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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?

Marva Jami
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3 Answers3

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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 $b$.

Arthur
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  • I didn't understand I haven't studied these things yet – Marva Jami Feb 23 '18 at 14:17
  • @MarvaJami What is it you don't understand? Can you be more specific? – Arthur Feb 23 '18 at 14:26
  • Which line has 45°?what does it mean 45° downward?how can you know that it is 45°? – Marva Jami Feb 23 '18 at 14:33
  • @MarvaJami The line $x+y+2 = 0$ makes an angle of $45^\circ$ with the $x$-axis. And by downward, I mean down and to the right (as opposed to up and to the right). This is also known as having a slope of $-1$, and I know that because the equation may be transformed into $y = -x-2$, where we can see the slope of $-1$ clearly. – Arthur Feb 23 '18 at 14:39
  • How did you found the other two line's angle? – Marva Jami Feb 23 '18 at 15:01
  • @MarvaJami There really isn't much choice in what the angles of the other lines can be if the three lines together are to make up the sides of an equilateral triangle. The angles between all the lines must be $60^\circ$, so the two new lines must make angles of $15^\circ$ and $75^\circ$ with the $x$-axis. – Arthur Feb 23 '18 at 15:13
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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 $(1,1)$.

Faraad Armwood
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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.
krirkrirk
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