2. The points $$ A(-2,3), B(1,8), C(6,11) $$ and $$ D(3,6) $$ are the vertices of a quadrilateral. a. Find the gradients of $$ A B $$ and $$ D C $$. What does this tell you about the lines $$ A B $$ and $$ D C $$ ? b. Find the equation of the diagonal joining $$ B $$ to $$ D $$. c. The diagonal $$ A C $$ has the equation $$ y=x+5 $$. Find the point of intersection of the two diagonals $$ A C $$ and $$ B D $$. d. Find the midpoint of $$ A C $$ and compare this point with the point of intersection of the diagonals $$ A C $$ and $$ B D $$.

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

---

### **a. Find the gradients of $$ AB $$ and $$ DC $$. What does this tell you about the lines $$ AB $$ and $$ DC $$?**

**Given Points:**
- $$ A(-2, 3) $$
- $$ B(1, 8) $$
- $$ C(6, 11) $$
- $$ D(3, 6) $$

**Gradient of $$ AB $$:**
$$
m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{8 - 3}{1 - (-2)} = \frac{5}{3}
$$

**Gradient of $$ DC $$:**
$$
m_{DC} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{6 - 11}{3 - 6} = \frac{-5}{-3} = \frac{5}{3}
$$

**Conclusion:**
Both lines $$ AB $$ and $$ DC $$ have the same gradient of $$ \frac{5}{3} $$. This implies that **lines $$ AB $$ and $$ DC $$ are parallel**.

---

### **b. Find the equation of the diagonal joining $$ B $$ to $$ D $$ (Line $$ BD $$).**

**Given Points:**
- $$ B(1, 8) $$
- $$ D(3, 6) $$

**Step 1: Calculate the Gradient of $$ BD $$**
$$
m_{BD} = \frac{6 - 8}{3 - 1} = \frac{-2}{2} = -1
$$

**Step 2: Use the Point-Slope Form to Find the Equation**
Using point $$ B(1, 8) $$:
$$
y - y_1 = m(x - x_1) \
y - 8 = -1(x - 1) \
y - 8 = -x + 1 \
y = -x + 9
$$

**Equation of Diagonal $$ BD $$:**
$$
\boxed{y = -x + 9}
$$

---

### **c. The diagonal $$ AC $$ has the equation $$ y = x + 5 $$. Find the point of intersection of the two diagonals $$ AC $$ and $$ BD $$.**

**Equations of Diagonals:**
- $$ AC: y = x + 5 $$
- $$ BD: y = -x + 9 $$

**Find Intersection Point:**
$$
x + 5 = -x + 9 \
2x = 4 \
x = 2
$$
$$
y = 2 + 5 = 7
$$

**Point of Intersection:**
$$
\boxed{(2,\ 7)}
$$

---

### **d. Find the midpoint of $$ AC $$ and compare this point with the point of intersection of the diagonals $$ AC $$ and $$ BD $$.**

**Given Points:**
- $$ A(-2, 3) $$
- $$ C(6, 11) $$

**Midpoint Formula:**
$$
\text{Midpoint} = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)
$$
$$
\text{Midpoint of } AC = \left( \frac{-2 + 6}{2}, \frac{3 + 11}{2} \right) = \left( \frac{4}{2}, \frac{14}{2} \right) = (2,\ 7)
$$

**Comparison:**
The midpoint of $$ AC $$ is **$$ (2,\ 7) $$**, which is **exactly the same as the point of intersection of the diagonals $$ AC $$ and $$ BD $$**. This indicates that the diagonals bisect each other at the midpoint, a property characteristic of **parallelograms**.

---

**Summary:**
- **a.** Both $$ AB $$ and $$ DC $$ have a gradient of $$ \frac{5}{3} $$, indicating they are parallel.
- **b.** The equation of diagonal $$ BD $$ is $$ y = -x + 9 $$.
- **c.** The diagonals $$ AC $$ and $$ BD $$ intersect at the point $$ (2, 7) $$.
- **d.** The midpoint of $$ AC $$ is $$ (2, 7) $$, which coincides with the intersection point of the diagonals, confirming that the diagonals bisect each other.
- **a.** Lines $$ AB $$ and $$ DC $$ are parallel because they have the same slope. - **b.** The equation of diagonal $$ BD $$ is $$ y = -x + 9 $$. - **c.** The diagonals $$ AC $$ and $$ BD $$ intersect at the point $$ (2, 7) $$. - **d.** The midpoint of $$ AC $$ is $$ (2, 7) $$, which is the same as the intersection point of the diagonals, showing that the diagonals bisect each other.

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