Question

Complete the coordinate proof of the theorem. Given: A B C D is a square. Prove: The diagonals of A B C D are perpendicular. Art: A square A B C D is graphed on a coordinate plane. The horizontal x-axis and vertical y-axis are solid. The vertex labeled as A lies on begin ordered pair 0 comma 0 end ordered pair. The vertex labeled as B lies on begin ordered pair a comma 0 end ordered pair. The vertices C and D are unlabeled. Diagonal A C and B D are drawn by dotted lines. Enter your answers in the boxes. The coordinates of square ABCD are A(0, 0) , B(a, 0) , C(, a), and D(0, ). The slope of AC⎯⎯⎯⎯⎯ , when simplified, is equal to . The slope of BD⎯⎯⎯⎯⎯, when simplified, is equal to −1 . The product of the slopes is equal to . Therefore, AC⎯⎯⎯⎯⎯ is perpendicular to BD⎯⎯⎯⎯⎯.

Answer 1

The slope of AC = 1

Product of both slopes AC and BD = -1

How to prove a quadrilateral?

Now, we know that properties of squares are;

- All sides are equal

- All angles are equal and are right angles

- Diagonals are perpendicular to each other

- Diagonals bisect each other.

Now, we are given that Slope of BD which is the diagonal = -1

Now, the other slope is AC and by virtue of properties of squares, we can say that; AC = -1/Slope of BD = -1/-1 = 1

Product of both slopes AC and BD = -1 * 1 = -1

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