4.3 Q-7

Question Statement

The points $A(-1,2)$, $B(3,-1)$, and $C(6, \ldots)$ are consecutive vertices of a rhombus. Find the fourth vertex and show that the diagonals of the rhombus are perpendicular to each other.

---

Background and Explanation

To solve this problem, we need to apply properties of rhombuses and parallel lines. A rhombus is a type of parallelogram where all sides are of equal length, and its diagonals bisect each other at right angles. Understanding slopes and how they relate to perpendicular lines will be crucial here.

---

Solution

We are asked to find the fourth vertex $D(a, b)$ of the rhombus, given three vertices $A(-1, 2)$, $B(3, -1)$, and $C(6, \ldots)$.

1. Calculate the slope of line $\overline{AB}$:

   The formula for the slope between two points $(x_1, y_1)$ and $(x_2, y_2)$ is:

$$m = \frac{y_2 - y_1}{x_2 - x_1}$$

For $A(-1, 2)$ and $B(3, -1)$:

$$m_{AB} = \frac{-1 - 2}{3 - (-1)} = \frac{-3}{4}$$

2. Set up the slope equation for line $\overline{CD}$:

   Since the opposite sides of a rhombus are parallel, the slope of $\overline{AB}$ will be equal to the slope of $\overline{CD}$.

   The slope of line $\overline{CD}$ is:

$$m_{CD} = \frac{b - 3}{a - 6}$$

Since $m_{AB} = m_{CD}$:

$$\frac{-3}{4} = \frac{b - 3}{a - 6}$$

Cross-multiply to get:

$$-3(a - 6) = 4(b - 3)$$

Simplifying:

$$-3a + 18 = 4b - 12$$ $$3a + 4b = 30 \tag{1}$$

3. Set up the slope equation for line $\overline{BC}$ and $\overline{AD}$:

   Similarly, the slope of line $\overline{BC}$ will be equal to the slope of line $\overline{AD}$ because opposite sides of a rhombus are parallel.

   The slope of line $\overline{BC}$ is:

$$m_{BC} = \frac{3 + 1}{6 - 3} = \frac{4}{3}$$

The slope of line $\overline{AD}$ is:

$$m_{AD} = \frac{b - 2}{a + 1}$$

Since $m_{BC} = m_{AD}$:

$$\frac{4}{3} = \frac{b - 2}{a + 1}$$

Cross-multiply to get:

$$4(a + 1) = 3(b - 2)$$

Simplifying:

$$4a + 4 = 3b - 6$$ $$4a - 3b = -10 \tag{2}$$

4. Solve the system of equations (1) and (2):

   We now have two equations:

   1. $3a + 4b = 30$
   2. $4a - 3b = -10$

   To solve this system, we will multiply equation (1) by 4 and equation (2) by 3 to eliminate $b$:

$$12a + 16b = 120 \tag{3}$$ $$12a - 9b = -30 \tag{4}$$

Subtract equation (4) from equation (3):

$$(12a + 16b) - (12a - 9b) = 120 - (-30)$$ $$25b = 150$$ $$b = 6$$

5. Find $a$:

   Substitute $b = 6$ into equation (1):

$$3a + 4(6) = 30$$ $$3a + 24 = 30$$ $$3a = 6$$ $$a = 2$$

Thus, the coordinates of the fourth vertex $D$ are $(2, 6)$.

6. Verify that the diagonals are perpendicular:

   The diagonals of a rhombus are perpendicular. To check this, we calculate the slopes of diagonals $\overline{AC}$ and $\overline{BD}$:

   • The slope of diagonal $\overline{AC}$ is:

$$m_{AC} = \frac{3 - 2}{6 - (-1)} = \frac{1}{7}$$

• The slope of diagonal $\overline{BD}$ is:

$$m_{BD} = \frac{6 - (-1)}{2 - 3} = \frac{7}{-1} = -7$$

The product of the slopes of two perpendicular lines is $-1$. Check:

$$m_{AC} \times m_{BD} = \frac{1}{7} \times (-7) = -1$$

Therefore, the diagonals are perpendicular.

---

Key Formulas or Methods Used

• Slope formula:

$$m = \frac{y_2 - y_1}{x_2 - x_1}$$

• The slopes of opposite sides of a rhombus are equal.
• The diagonals of a rhombus are perpendicular, meaning the product of their slopes equals $-1$.

---

Summary of Steps

1. Calculate the slope of line $\overline{AB}$.
2. Set up the equation for the slope of line $\overline{CD}$ and solve for $a$ and $b$.
3. Calculate the slope of line $\overline{BC}$ and $\overline{AD}$, and set up the equation for $a$ and $b$.
4. Solve the system of equations to find $a = 2$ and $b = 6$.
5. Verify that the diagonals are perpendicular by checking the product of their slopes.