Question

1. triangle abc has vertices a(1, 7), b(3, 2), and c(−2, −2). graph △abc and its image after a rotation of 270° clockwise about c.

Explanation:

Step1: Recall Rotation Rule

The rule for a \(270^\circ\) clockwise rotation about a point \((h,k)\) is: For a point \((x,y)\), first translate the point so that the center of rotation \((h,k)\) is at the origin by subtracting \((h,k)\), then apply the rotation \((x',y')=(y,-x)\) (since \(270^\circ\) clockwise is equivalent to \(90^\circ\) counter - clockwise), and then translate back by adding \((h,k)\). The formula for rotating a point \((x,y)\) about \((h,k)\) by \(270^\circ\) clockwise is \((x',y')=(h+(y - k),k-(x - h))\).

Step2: Rotate Point A about C

Given \(A(1,7)\) and \(C(-2,-2)\), where \(h=-2,k = - 2,x = 1,y = 7\).
First, calculate \(y - k=7-(-2)=9\) and \(x - h=1-(-2)=3\).
Then, \(x'=h+(y - k)=-2 + 9=7\) and \(y'=k-(x - h)=-2-3=-5\). So the image of \(A\) after rotation, \(A'=(7,-5)\).

Step3: Rotate Point B about C

Given \(B(3,2)\) and \(C(-2,-2)\), where \(h = - 2,k=-2,x = 3,y = 2\).
Calculate \(y - k=2-(-2)=4\) and \(x - h=3-(-2)=5\).
Then, \(x'=h+(y - k)=-2 + 4=2\) and \(y'=k-(x - h)=-2-5=-7\). So the image of \(B\) after rotation, \(B'=(2,-7)\).

Step4: Rotate Point C about C

The image of a point about itself after any rotation is the point itself. So \(C'=C=(-2,-2)\).

Step5: Graphing

• To graph \(\triangle ABC\), plot the points \(A(1,7)\), \(B(3,2)\), and \(C(-2,-2)\) and connect them.
• To graph the image \(\triangle A'B'C'\), plot the points \(A'(7,-5)\), \(B'(2,-7)\), and \(C'(-2,-2)\) and connect them.

Answer:

The vertices of \(\triangle ABC\) are \(A(1,7)\), \(B(3,2)\), \(C(-2,-2)\). The vertices of its image after a \(270^\circ\) clockwise rotation about \(C\) are \(A'(7,-5)\), \(B'(2,-7)\), \(C'(-2,-2)\). To graph, plot the original points and the rotated points and connect them respectively.