Question 1 A Rhombus has coordinates $$ \mathrm{A}(-3,1), \mathrm{B}(1,3), \mathrm{C}(3,7), \mathrm{D}(-1,5) $$. (i) Show that the diagonals of this rhombus are perpendicular (ii) Find the equation of the longest diagonal (iii) Using the information from (ii) above, find the coordinates of the point of intersection of the diagonals Hint: Plot the points on a set of axis To complete (i)-(iii), you need to complete the following steps - Find the gradient of each diagonal and show that they are perpendicular. - Find the length of each of the diagonals - Identify the longest diagonal and find its equation - Give the coordinates of the vertices of the rhombus

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Let's solve each part of the problem step-by-step.

### **Given:**
- **Vertices of the Rhombus:**
  - $$ \mathrm{A}(-3,1) $$
  - $$ \mathrm{B}(1,3) $$
  - $$ \mathrm{C}(3,7) $$
  - $$ \mathrm{D}(-1,5) $$

### **Part (i): Show that the diagonals of the rhombus are perpendicular**

**Step 1: Identify the Diagonals**

In a rhombus, the diagonals connect opposite vertices. Thus, the diagonals are:
- **Diagonal AC:** Connects points $$ \mathrm{A}(-3,1) $$ and $$ \mathrm{C}(3,7) $$
- **Diagonal BD:** Connects points $$ \mathrm{B}(1,3) $$ and $$ \mathrm{D}(-1,5) $$

**Step 2: Calculate the Slopes of the Diagonals**

- **Slope of Diagonal AC ($$ m_{AC} $$):**
  $$
  m_{AC} = \frac{y_C - y_A}{x_C - x_A} = \frac{7 - 1}{3 - (-3)} = \frac{6}{6} = 1
  $$

- **Slope of Diagonal BD ($$ m_{BD} $$):**
  $$
  m_{BD} = \frac{y_D - y_B}{x_D - x_B} = \frac{5 - 3}{-1 - 1} = \frac{2}{-2} = -1
  $$

**Step 3: Verify Perpendicularity**

Two lines are perpendicular if the product of their slopes is $$-1$$.

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

**Conclusion:** Since the product of the slopes is $$-1$$, diagonals AC and BD are perpendicular.

---

### **Part (ii): Find the equation of the longest diagonal**

**Step 1: Determine the Lengths of the Diagonals**

- **Length of Diagonal AC ($$ L_{AC} $$):**
  $$
  L_{AC} = \sqrt{(x_C - x_A)^2 + (y_C - y_A)^2} = \sqrt{(3 - (-3))^2 + (7 - 1)^2} = \sqrt{6^2 + 6^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2}
  $$

- **Length of Diagonal BD ($$ L_{BD} $$):**
  $$
  L_{BD} = \sqrt{(x_D - x_B)^2 + (y_D - y_B)^2} = \sqrt{(-1 - 1)^2 + (5 - 3)^2} = \sqrt{(-2)^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}
  $$

**Step 2: Identify the Longest Diagonal**

Comparing the lengths:
$$
6\sqrt{2} > 2\sqrt{2}
$$
Thus, **Diagonal AC** is the longest.

**Step 3: Find the Equation of Diagonal AC**

We already have the slope $$ m_{AC} = 1 $$ and a point on the line $$ \mathrm{A}(-3,1) $$.

Using the point-slope form:
$$
y - y_1 = m(x - x_1)
$$
$$
y - 1 = 1(x + 3) \Rightarrow y = x + 4
$$

**Conclusion:** The equation of the longest diagonal AC is $$ y = x + 4 $$.

---

### **Part (iii): Find the coordinates of the point of intersection of the diagonals**

**Observation:** In a rhombus, the diagonals bisect each other. Therefore, the point of intersection is the midpoint of both diagonals.

**Step 1: Calculate the Midpoint of Diagonal AC**

$$
\text{Midpoint} = \left( \frac{x_A + x_C}{2}, \frac{y_A + y_C}{2} \right) = \left( \frac{-3 + 3}{2}, \frac{1 + 7}{2} \right) = (0, 4)
$$

**Alternatively, Calculate the Midpoint of Diagonal BD:**

$$
\text{Midpoint} = \left( \frac{x_B + x_D}{2}, \frac{y_B + y_D}{2} \right) = \left( \frac{1 + (-1)}{2}, \frac{3 + 5}{2} \right) = (0, 4)
$$

**Conclusion:** The diagonals intersect at the point $$ (0, 4) $$.

---

### **Final Answers:**
1. **(i)** The diagonals AC and BD have slopes $$1$$ and $$-1$$, respectively. Since their product is $$-1$$, the diagonals are perpendicular.
2. **(ii)** The equation of the longest diagonal AC is $$ y = x + 4 $$.
3. **(iii)** The diagonals intersect at the point $$ (0, 4) $$.
The diagonals of the rhombus are perpendicular, the longest diagonal is $$ y = x + 4 $$, and the diagonals intersect at $$ (0, 4) $$.

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