Exercise 21 Page 417

We are asked to prove that the diagonals of a rhombus are perpendicular. We will write a coordinate proof.

Placement of the Rhombus in the Coordinate Plane

Since a rhombus is a parallelogram, we can use the following coordinates for the vertices.

• We start with A(0,0), B(a,b), and D(c,0).
• The y-coordinate of C is set as the y-coordinate of B so that BC is parallel to the x-axis.
• The x-coordinate of C is set so that BC and AD have the same (horizontal) length.

Formulas Needed

Let's use the Distance Formula to find the length of AB and AD. AB&=sqrt((a-0)^2+(b-0)^2)=sqrt(a^2+b^2) AD&=sqrt((c-0)^2+(0-0)^2)=sqrt(c^2)=c Since ABCD is a rhombus, we know that AB=AD. This gives us a relationship between the variables a, b, and c. sqrt(a^2+b^2)=c ⟹ a^2+b^2=c^2 We are asked to investigate the direction of the diagonals, so let's use the Slope Formula to find the slopes.

Diagonal	Slope (y_2-y_1/x_2-x_1)
	Substitution	Simplification
AC	b-0/a+c-0	m_(AC)=b/a+c
DB	b-0/a-c	m_(DB)=b/a-c

Finishing the Proof

We are asked to show that the diagonals are perpendicular. This is true if the slopes multiply to - 1, so let's multiply the expressions above and simplify the product. We know from above that a^2+b^2=c^2. This can be rearranged to a^2-c^2=- b^2, so let's substitute this in the denominator. Since the product of the slopes is - 1, diagonals AC and DB of rhombus ABCD are indeed perpendicular. This finishes the coordinate proof of Theorem 6-13.