Finding the Third Vertex Coordinates of an Equilateral Triangle

We are given two vertices of an equilateral triangle: A at (0, 0) and B at (2, 2). We need to find the difference in coordinates of the third vertex. Let the third vertex be C at (x, y).

In an equilateral triangle, all sides are equal in length. First, let's calculate the length of the side AB using the distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.

Calculating the Side Length of the Equilateral Triangle

Using points A(0, 0) and B(2, 2):

$\text{Side length AB} = \sqrt{(2 - 0)^2 + (2 - 0)^2} = \sqrt{2^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$.

So, the side length of the equilateral triangle is $2\sqrt{2}$. The square of the side length is $(2\sqrt{2})^2 = 8$.

Setting Up Equations for the Third Vertex

The distance from the third vertex C(x, y) to vertex A(0, 0) must be equal to the side length, $2\sqrt{2}$.

Distance AC $= \sqrt{(x - 0)^2 + (y - 0)^2} = \sqrt{x^2 + y^2}$.

Setting this equal to the side length:

$\sqrt{x^2 + y^2} = 2\sqrt{2}$

$x^2 + y^2 = (2\sqrt{2})^2 = 8$. (Equation 1)

Similarly, the distance from the third vertex C(x, y) to vertex B(2, 2) must also be equal to the side length, $2\sqrt{2}$.

Distance BC $= \sqrt{(x - 2)^2 + (y - 2)^2}$.

Setting this equal to the side length:

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

$(x - 2)^2 + (y - 2)^2 = (2\sqrt{2})^2 = 8$. (Equation 2)

Solving for the Coordinates (x, y)

Now we have a system of two equations with two variables:

1. $x^2 + y^2 = 8$
2. $(x - 2)^2 + (y - 2)^2 = 8$

Expand Equation 2:

$(x^2 - 4x + 4) + (y^2 - 4y + 4) = 8$

$x^2 + y^2 - 4x - 4y + 8 = 8$

Substitute $x^2 + y^2 = 8$ from Equation 1 into this expanded equation:

$8 - 4x - 4y + 8 = 8$

$16 - 4x - 4y = 8$

$8 = 4x + 4y$

Divide by 4:

$2 = x + y$

From this, we get $y = 2 - x$. Substitute this into Equation 1:

$x^2 + (2 - x)^2 = 8$

$x^2 + (4 - 4x + x^2) = 8$

$2x^2 - 4x + 4 = 8$

$2x^2 - 4x - 4 = 0$

Divide by 2:

$x^2 - 2x - 2 = 0$

Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ to solve for x. Here, $a=1$, $b=-2$, $c=-2$.

$x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-2)}}{2(1)}$

$x = \frac{2 \pm \sqrt{4 + 8}}{2}$

$x = \frac{2 \pm \sqrt{12}}{2}$

$x = \frac{2 \pm 2\sqrt{3}}{2}$

$x = 1 \pm \sqrt{3}$

Now find the corresponding y values using $y = 2 - x$:

• If $x = 1 + \sqrt{3}$, then $y = 2 - (1 + \sqrt{3}) = 2 - 1 - \sqrt{3} = 1 - \sqrt{3}$. The third vertex could be $(1 + \sqrt{3}, 1 - \sqrt{3})$.
• If $x = 1 - \sqrt{3}$, then $y = 2 - (1 - \sqrt{3}) = 2 - 1 + \sqrt{3} = 1 + \sqrt{3}$. The third vertex could be $(1 - \sqrt{3}, 1 + \sqrt{3})$.

Calculating the Difference of Coordinates

We need to find the absolute difference between the x and y coordinates for the third vertex, i.e., $|x - y|$.

• For the vertex $(1 + \sqrt{3}, 1 - \sqrt{3})$:

  $|x - y| = |(1 + \sqrt{3}) - (1 - \sqrt{3})| = |1 + \sqrt{3} - 1 + \sqrt{3}| = |2\sqrt{3}| = 2\sqrt{3}$.

• For the vertex $(1 - \sqrt{3}, 1 + \sqrt{3})$:

  $|x - y| = |(1 - \sqrt{3}) - (1 + \sqrt{3})| = |1 - \sqrt{3} - 1 - \sqrt{3}| = |-2\sqrt{3}| = 2\sqrt{3}$.

In both possible cases for the third vertex, the difference of coordinates is $2\sqrt{3}$.

The difference of coordinates of the third vertex is $2\sqrt{3}$.

Revision Table: Key Concepts for Equilateral Triangles and Coordinates

Additional Information: Geometric Interpretation

Geometrically, the two possible locations for the third vertex of an equilateral triangle, given two vertices A and B, lie on a line perpendicular to the segment AB. This line passes through the midpoint of AB. The distance from the midpoint to the third vertex is the height of the equilateral triangle. The height (h) of an equilateral triangle with side 's' is given by $h = \frac{\sqrt{3}}{2}s$.

In this case, the side length is $s = 2\sqrt{2}$. The height would be $h = \frac{\sqrt{3}}{2}(2\sqrt{2}) = \sqrt{6}$.

The midpoint of AB is $(\frac{0+2}{2}, \frac{0+2}{2}) = (1, 1)$.

The line connecting A(0,0) and B(2,2) has a slope of $\frac{2-0}{2-0} = 1$.

The line perpendicular to AB passing through the midpoint (1,1) has a slope of $-1$ (negative reciprocal). Its equation is $y - 1 = -1(x - 1)$, which simplifies to $y - 1 = -x + 1$, or $y = -x + 2$. This is the same relationship $x + y = 2$ we found algebraically!

The two possible third vertices lie on the line $y = -x + 2$ at a distance of $\sqrt{6}$ from the midpoint (1,1). This geometric perspective confirms the algebraic solution and shows why there are typically two possible locations for the third vertex.