Question

graph the triangle rotated 90° counterclockwise about the origin

Explanation:

Step1: Recall the rotation formula

For a point \((x,y)\) rotated \(90^{\circ}\) counter - clockwise about the origin, the new coordinates \((x',y')\) are given by the formula \(x'=-y\) and \(y' = x\).

Step2: Apply the formula to each vertex

• For vertex \(A(2,2)\):

\(x = 2\), \(y = 2\). Using \(x'=-y\) and \(y'=x\), we get \(x'=-2\), \(y' = 2\). So \(A'(-2,2)\).

• For vertex \(B(7,1)\):

\(x = 7\), \(y = 1\). Using \(x'=-y\) and \(y'=x\), we get \(x'=-1\), \(y' = 7\). So \(B'(-1,7)\).

• For vertex \(C(7,-4)\):

\(x = 7\), \(y=-4\). Using \(x'=-y\) and \(y'=x\), we get \(x' = 4\), \(y'=7\). So \(C'(4,7)\).

Step3: Plot the new points

Plot the points \(A'(-2,2)\), \(B'(-1,7)\), \(C'(4,7)\) on the coordinate plane and connect them to form the rotated triangle.

Answer:

Let the original vertices of the triangle be \(A(2,2)\), \(B(7,1)\), \(C(7, - 4)\). After a \(90^{\circ}\) counter - clockwise rotation about the origin, the new vertices are \(A'(-2,2)\), \(B'(-1,7)\), \(C'(4,7)\).