4.1 Q-9

Question Statement

Find $h$ such that the points $A(h, 1)$, $B(2, 7)$, and $C(-6, -7)$ are the vertices of a right triangle with the right angle at vertex $A$.

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Background and Explanation

For the points to form a right triangle at vertex $A$, the sum of the squares of the two sides meeting at $A$ must be equal to the square of the third side (the hypotenuse). This follows from the Pythagorean Theorem, which states:

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

Where:

• $a$ and $b$ are the lengths of the legs of the triangle.
• $c$ is the length of the hypotenuse.

The approach will involve calculating the squared distances between the points $A$, $B$, and $C$, and setting up an equation based on the Pythagorean Theorem.

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Solution

Step 1: Find the distances between the points

We need to find the distances between the points $A$, $B$, and $C$. The distance formula is given by:

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

We will calculate the squared distances (since we are not interested in the square roots for this problem):

• Distance from $A$ to $B$:

$$AB^2 = (2 - h)^2 + (7 - 1)^2 = (2 - h)^2 + 6^2 = (2 - h)^2 + 36$$

• Distance from $A$ to $C$:

$$AC^2 = (-6 - h)^2 + (-7 - 1)^2 = (-6 - h)^2 + (-8)^2 = (-6 - h)^2 + 64$$

• Distance from $B$ to $C$:

$$BC^2 = (-6 - 2)^2 + (-7 - 7)^2 = (-8)^2 + (-14)^2 = 64 + 196 = 260$$

Step 2: Apply the Pythagorean Theorem

To form a right triangle with the right angle at $A$, the sum of the squares of the distances from $A$ to $B$ and $A$ to $C$ must equal the square of the distance from $B$ to $C$. Therefore:

$$AB^2 + AC^2 = BC^2$$

Substitute the squared distances into the equation:

$$(2 - h)^2 + 36 + (-6 - h)^2 + 64 = 260$$

Simplifying:

$$(2 - h)^2 + (-6 - h)^2 + 100 = 260$$ $$(2 - h)^2 + (-6 - h)^2 = 160$$

Step 3: Expand and simplify the equation

Now, expand both squared terms:

$$(2 - h)^2 = h^2 - 4h + 4$$ $$(-6 - h)^2 = h^2 + 12h + 36$$

Substitute these into the equation:

$$h^2 - 4h + 4 + h^2 + 12h + 36 = 160$$

Combine like terms:

$$2h^2 + 8h + 40 = 160$$

Subtract 160 from both sides:

$$2h^2 + 8h - 120 = 0$$

Step 4: Solve the quadratic equation

Divide the entire equation by 2 to simplify:

$$h^2 + 4h - 60 = 0$$

Now factor the quadratic equation:

$$h^2 + 10h - 6h - 60 = 0$$ $$h(h + 10) - 6(h + 10) = 0$$

Factor out the common terms:

$$(h - 6)(h + 10) = 0$$

Step 5: Solve for $h$

Setting each factor equal to zero:

$$h - 6 = 0 \quad \text{or} \quad h + 10 = 0$$

Thus, we have:

$$h = 6 \quad \text{or} \quad h = -10$$

So, the possible values for $h$ are $h = 6$ and $h = -10$.

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Key Formulas or Methods Used

• Distance Formula: The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is:

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

• Pythagorean Theorem: In a right triangle:

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

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Summary of Steps

1. Use the distance formula to find the squared distances between points $A$, $B$, and $C$.
2. Apply the Pythagorean Theorem: $AB^2 + AC^2 = BC^2$.
3. Expand and simplify the resulting equation.
4. Solve the quadratic equation for $h$.
5. The possible values for $h$ are $h = 6$ and $h = -10$.