Question

a figure is fully contained in quadrant ii. the figure is transformed as shown.

• a reflection over the x - axis
• a reflection over the line y = x
• a 90° counterclockwise rotation about the origin.

in which quadrant does the resulting image lie?
a quadrant iv
b quadrant ii
c quadrant iii
d quadrant i

Explanation:

Step1: Analyze reflection over x - axis

A point \((x,y)\) in Quadrant II has \(x<0,y > 0\). After reflection over the \(x\) - axis, the coordinates become \((x,-y)\). Since \(x<0\) and \(-y<0\), the point moves to Quadrant III.

Step2: Analyze reflection over \(y = x\)

Take a point \((x,-y)\) from Quadrant III (\(x<0,-y<0\) implies \(y > 0\)). After reflection over \(y=x\), the coordinates become \((-y,x)\). Now, \(-y<0\) (because \(y > 0\)) and \(x<0\), so the point is still in Quadrant III? Wait, no, let's correct. Wait, after first reflection (over \(x\) - axis) of a point \((-a,b)\) (where \(a>0,b > 0\), since it's in Quadrant II) becomes \((-a,-b)\) (Quadrant III). Then reflection over \(y = x\) of \((-a,-b)\) is \((-b,-a)\). Both \(-b<0\) and \(-a<0\), so still Quadrant III.

Step3: Analyze \(90^{\circ}\) counter - clockwise rotation about origin

The rule for a \(90^{\circ}\) counter - clockwise rotation about the origin is \((x,y)\to(-y,x)\). Take the point \((-b,-a)\) (from Quadrant III, \(b>0,a > 0\)). Applying the rotation rule: \((-(-a),-b)=(a,-b)\). Since \(a>0\) and \(-b<0\), the point \((a,-b)\) lies in Quadrant IV.

Answer:

A. Quadrant IV