \(x=-3\) and \(y = 0\) into the rule \((x,y)\to(y, -x)\). The image \(S'\) is \((0,3)\).

• For \(T(-1,-3)\):

Substitute \(x=-1\) and \(y=-3\) into the rule \((x,y)\to(y, -x)\). The image \(T'\) is \((-3,1)\).

Answer:

The original polygon has vertices \(Q(-6,-3)\), \(R(-5,0)\), \(S(-3,0)\), \(T(-1,-3)\). After a \(270^{\circ}\) clock - wise rotation about the origin, the image polygon has vertices \(Q'(-3,6)\), \(R'(0,5)\), \(S'(0,3)\), \(T'(-3,1)\). Plot these original and image vertices on the coordinate plane.