Chapter 1.1, Problem 54AYU

An equilateral triangle is one in which all three sides are of equal length. If two vertices of an equilateral triangle are $\left(0,4\right)\text{and}\left(0,0\right)$, find the third vertex. How many of these triangles are possible?

To determine

To find: The third vertex of an equilateral triangle whose two vertices are $\left(0,4\right),\left(0,0\right)$

Answer to Problem 54AYU

The coordinates of the third vertex are $(\pm 2\sqrt{3},2)$.

Two such triangles are possible.

Explanation of Solution

Given:

Two vertices of an equilateral triangle ABC, $\text{A }\left(0,4\right),\text{ B }\left(0,0\right)$

Formula used:

Distance formula $\text{= d(P}₁\text{, P}₂\text{) = }\sqrt{{(x_2 - x_1)}^2 + {(y_2 - y_1)}^2}$

Calculation:

Length AB $= \sqrt{{(0 - 0)}^2 + {(0 - 4)}^2} = \sqrt{0 + {(4)}^2} = \sqrt{16} = 4$

Length BC $=$ length CA as ABC is an equilateral triangle.

$\sqrt{{(x - 0)}^2 + {(y - 0)}^2} = \sqrt{{(0 - x)}^2 + {(4 - y)}^2}$

$\sqrt{x^2 + y^2} = \sqrt{x^2 + 16 - 8y + y^2}$

Square both sides to solve for $x$ and $y$

$x^2 + y^2 = x^2 + 16 - 8y + y^2$

$x^2 + y^2 - x^2 = x^2 + 16 - 8y + y^2 - x^2$   Subtract $x^2$ from both sides

$y^2 = 16 - 8\text{y} + y^2$

$y^2 - y^2 = 16 - 8\text{y} + y^2 - y^2$   Subtract $y^2$ from both sides

$0=16-8y$

$8y = 16 - 8y + 8y$   Add $8y$ on both sides

$8y = 16$

$y = \frac{16}{8}$   Divide both sides by 8

$y = 2$

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

$4 = \sqrt{x^2 + {(2)}^2}$

$4 = \sqrt{x^2 + 4}$

${(4)}^2 = x^2 +4$   Squaring both sides

$16 = x^2 + 4$

$16-4= x^2 +4-4$

$12 = x^2$

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

Thus the coordinates of the third vertex is either $(-2\sqrt{3},2)$ or $(2\sqrt{3},2)$ and hence there are two such triangles are possible.