Question 1116014
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part (a)<br>
Diagonal AC contains the points (1,0) and (4,3); the equation of the line through those two points is y = x-1.<br>
Diagonal BD contains the points (3,1) and (2,2); the equation of the line through those two points is y = -x+4.<br>
The intersection of the two diagonals is the common point on the graphs of y = x-1 and y = -x+4; that point is (2.5,1.5).<br>
The midpoints of AC and BD are also both (2.5,1.5).<br>
The intersection of the two diagonals is the midpoint of both, so the diagonals bisect each other.<br>
part (b)<br>
The fact that the diagonals bisect each other means the quadrilateral is a parallelogram.<br>
In fact, the quadrilateral is a rhombus, because the two diagonals are perpendicular to each other (the product of their slopes is -1).<br>
But we were not asked to show that the diagonals are perpendicular; only knowing that the diagonals bisect each other only tells us that the quadrilateral is a parallelogram.