Question

triangle pqr is located entirely in quadrant ii. which transformation must result in an image of triangle pqr located entirely in quadrant iii?
o a. reflection over the y - axis
o b. reflection over the x - axis
o c. clockwise 90° rotation around the origin
o d. translation of 5 units down

Explanation:

• Reflection over the \(y\) - axis: Reflecting a point \((x,y)\) over the \(y\) - axis gives \((-x,y)\). A triangle in Quadrant II (where \(x<0,y > 0\)) would be reflected to Quadrant I (where \(x>0,y>0\)).
• Reflection over the \(x\) - axis: Reflecting a point \((x,y)\) over the \(x\) - axis gives \((x, - y)\). A triangle in Quadrant II (where \(x<0,y>0\)) would be reflected to Quadrant III (where \(x < 0,y<0\)).
• Clockwise \(90^{\circ}\) rotation around the origin: The rotation rule for a clockwise \(90^{\circ}\) rotation of a point \((x,y)\) around the origin is \((y,-x)\). A point \((x,y)\) in Quadrant II (\(x<0,y > 0\)) would be rotated to \((y,-x)\) which is in Quadrant IV (\(y>0,-x>0\)).
• Translation of 5 units down: Translating a point \((x,y)\) 5 units down gives \((x,y - 5)\). If \(y\) is large enough, the triangle may still be in Quadrant II (for example, if \(y=6\), then \(y-5 = 1>0\) and \(x<0\)).

Answer:

B. reflection over the \(x\) - axis