Question

1. what is the outcome of reflecting (\triangle abc), where (a(-2,2)), (b(5,1)), and (c(-3,-1)) over the (x -)axis and then rotating it (270^{circ}) cw?

a(a(2,-2)), (b(-1,5)), and (c(-1,-3))
b(a(-2,2)), (b(1,5)), and (c(1,3))
c(a(3,1)), (b(5,1)), and (c(2,-2))
d(a(-2,2)), (b(5,-1)), and (c(-3,-1))

Explanation:

Step1: Reflect over the x - axis

When reflecting a point \((x,y)\) over the \(x\) - axis, the rule is \((x,y)\to(x, - y)\).
For point \(A(-2,2)\): After reflection over the \(x\) - axis, \(A'(-2,-2)\)
For point \(B(5,1)\): After reflection over the \(x\) - axis, \(B'(5,-1)\)
For point \(C(-3,-1)\): After reflection over the \(x\) - axis, \(C'(-3,1)\)

Step2: Rotate \(270^{\circ}\) clockwise

The rule for a \(270^{\circ}\) clockwise rotation about the origin \((x,y)\to(y,-x)\)
For point \(A'(-2,-2)\):
Substitute \(x=-2\) and \(y = - 2\) into the rotation rule \((y,-x)\), we get \(A''(-2,2)\)
For point \(B'(5,-1)\):
Substitute \(x = 5\) and \(y=-1\) into the rotation rule \((y,-x)\), we get \(B''(-1,-5)\)
For point \(C'(-3,1)\):
Substitute \(x=-3\) and \(y = 1\) into the rotation rule \((y,-x)\), we get \(C''(1,3)\)

Answer:

A"(-2, 2), B"(1, 5), and C"(1, 3)