Equiletral triangle problem

Question

A vertex of an equiletral triangle is $A(2,3)$and the equation of opposite side is $x+y+2=0$ Find the equations of other two sides?

Answer 1

Your line has angle $45^\circ$ downwards (as we move to the right). That means that the other two lines have angle $15^\circ$ upwards and $75^\circ$ upwards.

An angle of $15^\circ$ means a slope of $\tan(15^\circ) = 2-\sqrt3$, so one side will have equation $y = (2-\sqrt3)x + a$ for some real number $a$. The other side will have slope $\tan(75^\circ) = 2+\sqrt3$, and therefore have equation $y = (2+\sqrt3)x + b$ for some real number $b$.

Answer 2

The answer is not too messy if you use vectors. Convince yourself that you need to travel from the vertex to a point on this line $L: x+y = -2$ via opposite direction of the $30^o$ counter-clockwise rotation of the normal vector to this line (which is $(1,1)$). To do this rotation just use the matrix,

$$ \begin{pmatrix} \cos(30^o) & - \sin(30^o) \\ \sin(30^o) & \cos(30^o) \end{pmatrix} = \begin{pmatrix} \frac{\sqrt{3}}{2} &- \frac{1}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2}\end{pmatrix}$$

and apply it to $(1,1)$ i.e,

$$\begin{pmatrix} \frac{\sqrt{3}}{2} &- \frac{1}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2}\end{pmatrix} \begin{pmatrix} 1 \\ 1\end{pmatrix} = \begin{pmatrix} \frac{\sqrt{3}-1}{2} \\ \frac{\sqrt{3}+1}{2}\end{pmatrix}$$

And now we try to find $t$ such that,

$$ (1,1) - t \left( \frac{\sqrt{3}-1}{2}, \frac{\sqrt{3}+1}{2}\right) = \left(1 + t\frac{1-\sqrt{3}}{2},1 +t \frac{-1- \sqrt{3}}{2} \right) \in L$$

Well it must be the case that,

$$1 + t\frac{1-\sqrt{3}}{2} +1 +t \frac{-1- \sqrt{3}}{2} = -2 \Rightarrow t = \frac{4}{\sqrt{3}}$$

And so another vertex is,

$$(1,1) - \frac{4}{\sqrt{3}}\left( \frac{\sqrt{3}-1}{2}, \frac{\sqrt{3}+1}{2}\right) = \left(1- \frac{2(\sqrt{3}-1)}{\sqrt{3}}, 1- \frac{2(1+ \sqrt{3})}{\sqrt{3}} \right)$$

Use the same idea to get the other vertex i.e travel in the opposite direction of the $30^o$ clockwise rotation of $(1,1)$.

Answer 3

Following my comment, the solution can be found using only a quadratic equation. Let $B(x_B, y_B)$ and $C(x_C, y_C)$ the points you are looking for.

1. Using $AB = AC$, prove that $x_B$ and $x_C$ are such that $P(x_B) = P(x_C)$, with $P(X) = 2X^2+6X+29$
2. Using $AB = BC$, prove that $P(x_B)$ must equal $\sqrt{2}(x_C-x_B)$
3. Solve the system.