Distance Formula Problem

Question

If two vertices of an equilateral triangle are $(1, -1)$ and $(-\sqrt{3}, - \sqrt{3})$, find the coordinates of the third vertex.

Step by Step procedure to get the answer.

Take $A=(1, -1)$, $B=(-\sqrt{3}, - \sqrt{3})$. Let the third vertex be $C=(x, y)$.

The distance d between the two points $A=(x_1, y_1)$ and $B=(x_2, y_2)$ is given by the formula $$ d= \sqrt{ (x_2-x_1)^2 + (y_2 - y_1)^2 } $$ So we get the distance $AB = 2 \sqrt{2}$ . Similarly we have to find the distance between BC & AC using distance formula where we get the equation in the form of x & y even though the distance is $2\sqrt{2}$ because it is equilateral triangle. But the correct solution is not able to get. Please help.

Answer 1

Distance between two given points will be the side of the triangle $a=\sqrt{(1+\sqrt{3})^2+(\sqrt{3}-1)^2}=2\sqrt{2}$. Now you just need to draw two circles of radius $a$ and with centers at the known vertices. The equations of the circles are:

$(x-1)^2+(y+1)^2=8$
$(x+\sqrt{3})^2+(y+\sqrt{3})^2=8$
The circles will intersect at two points and to find the coordinates you need to solve this system:$$ \begin{cases} (x-1)^2+(y+1)^2=(x+\sqrt{3})^2+(y+\sqrt{3})^2 \\[2ex] (x-1)^2+(y+1)^2=8 \end{cases}$$ $$ \begin{cases} x^2-2x+1+y^2+2y+1=x^2+2\sqrt{3}x+3+y^2+2\sqrt{3}y+3 \\[2ex] (x-1)^2+(y+1)^2=8 \end{cases}$$ $$ \begin{cases} x^2-2x+1+y^2+2y+1=x^2+2\sqrt{3}x+3+y^2+2\sqrt{3}y+3 \\[2ex] (x-1)^2+(y+1)^2=8 \end{cases}$$

$$ \begin{cases} y(2-2\sqrt{3})=x(2+2\sqrt{3})+4 \\[2ex] (x-1)^2+(y+1)^2=8 \end{cases}$$

$$ \begin{cases} y=\frac{x(1+\sqrt{3})+2}{1-\sqrt{3}} \\[2ex] (x-1)^2+(y+1)^2=8 \end{cases}$$Finally, we get a quadratic equation for $x$: $$x^2-2x+1+\Big(\frac{x(1+\sqrt{3})+2}{1-\sqrt{3}}+1\Big)^2=8 $$ The solutions are: $x=-1, x=2-\sqrt{3}$. We already have $y$ expressed via $x$ so just plug the $x$ values in that equation and find $y$.

Answer 2

Here is a solution using complex numbers. The two vertices are $A(a=1-i)$ and $B(b=-\sqrt{3}-i\sqrt{3})$ and $C(c)$ the third vertex . $a$ and $b$ are affix with respect to $A$ and $B$. Now we consider the complex $s=\frac{b-a}{b-c}$. If ABC is equilateral then $|s|=1$ and $\arg(s)=\pm\frac{\pi}{3}$ which mean: $s=e^{\pm\frac{\pi}{3}}$. Since $c=b-\frac{b-a}{s}=b-(b-a)\bar{s}$ we can easily compute $c$.

Answer 3

Equilateral triangle

$A=(1,-1)$

$B=(-\sqrt{3},-\sqrt{3})$

$C=(x,y)$

$|AB|=2\sqrt{2}$

height of an equilateral triangle:

$h=\frac{1}{2}\sqrt{3}a$

$a=|AB|$

$h=\frac{1}{2}\sqrt{3}.2\sqrt{2}=\sqrt{6}$

noting that

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

O = (0,0) origin Cartesian Cordinate System

BO is foot of the altitude

O is midpoint de CA

$\frac{xC+1}{2}=0⇒xC=-1$

$\frac{yC+(-1)}{2}=0⇒yC=1$ C=(-1,1)

Answer 4

After writing the equations of the two circles to find the points of intersection, I found the resulting quadratic equations to be too tedious for my patience. Instead, I determined the (parametric) vector from $(-\sqrt{3}, - \sqrt{3} )$ to $(1, -1)$ to be

$$A = \left(-\sqrt{3} + \left(1+\sqrt{3}\right) T, -\sqrt{3} + \left(-1+\sqrt{3}\right) T\right)$$ where $0 \le T \le 1$.

I then added $(\sqrt{3}, \sqrt{3})$ to this to translate it to the origin, then multiplied it by the rotation matrix $\begin{bmatrix} \cos & -\sin \\ \sin & \cos \end{bmatrix}$, using $\theta = +60^{\circ}$ and $T=1$. Lastly, I subtracted $(\sqrt{3}, \sqrt{3})$ to translate back. After MANY sign corrections, I got one of the two possible points to be $(-1,1)$, which checks out using the distance formula. No quadratics to solve, just vectors to add and rotate.

(Sorry about my lack of proper math notation formatting; if I find this site to be interesting, I will learn MathJax.)