Question

identifying points on a pre - image
a rectangle was rotated 270° about the origin.
the image is shown.
which points are vertices of the pre - image? check all
that apply.
k(-3,-1)
m(-3,2)
m(3,2)
n(-3,1)
n(3,1)

Explanation:

Step1: Recall the rotation rule

When a point \((x,y)\) is rotated \(270^{\circ}\) counter - clockwise about the origin, the transformation rule is \((x,y)\to(y, - x)\). Let the pre - image point be \((x,y)\) and the image point be \((x',y')\). Then \(x'=y\) and \(y'=-x\), so \(x = - y'\) and \(y=x'\).

Step2: Find the coordinates of \(K'\), \(L'\), \(M'\), \(N'\)

From the graph, \(K'=(1,3)\), \(L'=(2,3)\), \(M'=(2, - 3)\), \(N'=(1, - 3)\).
For a point \((x',y')\) (image), using \(x=-y'\) and \(y = x'\):

• For \(K'\): If \(x' = 1\) and \(y'=3\), then \(x=-3\) and \(y = 1\). The pre - image of \(K'\) is \((-3,1)\) (not \(K(-3,-1)\)).
• For \(L'\): If \(x'=2\) and \(y' = 3\), the pre - image is \((-3,2)\) (which is \(M(-3,2)\)).
• For \(M'\): If \(x'=2\) and \(y'=-3\), the pre - image is \((3,2)\) (not \(M(3,2)\) as per the rotation of the rectangle structure).
• For \(N'\): If \(x'=1\) and \(y'=-3\), the pre - image is \((3,1)\) (which is \(N(3,1)\)).

Answer:

\(M(-3,2)\), \(N(3,1)\)