Find the coordinate of third point of equilateral triangle.

Question

I have two points A and B whose coordinates are $(3,4)$ and $(-2,3)$ The third point is C. We need to calculate its coordinates.

I think there will be two possible answers, as the point C could be on the either side of line joining A and B.

Now I put AB = AC = BC.

We calculate AB by distance formula : $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

= $ \sqrt{5^2 + 1^2} = \sqrt{26}$

I plug this value into the distance of AC and BC and finally equate to get :

$5x + y = 6$

Now what can I do? There are two variables, I am getting equation of a line! How can I solve this?

Answer 1

Call the position of point $C$ by the coords $(a, b)$. The equations for $C$ are then

$$ \sqrt{(a-3)^2 + (b - 4)^2} = \sqrt{26} \\ \sqrt{(a+2)^2 + (b - 3)^2} = \sqrt{26} $$ Squaring both, we get $$ (a-3)^2 + (b - 4)^2 = 26 \\ (a+2)^2 + (b - 3)^2 = 26 $$ $$ a^2 - 6a + 9 + b^2 - 8b + 16= 26 \\ a^2 + 4a + 4 + b^2 - 6b + 9= 26 $$ Subtracting these two gives $$ -10a + 5 - 2b + 7 = 0 $$ or $$ 6 = 5a + b $$ which is a line both points must lie on. Writing this as $$ b = 6 - 5a $$ we can substitute in either equation. Let's got with the second:

$$ a^2 + 4a + 4 + b^2 - 6b + 9= 26 $$ becomes $$ a^2 + 4a + 4 + (6-5a)^2 - 6(6-5a) + 9= 26 $$ which is a quadratic that can now be solved for the two possible values of $a$.

(Once you do so, you use $b = 6 - 5a$ to find the corresponding $b$-values.)

Answer 2

The solution using a quadratic equation is a useful tool that will work for any non-degenerate triangle with known sides. In this particular case, however, you can exploit a known property of the equilateral triangle: if $M$ is the midpoint of side $AB$ then $\triangle AMC$ is a right triangle with a $60$-degree angle at $A$, and $CM = \sqrt3 \, AM$. We can combine this with some techniques that borrowed from other useful notions of coordinate geometry.

To find $M$, take the arithmetic means of the coordinates of $A =(3,4)$ and $B =(-2,3)$:

$$x_M = \frac12(x_A + x_B) = \frac12(3-2) = \frac12.$$ $$y_M = \frac12(y_A + y_B) = \frac12(4+3) = \frac72.$$

So $M=\left(\frac12, \frac72\right)$.

Now to get from $M$ to $A$ you increase $x$ by $\frac52$ and $y$ by $\frac12$, so that line has slope $\frac15$. The line $CM$ must have slope $-5$ to make a right angle with $AM$, that is, for a $\frac12$ change in $x$ we get a $-\frac52$ change in $y$. (In fact, the line with slope $-5$ through $\left(\frac12, \frac72\right)$ is the line $5x + y = 6$ that you found by other means.)

So the point $D = \left(x_m + \frac12, y_M - \frac52\right) = \left(\frac12 + \frac12, \frac72 - \frac52\right) = (1,1)$ is somewhere on the line $CM$ (not necessarily between $C$ and $M$) and $DM = AM$. But $C$ is just $\sqrt3$ times as far from $M$, so scale up the changes in $x$ and $y$ accordingly:

$$C = \left(\frac12 + \frac12\sqrt3, \frac72 - \frac52\sqrt3\right).$$

And of course we could also go an equal distance from $M$ in the opposite direction:

$$C' = \left(\frac12 - \frac12\sqrt3, \frac72 + \frac52\sqrt3\right).$$

Not surprisingly, these are the same results you get by solving the quadratic equation.

We could be a bit more sophisticated about all this (and save some words) by treating the directed segment $MA$ as a vector and performing transformations on it that rotate it by $90$ degrees (in either direction) and scale it by $\sqrt3$. But if we know how to write a general rotation-and-scaling transformation as a $2\times2$ matrix then we might as well take $AB$ as a vector and rotate it $60$ degrees with a scale factor of $1$ to determine the vector $AC$. So this method also generalizes in a useful way when you know an applicable angle of the triangle.

Answer 3

If you like vector approach:

Displace (shift/translate) point A by vector(-3,-4) to come to the origin.

Multiply the new radius vector $ab$ with $ e^ {i \pi/3} , e^ {-i \pi/3}$ (once clockwise and once anticlockwise) to obtain new points $ C_1 $ and $ C_2 $. The multiplying factor is $ (1/2 \pm i \sqrt 3/2) $.

Displace these points back to original positions translating by $ (3,4) $.

If you multiply thrice, all points of a hexagon would also be reached.

Answer 4

Calculate slope of AB (m), distance AB = s Point A is at (h,k) tan-1 m = angle N s x sin (60 + angle N) = y s x cos (60 + angle N) = x Third point is at x+h, y+k Use (180 + 60 + angle N) for second answer and h,k is now at point B