Question

triangle rst has vertices r(2, 0), s(4, 0), and t(1, -3). the image of triangle rst after a rotation has vertices r(0, -2), s(0, -4), and t(-3, -1). which rule describes the transformation?
o r0, 90°
o r0, 180°
o r0, 270°
o r0, 360°

Explanation:

Step1: Recall rotation rules

• For a rotation of \(90^{\circ}\) clockwise (\(R_{0,270^{\circ}}\) counter - clockwise) about the origin \((x,y)\to(y, - x)\).
• For a rotation of \(90^{\circ}\) counter - clockwise (\(R_{0,90^{\circ}}\)) about the origin \((x,y)\to(-y,x)\).
• For a rotation of \(180^{\circ}\) about the origin \((x,y)\to(-x,-y)\).
• For a rotation of \(360^{\circ}\) about the origin \((x,y)\to(x,y)\).

Step2: Apply the rules to a vertex

Take vertex \(R(2,0)\).

• If we use the rule for \(270^{\circ}\) counter - clockwise rotation (\(R_{0,270^{\circ}}\)):

Substitute \(x = 2\) and \(y=0\) into the formula \((x,y)\to(y,-x)\).
We get \((2,0)\to(0,- 2)\) which is \(R'\).

• Check another vertex \(S(4,0)\). Using \((x,y)\to(y,-x)\), we have \((4,0)\to(0,-4)\) which is \(S'\).
• Check vertex \(T(1,-3)\). Using \((x,y)\to(y,-x)\), we substitute \(x = 1\) and \(y=-3\) and get \((-3,-1)\) which is \(T'\).

Answer:

\(R_{0,270^{\circ}}\)