4.1 Q-6

Question Statement

Find the value of $h$ such that the points $A(\sqrt{3}, -1)$, $B(0, 2)$, and $C(h, -2)$ are the vertices of a right triangle with the right angle at vertex $A$.

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Background and Explanation

This problem involves the use of the Pythagorean Theorem for right-angled triangles. The Pythagorean theorem states that for a right-angled triangle with vertices at $A$, $B$, and $C$ (with right angle at $A$), the following equation holds:

$$\text{AB}^2 + \text{AC}^2 = \text{BC}^2$$

We will apply this to solve for $h$.

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Solution

Step 1: Apply the Pythagorean Theorem

We are given that the triangle has a right angle at vertex $A$, so by the Pythagorean Theorem, we have:

$$\text{AB}^2 + \text{AC}^2 = \text{BC}^2$$

Step 2: Calculate the Distance Between Points

• Distance AB:

$$\text{AB} = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2}$$

Substituting the coordinates of $A(\sqrt{3}, -1)$ and $B(0, 2)$:

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

• Distance AC:

$$\text{AC} = \sqrt{(x_C - x_A)^2 + (y_C - y_A)^2}$$

Substituting the coordinates of $A(\sqrt{3}, -1)$ and $C(h, -2)$:

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

• Distance BC:

$$\text{BC} = \sqrt{(x_C - x_B)^2 + (y_C - y_B)^2}$$

Substituting the coordinates of $B(0, 2)$ and $C(h, -2)$:

$$\text{BC} = \sqrt{(h - 0)^2 + (-2 - 2)^2} = \sqrt{h^2 + (-4)^2} = \sqrt{h^2 + 16}$$

Step 3: Set Up the Equation

Now, we use the Pythagorean theorem:

$$\text{AB}^2 + \text{AC}^2 = \text{BC}^2$$

Substitute the distances into the equation:

$$(2\sqrt{3})^2 + \left[(h - \sqrt{3})^2 + 1\right] = (h^2 + 16)$$

Simplifying:

$$12 + (h - \sqrt{3})^2 + 1 = h^2 + 16$$ $$13 + (h - \sqrt{3})^2 = h^2 + 16$$

Step 4: Expand and Solve for $h$

First, expand the squared term $(h - \sqrt{3})^2$:

$$(h - \sqrt{3})^2 = h^2 - 2h\sqrt{3} + 3$$

Substitute this into the equation:

$$13 + h^2 - 2h\sqrt{3} + 3 = h^2 + 16$$

Simplifying further:

$$16 + h^2 - 2h\sqrt{3} = h^2 + 16$$

Cancel out $h^2$ from both sides:

$$16 - 2h\sqrt{3} = 16$$

Now, subtract 16 from both sides:

$$-2h\sqrt{3} = 0$$

Step 5: Solve for $h$

Solving for $h$:

$$h = 0$$

Thus, the value of $h$ is 0.

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Key Formulas or Methods Used

• Pythagorean Theorem:

$$\text{AB}^2 + \text{AC}^2 = \text{BC}^2$$

• Distance Formula:

$$\text{Distance between two points} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

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Summary of Steps

1. Apply the Pythagorean Theorem: $\text{AB}^2 + \text{AC}^2 = \text{BC}^2$.
2. Calculate the distances AB, AC, and BC using the distance formula.
3. Set up the equation using the calculated distances.
4. Simplify the equation and solve for $h$.
5. The solution gives $h = 0$.

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