Question

graph the image of δijk after the following sequence of transformations: reflection across the x-axis rotation 270° counterclockwise around the origin

Explanation:

Step1: Identify original coordinates

From graph: \(I(6,10)\), \(J(4,4)\), \(K(-4,10)\)

Step2: Reflect across x-axis

Rule: \((x,y)\to(x,-y)\)
\(I'(6,-10)\), \(J'(4,-4)\), \(K'(-4,-10)\)

Step3: Rotate 270° counterclockwise

Rule: \((x,y)\to(y,-x)\)
\(I''(-10,-6)\), \(J''(-4,-4)\), \(K''(-10,4)\)

Step4: Plot final points

Connect \(I''(-10,-6)\), \(J''(-4,-4)\), \(K''(-10,4)\)

Answer:

The image of \(\Delta IJK\) after transformations has vertices at \((-10,-6)\), \((-4,-4)\), and \((-10,4)\) (plot these points to form the triangle).