Question

coordinate proof the rectangle with points at (-8,-6), (8,2), (10,2), (-6,-10)

Answer 1

To prove that the given points form a rectangle, we need to show that the opposite sides are equal in length and that the adjacent sides are perpendicular. Let's label the points for convenience:

- \( A = (-8, -6) \)
- \( B = (8, 2) \)
- \( C = (10, 2) \)
- \( D = (-6, -10) \)

### Step 1: Calculate the lengths of the sides

**Length of AB:**
\[
AB = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2} = \sqrt{(8 - (-8))^2 + (2 - (-6))^2}
= \sqrt{(8 + 8)^2 + (2 + 6)^2}
= \sqrt{16^2 + 8^2}
= \sqrt{256 + 64}
= \sqrt{320}
= 8\sqrt{5}
\]

**Length of BC:**
\[
BC = \sqrt{(x_C - x_B)^2 + (y_C - y_B)^2} = \sqrt{(10 - 8)^2 + (2 - 2)^2}
= \sqrt{(2)^2 + (0)^2}
= \sqrt{4}
= 2
\]

**Length of CD:**
\[
CD = \sqrt{(x_D - x_C)^2 + (y_D - y_C)^2} = \sqrt{(-6 - 10)^2 + (-10 - 2)^2}
= \sqrt{(-16)^2 + (-12)^2}
= \sqrt{256 + 144}
= \sqrt{400}
= 20
\]

**Length of DA:**
\[
DA = \sqrt{(x_A - x_D)^2 + (y_A - y_D)^2} = \sqrt{(-8 - (-6))^2 + (-6 - (-10))^2}
= \sqrt{(-2)^2 + (4)^2}
= \sqrt{4 + 16}
= \sqrt{20}
= 2\sqrt{5}
\]

### Step 2: Check pairs of equal lengths

From the calculations:
- \( AB = 8\sqrt{5} \)
- \( BC = 2 \)
- \( CD = 20 \)
- \( DA = 2\sqrt{5} \)

Now we need to verify if opposite sides are equal:
- \( AB \neq CD \)
- \( DA \neq BC \)

### Step 3: Check the perpendicularity of adjacent sides

To check the perpendicularity, we can calculate the slopes of the sides.

**Slope of AB:**
\[
\text{slope of } AB = \frac{y_B - y_A}{x_B - x_A} = \frac{2 - (-6)}{8 - (-8)} = \frac{8}{16} = \frac{1}{2}
\]

**Slope of BC:**
\[
\text{slope of } BC = \frac{y_C - y_B}{x_C - x_B} = \frac{2 - 2}{10 - 8} = \frac{0}{2} = 0
\]

**Slope of CD:**
\[
\text{slope of } CD = \frac{y_D - y_C}{x_D - x_C} = \frac{-10 - 2}{-6 - 10} = \frac{-12}{-16} = \frac{3}{4}
\]

**Slope of DA:**
\[
\text{slope of } DA = \frac{y_A - y_D}{x_A - x_D} = \frac{-6 - (-10)}{-8 - (-6)} = \frac{4}{-2} = -2
\]

**Check for perpendicularity:**
Two lines are perpendicular if the product of their slopes is \(-1\).

- For \( AB \) and \( BC \):
\[
\frac{1}{2} \times 0 \neq -1 \quad \text{(not perpendicular)}
\]

- For \( BC \) and \( CD \):
\[
0 \times \frac{3}{4} \neq -1 \quad \text{(not perpendicular)}
\]

- For \( CD \) and \( DA \):
\[
\frac{3}{4} \times -2 \neq -1 \quad \text{(not perpendicular)}
\]

### Conclusion:
Since the calculated lengths of sides do not satisfy the equality for opposite sides and the slopes do not show perpendicularity, the points \( A = (-8, -6) \), \( B = (8, 2) \), \( C = (10, 2) \), and \( D = (-6, -10) \) do not form a rectangle.