Final answer:
WXYZ is not a square because, despite WX being perpendicular to WZ, XY and YZ are not perpendicular to each other as required for the sides of a square.
Step-by-step explanation:
The student's question asks whether a quadrilateral with the given slopes for its sides is a square. In the quadrilateral WXYZ, the slope of WX is 4/3, the slope of XY is 0 (indicating a horizontal line), the slope of YZ is -3/4, and the slope of WZ is -4/3.
For a quadrilateral to be a square, it must not only have four equal sides but also the sides must be perpendicular to each other where they meet; this would mean the slopes of adjacent sides must be negative reciprocals of each other. Here, the slope of WX is the negative reciprocal of WZ, which indicates that these sides are indeed perpendicular.
However, the slope of XY is 0 and the slope of YZ is -3/4; a slope of 0 means the line is horizontal, and -3/4 is not the negative reciprocal of 0, as 0 does not have a reciprocal. Therefore, sides XY and YZ are not perpendicular, and thus WXYZ cannot be a square. The statement in the question is false.