Question

what are the vertices of the image ( r(90^{circ}, o)(\triangle bcd) )?

( b(-3,-3), c(-1,4), d(2,0) )
( b(3,3), c(1,-4), d(-2,0) )
( b(3,-1), c(1,4), d(0,2) )
( b(-3,-3), c(-1,-4), d(0,-2) )

Explanation:

Step1: Recall the rotation rule

When rotating a point \((x,y)\) \(90^{\circ}\) counter - clockwise about the origin \((0,0)\), the transformation rule is \((x,y)\to(-y,x)\).

Step2: Apply the rule to point \(B(-3,3)\)

For \(B(-3,3)\), using the rule \((x,y)\to(-y,x)\), we substitute \(x = - 3\) and \(y = 3\). Then \(x'=-3\) and \(y'=-3\), so \(B'(-3,-3)\).

Step3: Apply the rule to point \(C(4,1)\)

For \(C(4,1)\), substitute \(x = 4\) and \(y = 1\) into \((x,y)\to(-y,x)\). Then \(x'=-1\) and \(y' = 4\), so \(C'(-1,4)\).

Step4: Apply the rule to point \(D(0,-2)\)

For \(D(0,-2)\), substitute \(x = 0\) and \(y=-2\) into \((x,y)\to(-y,x)\). Then \(x' = 2\) and \(y'=0\), so \(D'(2,0)\).

Answer:

\(B'(-3,-3),C'(-1,4),D'(2,0)\) (the first option)