Answer: To determine whether quadrilateral ABCD is a rectangle, rhombus, or square, we need to examine the properties of its diagonals. Specifically, we need to determine whether the diagonals are perpendicular and/or bisect each other.
First, let's find the equations of the diagonals. The two diagonals of quadrilateral ABCD are the line segments connecting opposite vertices:
Diagonal AC connects (−6,5) and (2,7). Its equation is y = (1/4)x + 31/8.
Diagonal BD connects (−1,2) and (−3,10). Its equation is y = -2x + 4.
To see if the diagonals are perpendicular, we need to check if their slopes are negative reciprocals of each other:
(1/4) * (-1/2) = -1/8
Since the product of the slopes is not -1, the diagonals are not perpendicular, so we can eliminate the possibility of the quadrilateral being a square.
To see if the diagonals bisect each other, we need to find the midpoints of the diagonals:
The midpoint of AC is ((-6+2)/2, (5+7)/2) = (-2, 6).
The midpoint of BD is ((-1-3)/2, (2+10)/2) = (-2, 6).
We can see that the midpoints are the same, which means that the diagonals bisect each other. This property is true for all parallelograms, so we cannot use this property to distinguish between the remaining possibilities of rectangle and rhombus.
To determine whether the quadrilateral is a rectangle or a rhombus, we need to use the distance formula to find the lengths of the sides:
AB = √[(1-5)^2 + (2-5)^2] = √20
BC = √[(2-(-1))^2 + (7-2)^2] = √50
CD = √[(−3-2)^2 + (10-7)^2] = √34
DA = √[(−6-(-3))^2 + (5-10)^2] = √34
Now we can compare the lengths of the sides to determine the type of quadrilateral:
A rectangle has two pairs of opposite sides that are congruent and parallel. In this case, AB is not congruent to CD, so ABCD is not a rectangle.
A rhombus has four congruent sides. In this case, all four sides have different lengths, so ABCD is not a rhombus.
Therefore, we can conclude that quadrilateral ABCD is neither a rectangle nor a rhombus.
Explanation: