Question

under a rotation about the origin, ( a(5,-1) ) is mapped to the point ( a^{prime}(1,5) ). what is the image of ( b(-4,6) ) under this rotation? use words and coordinates to explain.
( b^{prime}(quad,quad) )
the rotation maps ( (x,y) ) to ( (quad,quad) ), so it is a ( 90 ) counterclockwise or ( 270 ) clockwise rotation.
therefore, the image of ( (-4,6) ) is ( (quad,quad) ).

Explanation:

Step1: Determine the rotation rule

Given \(A(5,-1)\) is mapped to \(A'(1,5)\).
For a \(90^{\circ}\) counter - clockwise rotation, the rule is \((x,y)\to(-y,x)\).
For a \(270^{\circ}\) clockwise rotation, the rule is also \((x,y)\to(-y,x)\) (since \(270^{\circ}\) clockwise \(= 90^{\circ}\) counter - clockwise).
Let's check:
If we apply the rule \((x,y)\to(-y,x)\) to \(A(5,-1)\), we get \((-(-1),5)=(1,5)\)

Step2: Apply the rotation rule to point \(B(-4,6)\)

Using the rule \((x,y)\to(-y,x)\)
For \(B(-4,6)\), \(x = - 4\) and \(y = 6\)
The \(x\) - coordinate of \(B'\) is \(-y=-6\)
The \(y\) - coordinate of \(B'\) is \(x=-4\)

Answer:

\(B'(-6,-4)\)
The rotation maps \((x,y)\) to \((-y,x)\), so it is a \(90^{\circ}\) counter - clockwise or \(270^{\circ}\) clockwise rotation.
Therefore, the image of \((-4,6)\) is \((-6,-4)\)