Question

1. on a grid, draw several different rhombuses. use slopes to determine the relationship between the diagonals.

Explanation:

Step1: Draw a Rhombus (Example 1)

Let's take vertices \( A(0,0) \), \( B(4,0) \), \( C(6,2) \), \( D(2,2) \).
- Diagonal 1: \( AC \). Slope of \( AC \): \( m_{AC}=\frac{2 - 0}{6 - 0}=\frac{2}{6}=\frac{1}{3} \).
- Diagonal 2: \( BD \). Slope of \( BD \): \( m_{BD}=\frac{2 - 0}{2 - 4}=\frac{2}{-2}=- 1 \). Wait, no, wait—wait, rhombus diagonals should bisect at right angles? Wait, maybe my example is wrong. Let's take a better rhombus: \( A(0,0) \), \( B(3,0) \), \( C(4,2) \), \( D(1,2) \).
- Diagonal \( AC \): from \( (0,0) \) to \( (4,2) \), slope \( m_{AC}=\frac{2 - 0}{4 - 0}=\frac{1}{2} \).
- Diagonal \( BD \): from \( (3,0) \) to \( (1,2) \), slope \( m_{BD}=\frac{2 - 0}{1 - 3}=\frac{2}{-2}=-1 \). Wait, no, maybe a rhombus with sides as vectors. Let's use a rhombus with vertices \( A(1,1) \), \( B(4,1) \), \( C(5,4) \), \( D(2,4) \).
- Diagonal \( AC \): from \( (1,1) \) to \( (5,4) \), slope \( m_{AC}=\frac{4 - 1}{5 - 1}=\frac{3}{4} \).
- Diagonal \( BD \): from \( (4,1) \) to \( (2,4) \), slope \( m_{BD}=\frac{4 - 1}{2 - 4}=\frac{3}{-2}=-\frac{3}{2} \). Wait, product of slopes: \( \frac{3}{4}\times(-\frac{3}{2})=-\frac{9}{8} eq - 1 \). Oops, wrong rhombus. Let's take a rhombus with perpendicular diagonals. Let's use \( A(0,0) \), \( B(2,0) \), \( C(3,1) \), \( D(1,1) \). Wait, no, let's use a rhombus with diagonals along axes? No, better: recall that in a rhombus, diagonals are perpendicular. So let's construct a rhombus with diagonals: diagonal 1 from \( (0,0) \) to \( (4,0) \) (horizontal), diagonal 2 from \( (2,1) \) to \( (2, - 1) \) (vertical). Wait, no, rhombus vertices: midpoint of diagonals is same. So midpoint of \( (0,0) \) and \( (4,0) \) is \( (2,0) \); midpoint of \( (2,1) \) and \( (2, - 1) \) is \( (2,0) \). So vertices: \( (0,0) \), \( (4,0) \), \( (2,1) \), \( (2, - 1) \)? No, that's a kite? Wait, no, rhombus has all sides equal. Let's calculate distances: \( AB \): from \( (0,0) \) to \( (4,0) \): length 4. \( BC \): from \( (4,0) \) to \( (2,1) \): \( \sqrt{(2)^2+(1)^2}=\sqrt{5} \). Not equal. Oops. Let's use a rhombus with vertices \( (0,0) \), \( (1,2) \), \( (3,3) \), \( (2,1) \). Wait, maybe better to use a rhombus with sides as \( (3,0) \) and \( (1,2) \). So vertices: \( A(0,0) \), \( B(3,0) \), \( C(4,2) \), \( D(1,2) \). Now, diagonals: \( AC \) from \( (0,0) \) to \( (4,2) \), slope \( \frac{2}{4}=\frac{1}{2} \). \( BD \) from \( (3,0) \) to \( (1,2) \), slope \( \frac{2 - 0}{1 - 3}=\frac{2}{-2}=-1 \). Product of slopes: \( \frac{1}{2}\times(-1)=-\frac{1}{2} eq - 1 \). Wait, I'm making a mistake. Let's recall the formula: in a rhombus, diagonals are perpendicular, so their slopes should multiply to -1. Let's take a rhombus with vertices \( (0,0) \), \( (2,1) \), \( (3,3) \), \( (1,2) \). Wait, distance \( AB \): \( \sqrt{(2)^2+(1)^2}=\sqrt{5} \), \( BC \): \( \sqrt{(1)^2+(2)^2}=\sqrt{5} \), \( CD \): \( \sqrt{(-2)^2+(-1)^2}=\sqrt{5} \), \( DA \): \( \sqrt{(-1)^2+(-2)^2}=\sqrt{5} \). Good, rhombus. Diagonals: \( AC \) from \( (0,0) \) to \( (3,3) \), slope \( \frac{3}{3}=1 \). \( BD \) from \( (2,1) \) to \( (1,2) \), slope \( \frac{2 - 1}{1 - 2}=\frac{1}{-1}=-1 \). Now, product of slopes: \( 1\times(-1)=-1 \). Ah, there we go! So diagonals are perpendicular.

Step2: Generalize (Using Slope Property)

For any rhombus, the diagonals are perpendicular. To show this with slopes: let the diagonals be \( d_1 \) and \( d_2 \). The slope of \( d_1 \) is \( m_1 \), slope of \( d_2 \) is \( m_2 \). In a rhombus, diagonals bisect each other at right angles, so \( m_1\times m_2=-1 \) (property of perpendicular lines: product of slopes is -1, unless one is vertical and the other horizontal).

Step3: Draw More Rhombuses (Example 2)

Take vertices \( A(-1,-1) \), \( B(2,-1) \), \( C(3,2) \), \( D(0,2) \). Diagonal \( AC \): from \( (-1,-1) \) to \( (3,2) \), slope \( \frac{2 - (-1)}{3 - (-1)}=\frac{3}{4} \). Diagonal \( BD \): from \( (2,-1) \) to \( (0,2) \), slope \( \frac{2 - (-1)}{0 - 2}=\frac{3}{-2}=-\frac{3}{2} \). Wait, product: \( \frac{3}{4}\times(-\frac{3}{2})=-\frac{9}{8} eq - 1 \). Wait, no, this is not a rhombus. Wait, distance \( AB \): \( 3 \), \( BC \): \( \sqrt{(1)^2+(3)^2}=\sqrt{10} \), not equal. Oops, wrong vertices. Let's take \( A(0,0) \), \( B(1,3) \), \( C(4,4) \), \( D(3,1) \). Distance \( AB \): \( \sqrt{1 + 9}=\sqrt{10} \), \( BC \): \( \sqrt{9 + 1}=\sqrt{10} \), \( CD \): \( \sqrt{9 + 1}=\sqrt{10} \), \( DA \): \( \sqrt{9 + 1}=\sqrt{10} \). Good, rhombus. Diagonal \( AC \): from \( (0,0) \) to \( (4,4) \), slope \( 1 \). Diagonal \( BD \): from \( (1,3) \) to \( (3,1) \), slope \( \frac{1 - 3}{3 - 1}=\frac{-2}{2}=-1 \). Product: \( 1\times(-1)=-1 \). Perfect, perpendicular.

Step4: Conclusion

By calculating slopes of diagonals for multiple rhombuses, we see that the product of their slopes is \( - 1 \) (or one is vertical, the other horizontal, where one slope is undefined and the other is 0, also satisfying perpendicularity). Thus, the diagonals of a rhombus are perpendicular.

Answer:

The diagonals of a rhombus are perpendicular (their slopes multiply to \(-1\), or one is vertical and the other horizontal).