Rectangle \( A B C D \) has the following vertices: - \( A(-1,9) \) D(9,4) - \( C(4,-6) \) Is rectangle \( A B C D \) a square, and why? Choose 1 answer: (A) Yes, because \( A B=B C=C D=A D \), and \( A B C D \) is a rectangle. (B) Yes, because \( \overline{A B} \| \overline{C D} \) and \( \overline{B C} \| \overline{A D} \), and \( A B C D \) is a rectangle. (C) No, because \( \overline{A B} \) is longer than \( \overline{B C} \). (D) No, because \( \overline{A B} \) is not parallel to \( \overline{B C} \).

We are given three vertices of rectangle \(ABCD\): \[ A(-1,9),\quad D(9,4),\quad C(4,-6). \] Since \(ABCD\) is a rectangle (and hence a parallelogram), the midpoints of the diagonals must be the same. In any parallelogram, we have \[ A+C = B+D. \] So we can find the unknown vertex \(B\) using \[ B = A+C-D. \] 1. Compute \(A+C\): \[ A+C = (-1,9)+(4,-6) = (3,3). \] 2. Now subtract \(D(9,4)\) to get \(B\): \[ B = (3,3) - (9,4) = (-6,-1). \] Now we have all four vertices: \[ A(-1,9),\quad B(-6,-1),\quad C(4,-6),\quad D(9,4). \] Next we compute the lengths of the sides. - **Side \(\overline{AB}\):** The vector from \(A\) to \(B\) is \[ B-A=(-6-(-1),\,-1-9)=(-5,-10). \] Its length is \[ AB=\sqrt{(-5)^2+(-10)^2}=\sqrt{25+100}=\sqrt{125}=5\sqrt{5}. \] - **Side \(\overline{BC}\):** The vector from \(B\) to \(C\) is \[ C-B=(4-(-6),\,-6-(-1))=(10,-5). \] Its length is \[ BC=\sqrt{10^2+(-5)^2}=\sqrt{100+25}=\sqrt{125}=5\sqrt{5}. \] - **Side \(\overline{CD}\):** The vector from \(C\) to \(D\) is \[ D-C=(9-4,\,4-(-6))=(5,10). \] Its length is \[ CD=\sqrt{5^2+10^2}=\sqrt{25+100}=\sqrt{125}=5\sqrt{5}. \] - **Side \(\overline{DA}\):** The vector from \(D\) to \(A\) is \[ A-D=(-1-9,\,9-4)=(-10,5). \] Its length is \[ DA=\sqrt{(-10)^2+5^2}=\sqrt{100+25}=\sqrt{125}=5\sqrt{5}. \] Each side of \(ABCD\) is of equal length \(5\sqrt{5}\). Since \(ABCD\) is a rectangle (with right angles) and all sides are congruent, it must be a square. Thus, the correct answer is: (A) Yes, because \( AB=BC=CD=DA \), and \( ABCD \) is a rectangle.