4.1 Q-4

Question Statement

Prove the following:

1. Points $A(0, 2)$, $B(\sqrt{3}, 1)$, and $C(0, -2)$ are vertices of a right triangle.
2. Points $A(3, 1)$, $B(-2, -3)$, and $C(2, 2)$ are vertices of an isosceles triangle.
3. Points $A(5, 2)$, $B(-2, 3)$, $C(-3, -4)$, and $D(4, -5)$ are vertices of a parallelogram. Determine whether it is a square.

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Background and Explanation

To solve problems involving the classification of triangles and parallelograms:

• Use the distance formula $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ to calculate the length of sides.
• For a right triangle, the square of the longest side (hypotenuse) equals the sum of the squares of the other two sides.
• For an isosceles triangle, at least two sides must have the same length.
• A parallelogram is classified as a square if all its sides are equal and its diagonals are also equal.

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Solution

Part 1: Verifying a Right Triangle

We calculate the distances between the given points:

1. Distance $AB$:

$$|AB| = \sqrt{(\sqrt{3} - 0)^2 + (1 - 2)^2} = \sqrt{(\sqrt{3})^2 + (-1)^2} = \sqrt{3 + 1} = \sqrt{4} = 2$$

2. Distance $BC$:

$$|BC| = \sqrt{(0 - \sqrt{3})^2 + (-2 - 1)^2} = \sqrt{(\sqrt{3})^2 + (-3)^2} = \sqrt{3 + 9} = \sqrt{12}$$

3. Distance $AC$:

$$|AC| = \sqrt{(0 - 0)^2 + (-2 - 2)^2} = \sqrt{0 + 16} = 4$$

4. Check for the Pythagorean theorem:

$$|AC|^2 = |AB|^2 + |BC|^2 \quad \text{(i.e., $16 = 4 + 12$) is true.}$$

Thus, $A, B, C$ form a right triangle.

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Part 2: Verifying an Isosceles Triangle

We calculate the distances between the given points:

1. Distance $AB$:

$$|AB| = \sqrt{(-2 - 3)^2 + (-3 - 1)^2} = \sqrt{(-5)^2 + (-4)^2} = \sqrt{25 + 16} = \sqrt{41}$$

2. Distance $BC$:

$$|BC| = \sqrt{(2 - (-2))^2 + (2 - (-3))^2} = \sqrt{(4)^2 + (5)^2} = \sqrt{16 + 25} = \sqrt{41}$$

3. Distance $AC$:

$$|AC| = \sqrt{(2 - 3)^2 + (2 - 1)^2} = \sqrt{(-1)^2 + (1)^2} = \sqrt{1 + 1} = \sqrt{2}$$

Since $|AB| = |BC|$, the triangle is isosceles.

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Part 3: Verifying a Square Parallelogram

We calculate the distances between the given points:

1. Distance $AB$:

$$|AB| = \sqrt{(-2 - 5)^2 + (3 - 2)^2} = \sqrt{(-7)^2 + (1)^2} = \sqrt{49 + 1} = \sqrt{50}$$

2. Distance $AD$:

$$|AD| = \sqrt{(4 - 5)^2 + (-5 - 2)^2} = \sqrt{(-1)^2 + (-7)^2} = \sqrt{1 + 49} = \sqrt{50}$$

3. Distance $BC$:

$$|BC| = \sqrt{(-3 - (-2))^2 + (-4 - 3)^2} = \sqrt{(-1)^2 + (-7)^2} = \sqrt{1 + 49} = \sqrt{50}$$

4. Distance $CD$:

$$|CD| = \sqrt{(4 - (-3))^2 + (-5 - (-4))^2} = \sqrt{(7)^2 + (-1)^2} = \sqrt{49 + 1} = \sqrt{50}$$

5. Diagonals $AC$ and $BD$:

$$|AC| = \sqrt{(-3 - 5)^2 + (-4 - 2)^2} = \sqrt{(-8)^2 + (-6)^2} = \sqrt{64 + 36} = 10$$ $$|BD| = \sqrt{(4 - (-2))^2 + (-5 - 3)^2} = \sqrt{(6)^2 + (-8)^2} = \sqrt{36 + 64} = 10$$

Since all sides and diagonals are equal, $ABCD$ is a square.

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Key Formulas or Methods Used

1. Distance formula:

$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

2. Pythagorean theorem:

$$c^2 = a^2 + b^2$$

3. Properties of triangles and parallelograms.

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Summary of Steps

1. Calculate the distances between the points using the distance formula.
2. For a right triangle:
   • Verify the Pythagorean theorem.
3. For an isosceles triangle:
   • Check if at least two sides are equal.
4. For a parallelogram:
   • Verify if opposite sides are equal.
   • Check if diagonals are equal to determine if it is a square.