To show that the points A(0, 6), B(2, 1), and C(7, 3) are three corners of a square ABCD, we will follow these steps:
### Step 1: Calculate the slopes of AB and BC
The slope of a line through two points (x1, y1) and (x2, y2) is given by the formula:
\[
\text{slope} = \frac{y_2 - y_1}{x_2 - x_1}
\]
**Calculating the slope of AB:**
- Points A(0, 6) and B(2, 1)
\[
\text{slope of AB} = \frac{1 - 6}{2 - 0} = \frac{-5}{2}
\]
**Calculating the slope of BC:**
- Points B(2, 1) and C(7, 3)
\[
\text{slope of BC} = \frac{3 - 1}{7 - 2} = \frac{2}{5}
\]
### Step 2: Check if AB is perpendicular to BC
Two lines are perpendicular if the product of their slopes is -1.
\[
\text{slope of AB} \times \text{slope of BC} = \left(\frac{-5}{2}\right) \times \left(\frac{2}{5}\right) = -1
\]
Since the product is -1, AB is perpendicular to BC.
### Step 3: Calculate the slope of AC
**Calculating the slope of AC:**
- Points A(0, 6) and C(7, 3)
\[
\text{slope of AC} = \frac{3 - 6}{7 - 0} = \frac{-3}{7}
\]
### Step 4: Calculate the slope of BD
Since the diagonals of a square are perpendicular, we can find the slope of BD using the relationship:
\[
\text{slope of AC} \times \text{slope of BD} = -1
\]
Let the slope of BD be \( m_{BD} \):
\[
\left(\frac{-3}{7}\right) \times m_{BD} = -1 \implies m_{BD} = \frac{7}{3}
\]
### Step 5: Find the coordinates of the fourth vertex D
To find the coordinates of vertex D, we will use the midpoint formula. The midpoint of AC should equal the midpoint of BD.
**Midpoint of AC:**
\[
\text{Midpoint of AC} = \left(\frac{0 + 7}{2}, \frac{6 + 3}{2}\right) = \left(\frac{7}{2}, \frac{9}{2}\right)
\]
**Let D be (x, y). The midpoint of BD:**
\[
\text{Midpoint of BD} = \left(\frac{2 + x}{2}, \frac{1 + y}{2}\right)
\]
Setting the midpoints equal:
\[
\frac{2 + x}{2} = \frac{7}{2} \quad \text{and} \quad \frac{1 + y}{2} = \frac{9}{2}
\]
From the first equation:
\[
2 + x = 7 \implies x = 5
\]
From the second equation:
\[
1 + y = 9 \implies y = 8
\]
Thus, the coordinates of point D are (5, 8).
### Final Answers:
1. The slope of diagonal BD is \( \frac{7}{3} \).
2. The coordinates of the fourth vertex D are (5, 8).
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