4.1 Q-12

Question Statement

Given two vertices of an equilateral triangle at $A(-3, 0)$ and $B(3, 0)$, find the third vertex of the triangle. How many such equilateral triangles are possible?

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Background and Explanation

In an equilateral triangle, all three sides are of equal length, and the angles between the sides are all $60^\circ$. To find the third vertex, we will use the distance formula to ensure that the distance from the third vertex to both $A$ and $B$ is equal to the distance between $A$ and $B$. This condition will help us find the possible coordinates of the third vertex.

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Solution

Step 1: Set up the distance formula

Let the third vertex be $C(x, y)$. We know that in an equilateral triangle, the distance from the third vertex $C$ to both $A$ and $B$ must be equal to the distance between $A$ and $B$.

The distance between $A$ and $B$ is:

$$|AB| = \sqrt{(3 - (-3))^2 + (0 - 0)^2} = \sqrt{(6)^2} = 6$$

Thus, we need to solve for $C(x, y)$ such that:

$$|AC| = |BC| = |AB| = 6$$

This gives us the following two distance equations:

1. Distance from $A$ to $C$:

$$\sqrt{(x + 3)^2 + y^2} = 6$$

2. Distance from $B$ to $C$:

$$\sqrt{(x - 3)^2 + y^2} = 6$$

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Step 2: Set up the equations

Now, let’s square both sides of each equation to eliminate the square roots.

For $|AC|$:

$$(x + 3)^2 + y^2 = 36$$

For $|BC|$:

$$(x - 3)^2 + y^2 = 36$$

Now we have the system of equations:

1. $(x + 3)^2 + y^2 = 36$
2. $(x - 3)^2 + y^2 = 36$

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Step 3: Simplify and solve for $x$

Subtract the second equation from the first:

$$(x + 3)^2 + y^2 - \left[(x - 3)^2 + y^2\right] = 0$$

Simplify:

$$(x + 3)^2 - (x - 3)^2 = 0$$

Using the difference of squares formula:

$$\left[(x + 3) - (x - 3)\right]\left[(x + 3) + (x - 3)\right] = 0$$

This simplifies to:

$$6 \cdot (2x) = 0$$

Therefore:

$$12x = 0 \quad \Rightarrow \quad x = 0 \tag{1}$$

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Step 4: Solve for $y$

Substitute $x = 0$ into one of the original distance equations. Using $(x + 3)^2 + y^2 = 36$:

$$(0 + 3)^2 + y^2 = 36$$

Simplify:

$$9 + y^2 = 36$$

Solve for $y^2$:

$$y^2 = 36 - 9 = 27$$

Taking the square root of both sides:

$$y = \pm \sqrt{27} = \pm 3\sqrt{3}$$

Thus, the possible coordinates for the third vertex are $(0, 3\sqrt{3})$ and $(0, -3\sqrt{3})$.

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Step 5: Conclusion

Since we have two possible values for $y$, there are two distinct equilateral triangles possible, corresponding to the two different positions for the third vertex.

Thus, the two possible triangles are formed with the third vertex at either $(0, 3\sqrt{3})$ or $(0, -3\sqrt{3})$.

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Key Formulas or Methods Used

• Distance Formula:

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

• Difference of Squares:

$$(a^2 - b^2) = (a - b)(a + b)$$

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Summary of Steps

1. Set up the distance equations for $AC$ and $BC$ to be equal to $AB$.
2. Simplify the equations by squaring both sides and subtracting them.
3. Solve for $x$, which gives $x = 0$.
4. Substitute $x = 0$ into the distance equation to solve for $y$, yielding $y = \pm 3\sqrt{3}$.
5. Conclude that two distinct triangles are possible with third vertex at $(0, 3\sqrt{3})$ or $(0, -3\sqrt{3})$.