Question

find a congruence transformation that maps \\( \triangle n p b \\) to \\( \triangle k h d \\).
which congruence transformation maps the preimage to the congruent image? select the correct choice below and fill in the answer boxes to complete you choice.
a. \\( t _ { ( square , square ) } circ r _ { ( 2 7 0 ^ { circ } , o ) } ( delta n p b ) \\)
b. \\( t _ { ( square , square ) } circ r _ { y - \text { axis } } ( delta n p b ) \\)
c. \\( t _ { ( square , square ) } circ r _ { ( 9 0 ^ { circ } , o ) } ( delta n p b ) \\)
d. \\( t _ { ( square , square ) } circ r _ { x - \text { axis } } ( delta n p b ) \\)

Explanation:

Step1: Determine the rotation

First, consider the rotation. If we rotate \(\triangle NPB\) \(270^{\circ}\) clockwise about the origin \((x,y)\to(y, - x)\).
Let's take a point \(N(- 4,4)\). After a \(270^{\circ}\) clockwise rotation about the origin \(O\), using the formula \((x,y)\to(y, - x)\), we get \(N'=(4,4)\).

Step2: Determine the translation

After rotation, we need to translate. The \(x\) - coordinate of the rotated point \(N'\) (from \(N\) after \(270^{\circ}\) rotation) needs to change from \(4\) to \(-6\) (the \(x\) - coordinate of \(K\)), and the \(y\) - coordinate of the rotated point \(N'\) needs to change from \(4\) to \(-6\) (the \(y\) - coordinate of \(K\)).
The translation rule is \(T_{(x,y)}\) where \(x=-6 - 4=-10\) and \(y=-6 - 4=-10\).

Answer:

A. \(T_{(-10,-10)}\circ r_{(270^{\circ},O)}(\triangle NPB)\)