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I am trying to figure out how to find the Cartesian coordinates of an equilateral triangle centered around (0,0).

TheMathNoob
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2 Answers2

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What does exactly mean "centered around (0,0)" ?

If it is what is drawn on the figure, the coordinates of the vertexes are given for sides length $=L$.

enter image description here

One little picture says more than a long speech!

JJacquelin
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Here's an outline for a "hands-on" approach. Pen and paper required.

Assuming the triangle is to be oriented with one of its points facing up on the $y$-axis, consider only one half of the triangle - say the left half. This will be a right triangle with a vertical side along the $y$-axis, centered about $(0,0)$. The length of the hypotenuse will be the length which is given, and the base will be half as long. Use what you know about right triangles to find the length of the vertical side, at which point you can draw the triangle. Then take note of the top and left vertices. The third vertex for your desired triangle is the vertex obtained by reflecting the left vertex over the $y$-axis.

Suppose $L$ is the length we are given, and let $x$ denote the length of the vertical line described above. Then by Pythagoras's theorem, we must have $L^2=x^2+(L/2)^2$. We deduce $$ x = \sqrt{L^2-(L/2)^2} = \frac{\sqrt{3}L}{2}. $$ Since this line segment sits on the $y$-axis, centered at $(0,0)$, we deduce that it runs from $(0,-x/2)=(0,-\sqrt{3}L/4)$ up to $(0,x/2)=(0,\sqrt{3}L/4)$. Now the base goes out horizontally a distance $L/2$ in each direction from the bottom point. Therefore the top vertex is $$ \left(0,\frac{\sqrt{3}L}{4}\right),$$ the left vertex is $$ \left(-L/2,-\frac{\sqrt{3}L}{4}\right),$$ and the right vertex is $$ \left(L/2,-\frac{\sqrt{3}L}{4}\right).$$

John Griffin
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  • $(-L/2,-L/2), (L/2,-L/2), (0,SQRT((L/2)^2+(L/2)^2))$ – TheMathNoob Aug 07 '17 at 04:05
  • nevermind.......... – TheMathNoob Aug 07 '17 at 04:08
  • So if $L$ is the length we are given, then in order to find the length of the vertical side, we need to solve $L^2=x^2+(L/2)^2$ for $x$. Now let this line sit on the $y$-axis centered at $(0,0)$, so that the top point is at $(0,x/2)$ and the bottom point (mid-point of base of equilateral triangle) is at $(0,-x/2)$. We are about 80% done at this point. – John Griffin Aug 07 '17 at 04:11
  • yes the hard vertex is $( 0, sqrt(L^2 + (L/2)^2) - L/2)$ – TheMathNoob Aug 07 '17 at 04:13
  • Well $x=\sqrt{L^2-(L/2)^2}$, so the top vertex should be $$\left(0,\frac{\sqrt{L^2-(L/2)^2}}{2}\right)=(0,\sqrt{3}L/4).$$ – John Griffin Aug 07 '17 at 04:15
  • The base is $L/2$ away from (0,0), so the lower coordinates are $(-L/2,-L/2)$ and $(L/2, -L/2)$ – TheMathNoob Aug 07 '17 at 04:19
  • here are the specifications of the assignment: An equilateral triangle has a horizontal base (two points have the same y-value). The apex has a higher y-value.The constructor just uses the constructor for a triangle – TheMathNoob Aug 07 '17 at 04:23
  • Please take a look at my edited answer. I worked the details out. I'm not sure how you are getting $y$-values of $L/2$ and $\sqrt{L^2+(L/2)^2}-L/2$, as that surely wouldn't leave the triangle centered at $(0,0)$. – John Griffin Aug 07 '17 at 04:44